Haichen Zhu

Dan Tong

Lu Zhang

Shijie Wang

Weiwen Wu

Hui Tang

Yang Chen

Limin Luo

Jian Zhu

Baosheng Li

Temporally Downsampled Cerebral CT Perfusion Image Restoration Using Deep Residual Learning

Keywords: Acute Ischemic Stroke, Low-dose CT, CT Perfusion Imaging, Deep residual CNN

Purpose Acute ischemic stroke is one of the most causes of death all over the world. Onset to treatment time is critical in stroke diagnosis and treatment. Considering the time consumption and high price of MR imaging, CT perfusion (CTP) imaging is strongly recommended for acute stroke. However, too much CT radiation during CTP imaging may increase the risk of health problems. How to reduce CT radiation dose in CT perfusion imaging has drawn our great attention.

Method

In this study, the original 30-pass CTP images are downsampled to 15 passes in time sequence, which equals to 50% radiation dose reduction. Then, a residual deep convolutional neural network (CNN) model is proposed to restore the downsampled 15-pass CTP images to 30 passes to calculate the parameters such as CBF (Cerebral Blood Flow), CBV (Cerebral Blood Volume), MTT (Mean Transit Time), TTP (Time To Peak) for stroke diagnosis and treatment. The deep restoration CNN network is implemented simply and effectively with 16 successive convolutional layers which form a wide enough receptive field for input image data. 18 patients' CTP images are employed as training set and the other 6 patients' CTP images are treated as test dataset in this study.

Results

Experiments demonstrate that our CNN network can restore high-quality CTP images in terms of structural similarity index (SSIM) and peak signal-to-noise ratio (PSNR). The average SSIM and PSNR for test images are 0.981 and 56.25, and the SSIM and PSNR of regions of interest (ROI) are 0.915 and 42.44 respectively, showing promising quantitative level. In addition, we compare the perfusion maps calculated from the restored images and from the original images, and the average perfusion results of them are extremely close. Areas of hypoperfusion of 6 test cases could be detected with comparable accuracy by radiologists.

Conclusion

The trained model can restore the temporally downsampled 15-pass CTP to 30 passes very well. According to the contrast test, sufficient information cannot be restored with e.g., simple interpolation method and Deep Convolutional Generative Adversarial Network (DCGAN), but can be restored with the proposed CNN model. This method can be an optional way to reduce radiation dose during CTP imaging.

Introduction

As one of leading causes of death, stroke may bring serious long-term disability. According to the statistics, there are about 2.5 million new stroke cases in China every year, and about 1.7 million patients die from stroke [START_REF] Wang | China stroke prevention report 2016 summary[END_REF]. In acute stroke treatment, time is life. It is strongly recommended to conduct a rapid diagnosis of suspected stroke patients on the basis of the diagnostic procedure, complete the basic assessment of brain CT and start treatment within 60 minutes after reaching the emergency room as much as possible, and perform intravenous thrombolysis within 6 hours by Chinese Medical Association Cerebrovascular Disease Group [START_REF] Medical | Guidelines for the diagnosis and treatment of acute ischemic stroke in China[END_REF]. Considering the drawbacks of imaging speed and expense of MRI, CT perfusion (CTP) imaging is widely used in acute stroke diagnosis in China. However, CT radiation dose has always been concerned. How to reduce radiation exposure while maintaining image quality is a key point in CT imaging, which is known as the principle of "as low as reasonably achievable (ALARA)" [START_REF] Krishnamoorthi | Effectiveness of a staged US and CT protocol for the diagnosis of pediatric appendicitis: reducing radiation exposure in the age of ALARA[END_REF]. To reduce radiation dose and related risk of cancers, low dose CT (LDCT) has been of great research interest in CT imaging. Generally speaking, there are three typical approaches to reduce CT radiation: reducing tube current [START_REF] Liu | 3D Feature Constrained Reconstruction for Low Dose CT Imaging[END_REF], sparse-view sampling [START_REF] Bian | Optimization-based image reconstruction from sparse-view data in offset-detector CBCT[END_REF], and region-of-interest (ROI) scanning [START_REF] Cho | Region-of-interest image reconstruction with intensity weighting in circular cone-beam CT for image-guided radiation therapy[END_REF]. Current brain CTP consists of about 30-pass CT images, which means scanning brain for 30 times within about 50 seconds after injecting iodinated contrast agents. Inspired by the sparse-view low-dose CT imaging, we cut down the original 30 passes to 15 passes in time sequence since it is the easiest way to reduce scan radiation for our study. Then some brain perfusion parameters such as CBF (Cerebral Blood Flow), CBV (Cerebral Blood Volume), MTT (Mean Transit Time) and TTP (Time To Peak) were calculated for clinical diagnosis. However, the missing data would lead to miscalculating brain perfusion parameters.

Hence, we intended to recover 30-pass high-quality CTP images from the downsampled CTP images.

Streak artifacts usually occur in sparsely sampled low-dose CT reconstruction when using analytical methods. Interpolating the missing view can be an optional way to reduce artifacts of reconstructed sparse-view CT images [START_REF] Kalke | Sinogram interpolation method for sparse-angle tomography[END_REF][START_REF] Zhang | Directional sinogram interpolation for sparse angular acquisition in cone-beam computed tomography[END_REF]. Recently, deep learning methods have been available for extensive computer vision tasks, such as image classification [START_REF] He | Deep residual learning for image recognition[END_REF], semantic segmentation [START_REF] Chen | Deeplab: Semantic image segmentation with deep convolutional nets, atrous convolution, and fully connected crfs[END_REF][START_REF] Ronneberger | U-net: Convolutional networks for biomedical image segmentation[END_REF], super-resolution generation [START_REF] Dong | Image super-resolution using deep convolutional networks[END_REF] and medical image reconstruction [START_REF] Chen | LEARN: Learned experts' assessment-based reconstruction network for sparse-data CT[END_REF]. As for low-dose CT imaging, H. Lee [START_REF] Lee | High quality imaging from sparsely sampled computed tomography data with deep learning and wavelet transform in various domains[END_REF] and D.

Lee [START_REF] Lee | View-interpolation of sparsely sampled sinogram using convolutional neural network[END_REF] proposed some effective convolutional neural networks (CNN) to interpolate sparsely sampled CT data and reduce the streak artifacts in sinogram and image domain. More recently, a generative adversarial network (GAN) containing convolutional layer was trained to estimate routine-dose CT images from low-dose CT images [START_REF] Wolterink | Generative adversarial networks for noise reduction in low-dose CT[END_REF]. Though the results showed improved performance, the application of comprehensive networks are greatly limited by the high computation cost related to the massive back-projections in network training. Besides, these networks were not developed for CTP image reconstruction task.

In this study, we proposed a simple three-dimensional convolutional neural network to interpolate the missing data to restore temporally downsampled CTP images in time sequence. Then the restored 30-pass CTP images were input to the perfusion tool PMA (ASIST-Japan) [START_REF] Kudo | Differences in CT perfusion maps generated by different commercial software: quantitative analysis by using identical source data of acute stroke patients[END_REF] to calculate CBF, CBV, MTT and TTP for stroke diagnosis. Section II briefly introduce the residual deep learning model. Section III presents the experiment results. In Section IV, we discuss some related issues, and in Section V, we make a conclusion for this work.

II. Methods and Materials

II.A Deep CNN model

Deep convolutional neural networks (DCNN) using back propagation algorithm, sometimes also containing deconvolutional layers, is an outstanding tool for image analysis and image reconstruction [START_REF] Wang | A Perspective on Deep Imaging[END_REF][START_REF] Long | Fully convolutional networks for semantic segmentation[END_REF][START_REF] Krizhevsky | Imagenet classification with deep convolutional neural networks[END_REF]. Considering our CTP data has three dimensions: height, weight and time sequence, we constructed a simple CNN model with 3-D convolutional layers and activation layers. Different from LSTM which is applied in time series analysis [START_REF] Petnehá | Recurrent neural networks for time series forecasting[END_REF], the 3-D convolutional operation can extract information of the images along the direction of width and height as well as that of time series. Thus CNN can probably reconstruct CTP data from temporal information as well as voxel-wise information in image plain. Because rectified linear unit (ReLU) has excellent performance in computer vision tasks and fast convergence speed in training [START_REF] Krizhevsky | Imagenet classification with deep convolutional neural networks[END_REF], it is the only activation function for our model. Meanwhile, pooling layers are not adopted to keep the output image size identical to input image size.

The proposed CNN model with 8 successive convolutional blocks were implemented by Fig. 1 The structure of residual CNN model. Learning the differences between input data and ground truth, which is commonly called residual, has faster convergence speed when training CNN [START_REF] Lee | View-interpolation of sparsely sampled sinogram using convolutional neural network[END_REF][START_REF] Yang | Improving low-dose ct image using residual convolutional network[END_REF]. This is because it is difficult to map the downsampled image data to ground truth with conventional CNN structure (eg. AlexNet and VGG) directly, since CNNs are usually initialized with small weights which will result in a large loss between the CNN output and ground truth to optimize. However, the residuals, which can be considered as the noise in input data, are much closer to 0 and easier to learn since CNN is initiated with an initializer of ()

O(x) resize f(x)+ x  (eq.1)
where x is the input, f(x) is convolutional operations and O(x) is output. In our task, the input data was divided into smaller patches of size 32*32*15. According to our computation, the receptive field of the 15 successive 5*5*5 convolutional layers is 61, which is large enough to cover the input image. When computing pixel-wise loss, the loss of each pixel was taken into account, which contributed to increasing training data volume significantly. Pixel-wise mean absolute error (L1 loss) was adopted in our proposed model

II.B Data preparation and network training

In this study, the CTP images were collected from eight eligible slice locations of 24 acute stroke patients, as demonstrated in Fig. 3 The proposed CNN model was trained by adaptive moment estimation (Adam) optimization [START_REF] Kingma | Adam: A method for stochastic optimization[END_REF].

The initiated learning rate was set to 1e-4 and then decayed in each epoch. The batch size was set to 32.

It took about 35 hours to train 10 epochs to acquire stable convergence on our PC (Core i5 CPU, 1080Ti GPU, 16G RAM), of which the loss would not decline.

III. Result Analysis

III.A Restoration results

After training, the test dataset was input into the trained CNN model. Original 30-pass CTP images were used as ground truth for verifying the validity of our interpolation CNN model. We also used cubic interpolation and deep convolutional GAN (DCGAN) to restore 15-pass CTP images for comparison.

Cubic interpolation is a conventional interpolation method and easy to realize. DCGAN is a proven effective generative model for feature representation, image generation and image analysis [START_REF] Radford | Unsupervised Representation Learning with Deep Convolutional Generative Adversarial Networks[END_REF]. Besides, 30-pass CTP images were also downsampled to 10 passes and then processed using the retrained CNN of proposed structure. Fig. 4 shows the 12 th -pass and 18 th -pass ground truth CTP images and restored downsampled images of one selected patient using different algorithms from test dataset. The concentration of contrast agents, which is near the peak at 12 th pass, decreases a lot at 18 th pass in a brain blood circulation period. The enlarged images of red box are shown in the top-left corner. From Fig. 4, it can be seen that 15-pass CNN method can obtain high-quality CTP images with finer image details and clearer image edges, especially vascular details which is critical to calculate perfusion parameters. However, results of 10-pass CNN (c) are a bit fuzzy and lose some highlighted vessel pixels. Results of DCGAN (e) still exist some noise since that training of GAN is not stable [START_REF] Gulrajani | Improved training of wasserstein gans[END_REF]. Fig. 5 gives the 30-pass time density curve (TDC) of anterior cerebral artery from the selected patient. From Fig. 5, we can notice that curves generated by our 15-pass CNN is the closest to the ground truth, especially when the contrast density is high-level from the 10 th pass to 15 th pass. Curve of 10-pass CNN (yellow) loses a lot of information during 10 th and 15 th pass.

Curve of cubic (purple) and DCGAN (green) show lower peak values and delay of reaching peak in time sequence. That will lead to higher TTP value when calculating perfusion parameters.

(1a) (1b) (1c) (1d) (1e) (2a) (2b) (2c) (2d) (2e)
Fig. 5 The time density curves of anterior cerebral artery from selected patient.

III.B Quantitative analysis

We evaluated the restored image quality quantitatively with structural similarity index (SSIM) and peak signal-to-noise ratio (PSNR). Definition of SSIM is demonstrated in eq. 2, (2

)(2 ) ( , ) ( )( ) x y 1 xy x y 1 x y cc SSIM x y c c              (eq.2)
where x and y are two images of same size,  and  are average and covariance,

III.C Perfusion parameters results

For acute stroke diagnosis and treatment, we care much about perfusion parameter maps. After restoring downsampled CTP images with CNN, we input the restored CTP to calculate perfusion maps with PMA software. Fig. 6 gives the CBF, CBV, MTT, TTP maps of the selected patient at slice location 2 presents the average perfusion values of four methods. Table

(1a) (1b) (1c) (1d) (2a) (2b) (2c) (2d) (3a) (3b) (3c) (3d)
3 gives the root mean square error (RMSE) and mean absolute percentage error (MAPE) between different method results and ground truth of six test patients' data. From Table 2 and Table 3, we can see for imaging since that running on a PC with Core i5 CPU. However, running time for each patient with DCGAN is over 77s because of too many parameters. Cubic interpolation has the fastest speed, with processing time of 6s each, as shown in Table4, but has low imaging quality.

III.D Model efficiency analysis

IV. Discussion

Perfusion imaging is a common method for acute stroke examination. It usually takes about 50s for blood to circulate the whole body. Most imaging workstations, like GE and Philips, scan brain 30 times within 50s after injecting contrast agents consequently. The 30-pass CTP scan will significantly increase the radiation dose. Hence, it inspires us to reduce the scan frequency. We first downsampled the original CTP to 15 passes to reduce a half radiation dose. However if we calculated with the scan-reduced CTP images directly, the perfusion parameter value would be inaccurate as illustrated in Fig. After restoring 30-pass CTP images from downsampled 15-pass CTP, perfusion parameters can be computed. However, different perfusion tools will acquire different results, even from identical source data [START_REF] Kudo | Differences in CT perfusion maps generated by different commercial software: quantitative analysis by using identical source data of acute stroke patients[END_REF][START_REF] Austein | Comparison of perfusion CT software to predict the final infarct volume after thrombectomy[END_REF]. Some softwares may overestimate or underestimate ischemic core. This is probably because of differences in tracer delay sensitivity and postprocessing algorithms. The common perfusion postprocessing algorithms can be classified into three categories: Gamma Fitting, Maximum Slope and Deconvolution. Kudo et al. [START_REF] Kudo | Accuracy and reliability assessment of CT and MR perfusion analysis software using a digital phantom[END_REF] assessed the accuracy and reliability using a digital phantom with 13 perfusion postprocessing algorithms. Experiment results illustrate that the single value decomposition (SVD) of PMA can achieve the closest CBF, CBV and MTT values to ground truth. Hence, PMA was employed to calculate CBF and other perfusion maps in this study. We calculated the mean perfusion values from 15-pass CNN, 10-pass CNN, cubic interpolation and DCGAN output using PMA. And the errors prove that 15-pass CNN has the best results.

V. Conclusion

In this work, we proposed an optional way to reduce CTP imaging radiation dose, that is downsampling 30-pass images to 15 passes in temporal domain and then restoring them to 30 passes with the deep residual CNN model. Finally, with the help of the radiologists of the First Hospital of Jilin University in China, we can find out the coincident hypoperfusion areas according to perfusion parameters calculated from restored CTP and original CTP. Experiment results have presented its effectiveness. Moreover, the post-processing time is also within acceptable limits. In the future, we will focus on reducing radiation dose not only in time sequence, but also in spatial domain [START_REF] Liu | Discriminative feature representation to improve projection data inconsistency for low dose CT imaging[END_REF]. How to cut down the CTP scan frequency and reduce the imaging tube current at the same time and then process with a CNN model to recover high-quality CTP images is our next research program.

TensorFlow (Version 1 .

 1 [START_REF] Chen | Deeplab: Semantic image segmentation with deep convolutional nets, atrous convolution, and fully connected crfs[END_REF]. Each convolutional block consists of two 3-D convolutional layer followed by ReLU activation layers. The first 7 blocks in blue have 64-channel feature maps while the last block in red has 1-channel output, as shown in Fig.1. A convolution based plain network has connections only between adjacent layers, as shown in Fig.2(a). It will increase the training difficulty when the network becomes deeper due to gradient diffusion. In ResNet, a simple identity mapping directly connecting the input and output layers and bypass connection are added into the network[START_REF] He | Deep residual learning for image recognition[END_REF][START_REF] Chen | Low-dose CT with a residual encoder-decoder convolutional neural network[END_REF]. In the end, the 7 blue blocks were implemented with network structure of Fig.2(b) and the last red block were implemented as Fig. 2(c). For each convolutional layer, kernel size was set to 5*5*5, stride was set to 1 and padding zeros at the edges of each feature map to keep image size consistent. Original CTP images of 30 passes were treated as ground truth. The downsampled 15-pass CTP was input to CNN model to extract feature maps and finally feature maps were resized to 30 passes as restored CTP output. Here the resize operation is easily realized with a deconvolutional layer with stride (1,1,2).

Fig. 2

 2 Fig. 2 Plain network and residual network: (a) plain network, and (b) residual network, (c) revised residual network. truncated normal distribution, where mean and standard deviation are 0.0 and 1.0 respectively. The proposed residual CNN model can be expressed by eq. 1, () O(x) resize f(x)+ x  (eq.1)

  . The scan protocol is detailed as follows: tube voltage, 80 kV; tube current, 250mA; slice thickness, 5mm. Each slice location has 30 CTP images with 512*512 pixels corresponding to 30 pass in time sequence. 18 of 24 patients were randomly selected as training set and the other 6 patients were used as test set. The original 512*512*30 data volume for each slice location was first downsampled to the dataset of size 512*512*15 to simulate reducing 15 scans. We subdivided the data volume to 32*32*N (N is 15 for training and 30 for test) 3-D patches along width and height directions as the input of CNN network, since batch size needs to be set higher when training the network with small patches [24]. Cropping stride was set to 16 and 32 for training and test dataset respectively.

Fig. 3

 3 Fig. 3 Eight slice locations of selected CTP scans.

Fig. 4

 4 Fig. 4 The first row and second row are the 12 th pass and 18 th pass images of selected patient at slice location 64. (a) is the original image, (b) is result of our 15-pass CNN, (c) is result of 10-pass CNN, (d) is result of 15-pass cubic interpolation, (e) is result of 15-pass DCGAN.

Fig. 6

 6 Fig. 6 Perfusion maps of the selected patient. Three rows are maps calculated from ground truth CTP, 15-pass CNN restored results, and unprocessed 15-pass CTP respectively. (a) is CBF, (b) is CBV, (c) is MTT, (d) is TTP. 64.Then two radiologists of our group reviewed the perfusion maps of different methods together and gave conclusions by consensus. According to Fig. 6, the restored CTP calculated perfusion maps share high similarity with ground truth CTP calculated maps. Both the first two rows indicate CBF descending, CBV ascending, MTT ascending, TTP ascending in right temporal lobe. In the third row, we gave the perfusion maps of unprocessed 15-pass images. In fact, the maps show imaging quality of unprocessed 15-pass images is worse than restored CTP images, which contain more scattered points. As indicated by the white arrows, perfusion maps of unprocessed images cannot reveal CBF severely degraded areas. Besides, the perfusion value ranges of unprocessed images greatly differs from that of ground truth. The maps of the third row are misleading for radiologists' diagnosis. Hence, restoration is essential to improve the quality of perfusion parameter imaging. We tested 6 patients' restored CTP images of test dataset, and radiologists can get the same diagnostic conclusions and assessment as original ground truth CTP maps. Although the other three methods (10-pass CNN, cubic interpolation, DCGAN) could show the similar hypoperfusion areas to ground truth, they would lead to the considerable errors because of larger TTP and smaller CBF, CBV, MTT values. This would mislead radiologists' diagnosis such as changed hypoperfusion area and inaccurate hypoperfusion level of test patients. Take the MTT map for example, Fig. 7 gives five maps of ground truth and the results of different algorithms. It is clear to see that the result of our 15-pass CNN is the closest to ground truth and has the consistent MTT range ([0, 7.6]), while maps of the other three methods have a lot of scattered points and larger MTT range (10-pass CNN: [0, 7.8], cubic: [0, 8.0], DCGAN: [0, 8.2]). It suggests that the 15-pass CNN can attain finer CTP image details and less artifacts which was stated in Section III.A.Table 2 presents the average perfusion values of four methods. Table

  that the parameter values of our 15-pass CNN are closest to the parameters values calculated from ground truth although they have small errors. In comparison, results of the other three methods have larger errors. Considering both RMSE and MAPE, our 15-pass CNN outperforms other three methods. Meanwhile, DCGAN has the least CBF error, displaying the restoration of generative model.

Fig. 7

 7 Fig. 7 MTT maps of the selected patient calculated from different results. (a) is ground truth, (b) is results from 15-pass CNN, (c) is 10-pass CNN, (d) is cubic interpolation, (e) is DCGAN

Table 1

 1 The SSIM and PSNR of global images and ROI using different methods (Mean ± SD)

	1 c k L 1 * 	and

default, and L is the dynamic range of the pixel-values. SSIM ranges from 0 to 1, and 1 means the highest similarity. PSNR is defined as eq.3, Table

1

gives the average SSIM and PSNR of six test patients, each patient has eight slice location data. The global image means we calculated SSIM and PSNR with the whole restoration images of size 512x512. Although these four methods can restore the downsampled CTP effectively, with SSIM all over 0.97and PSNR over 50, the 15-pass CNN model acquire the highest SSIM and PSNR, which explains the best performance on restoring downsampled CTP images. Besides, our radiologists also selected regions of interest (ROI) of size 80x80 from the hypoperfusion areas to calculate mean ROI SSIM and PSNR. Results are given in Table

1

too. Among these four methods, the 10-pass CNN shows the lowest SSIM and PSNR, illustrating that it is hard to restore CTP images if omitting too many passes.

Table 2

 2 The perfusion parameters using different algorithms of selected patient (Mean ± SD)

		Ground Truth	15-pass CNN	10-pass CNN	Cubic	DCGAN
	CBF	35.08±6.95	34.50±5.16	33.59±6.73	33.98±6.64	33.59±6.76
	CBV	2.41±0.46	2.39±0.42	2.21±0.33	2.28±0.46	2.27±0.49
	MTT	4.32±0.55	4.33±0.34	4.22±0.38	4.22±0.51	4.27±0.57
	TTP	8.71±0.37	8.76±0.11	9.26±0.46	9.24±0.11	9.25±0.12

Table 3

 3 The RMSE and MAPE of perfusion parameters using different algorithms

			RMSE			MAPE
		15-pass	10-pass	Cubic	DCGAN	15-pass	10-pass	Cubic	DCGAN
		CNN	CNN			CNN	CNN
	CBF	3.245	3.208	2.674	2.531	6.73%	7.66%	6.12%	5.70%
	CBV	0.121	0.280	0.185	0.188	4.44%	8.76%	5.73%	6.53%
	MTT	1.653	3.321	1.801	1.759	4.58%	8.25%	3.39%	3.09%
	TTP	2.399	5.806	2.897	2.908	3.49%	8.61%	5.40%	5.45%

Table 4

 4 The parameter amount and mean processing time of different modelsThe massive parameters contained in DCGAN make it hard to train. Due to the same topological structures of 15-pass CNN and 10-pass CNN, their parameter amounts are almost same. The parameter amounts of classical VGG-16[START_REF] Andrew | Very deep convolutional networks for large-scale image recognition[END_REF] and ResNet-50[START_REF] He | Deep residual learning for image recognition[END_REF] are 138 million and 25 million, so our 15-pass CNN is much simpler by contrast. After training the CNN model, it took about 220 seconds to generate all six patients' restoration CTP images. The mean time for each patient is 36.6s, which is fast enough

	Model	Parameter amount (million)	Processing time (second)
	15-pass CNN	7.1849	36.6
	10-pass CNN	7.1848	32.0
	Cubic	-	6
	DCGAN	64.92	77.2
	In this section, we mainly discuss the model running speed and memory use. For deep neural
	networks, the model parameter amount is most concerned. In this study, the parameter amount of 15-
	pass CNN is about 7.1849 million, while 10-pass CNN is 7.1848 million and DCGAN is 64.92 million.

  [START_REF] Cho | Region-of-interest image reconstruction with intensity weighting in circular cone-beam CT for image-guided radiation therapy[END_REF]. Then we used a trained CNN model to restore 15-pass CTP images to 30-pass images for calculating perfusion maps.Residual learning is necessary in this task. In our experiments, if we did not add the residual skip connection in CNN structure, the L1-loss value would be very large during the first several training iterations. However, in residual structure, the CNN just needs to learn the difference between downsampled CTP and ground truth. It will accelerate the convergence speed of deep neural networks.According to our experiments, the loss can decline quickly with residual skip connection. It only needs less than 200 iterations to bring down loss below 1 with initial learning rate of 1e-4. In comparison, the

loss would descend slowly and fluctuate if without residual skip connection. At last, we just trained our CNN model 10 epochs efficiently. Residual learning has shown its excellent performance in the restoration CNN.

Funding

This work was supported in part by the State's Key Project of Research and Development Plan under Grant 2017YFA0104302, Grant 2017YFC0109202 and 2017YFC0107900, National Natural Science Foundation under Grant 81530060, 31571001 and 81471752, and the R&D Projects in Key Technology Areas of Guangdong Province under Grant 2018B030333001.

Compliance with ethical standards

Conflict of Interest

The authors declare that they have no conflict of interest.

Ethical approval All procedures performed in studies involving human participants were in accordance with the ethical standards of the institutional and/or national research committee and with the 1964 Helsinki declaration and its later amendments or comparable ethical standards.

Informed consent Informed consent was obtained from all individual participants included in the study.


Daciberg Lima Gonçalves

John Guaschi
email: john.guaschi@unicaen.fr

Vinicius Casteluber Laass
email: vinicius.laass@ufba.br

The Borsuk-Ulam property for homotopy classes of maps between the torus and the Klein bottle

Let M be a topological space that admits a free involution τ , and let N be a topological space. A homotopy class β ∈ [M, N ] is said to have the Borsuk-Ulam property with respect to τ if for every representative map f : M → N of β, there exists a point x ∈ M such that f (τ (x)) = f (x). In this paper, we determine the homotopy classes of maps from the 2-torus T 2 to the Klein bottle K 2 that possess the Borsuk-Ulam property with respect to a free involution τ 1 of T 2 for which the orbit space is T 2 . Our results are given in terms of a certain family of homomorphisms involving the fundamental groups of T 2 and K 2 .

Introduction

In the early twentieth century, St. Ulam conjectured that if f : S n → R n is a continuous map, there exists x ∈ S n such that f (A(x)) = f (x), where A : S n → S n is the antipodal map. The confirmation of this result by K. Borsuk in 1933 [START_REF] Borsuk | Drei Sätze über die n-dimensionale Euklidische Sphäre[END_REF], known as the Borsuk-Ulam theorem, was the beginning of what it now referred to as Borsuk-Ulam type theorems or the Borsuk-Ulam property. More information about the history and some applications of the Borsuk-Ulam theorem may be found in [START_REF] Matoušek | Using the Borsuk-Ulam theorem[END_REF], for example.

One possible generalisation of the classical Borsuk-Ulam theorem is to substitute S n and R n by other spaces, and to replace the antipodal map by a free involution. A natural question is the following: does every continuous map collapse an orbit of the involution? More precisely, given topological spaces M and N such that M admits a free involution τ , we say that the triple (M, τ ; N ) has the Borsuk-Ulam property if for every continuous map f : M → N , there exists a point x ∈ M such that f (τ (x)) = f (x). For the cases where M is a compact surface without boundary admitting a free involution τ and N is either R 2 or a compact surface without boundary, the triples (M, τ ; N ) that have the Borsuk-Ulam property have been classified (see [START_REF] Gonçalves | The Borsuk-Ulam theorem for surfaces[END_REF] and [START_REF] Gonçalves | The Borsuk-Ulam theorem for maps into a surface[END_REF]). One generalisation of this property is to consider a local Borsuk-Ulam problem in the sense of the following definition: a homotopy class β ∈ [M, N ] has the Borsuk-Ulam property with respect to τ if for every representative f : M → N of β, there exists a point x ∈ M such that f (τ (x)) = f (x).

In [START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF], the Borsuk-Ulam problem for homotopy classes of maps between compact surfaces without boundary was studied, and the sets [T 2 , T 2 ] and [K 2 , K 2 ] whose elements possess the Borsuk-Ulam property were characterised. By [START_REF] Gonçalves | The Borsuk-Ulam theorem for maps into a surface[END_REF]Theorem 12], for any involution τ : T 2 → T 2 , the triple (T 2 , τ ; K 2 ) does not have the Borsuk-Ulam property. Using this information, in this paper we classify the homotopy classes of maps from T 2 to K 2 that have the Borsuk-Ulam property for the orientation-preserving free involution τ 1 of T 2 . Our approach, which we now describe, is similar to that used in [START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF]. First, as in [START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF]Theorems 12 and 19], we identify π 1 (T 2 , * ) and π 1 (K 2 , * ) with the free Abelian group Z ⊕ Z and the (non-trivial) semi-direct product Z ⋊ Z respectively. These identifications will be helpful in formulating the results and in making explicit computations.

To prove our results, we will make use of the following algebraic description given in [6, Corollary 2.1] of the set [T 2 , K 2 ] in terms of pointed homotopy classes and the corresponding fundamental groups. (1, 0) → (i, 2s 1 + 1) (0, 1) → (0, 2s 2 )

Type 2:

(1, 0) → (i, 2s 1 + 1) (0, 1) → (i, 2s 2 + 1)

Type 3:

(1, 0) → (0, 2s 1 ) (0, 1) → (i, 2s 2 + 1)

Type 4:

(1, 0) → (r 1 , 2s 1 ) (0, 1) → (r 2 , 2s 2 ),

where i ∈ {0, 1} and s 1 , s 2 ∈ Z for Types 1,2 and 3, and r 1 , r 2 , s 1 , s 2 ∈ Z and r 1 ≥ 0 for Type 4.

Remark 1.2. The bijection of Proposition 1.1 may be obtained using standard arguments in homotopy theory that are described in detail in [11, Chapter V, Corollary 4.4], and more briefly in [START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF]Theorem 4]. In our specific case, the bijection is defined as follows: given a homotopy class β ∈ [T 2 , K 2 ], there exists a pointed map f : (T 2 , * ) → (K 2 , * ) that gives rise to a representative of β if we omit the basepoints. The induced homomorphism f # : π 1 (T 2 , * ) → π 1 (K 2 , * ) is conjugate to exactly one of the elements of Hom(Z ⊕ Z, Z ⋊ Z) described in Proposition 1.1, which we denote by β # . Note that β # does not depend on the choice of f .

In order to solve the Borsuk-Ulam problem for homotopy classes, we now describe the relevant involution of T 2 . Consider the following short exact sequence:

1 → π 1 (T 2 ) = Z ⊕ Z i 1 -→ π 1 (T 2 ) = Z ⊕ Z θ 1 -→ Z 2 → 1 (1)
where:

i 1 :
(1, 0) -→ (2, 0) (0, 1) -→ (0, 1) θ 1 :

(1, 0) -→ 1 (0, 1) -→ 0.

By standard results in covering space theory, there exists a double covering c 1 : T 2 → T 2 whose induced homomorphism on the level of fundamental groups is i 1 . If τ 1 : T 2 → T 2 is the nontrivial deck transformation associated with c 1 , then τ 1 is a free involution. Further, τ 1 lifts to a homeomorphism τ 1 : R 2 → R 2 , where τ 1 (x, y) = (x + 1 2 , y) for all (x, y) ∈ R 2 . In this paper, we classify the elements of the set [T 2 , K 2 ] that possess the Borsuk-Ulam property with respect to τ 1 . This is achieved in the following theorem, which is the main result of this paper. if r 1 = 0.

Besides the introduction and an Appendix, this paper consists of three sections. In Section 2, we show how to reduce the number of homotopy classes to be studied with respect to the Borsuk-Ulam property. In Section 3, we study the normal closure of σ 2 in P 2 (K 2 ), which is a free group of infinite rank. A convenient basis for this subgroup is obtained in the Appendix. In Section 4, we prove Theorem 1.3.

The study of the free involution τ 2 of T 2 for which the associated orbit space is the Klein bottle is the subject of work in progress.

Some preliminary results

The following results will enable us to reduce the number of cases to be analysed in the proof of Theorems 1.3. Lemma 2.1. Let M and N be topological spaces, let τ : M → M be a free involution, and let Proof. Clearly the map H is a bijection whose inverse is given by H

H : N → N be a homeomorphism. Then the map H : [M, N ] → [M, N ] defined by H([f ]) = [H •f ] for all maps f : M → N is a bijection. Further, if β ∈ [M, N ] is a
-1 ([g]) = [H -1 • g].
To prove the second part of the statement, let β ∈ [M, N ] be a homotopy class that has the Borsuk-Ulam property with respect to τ , and let g ∈ H(β). Thus H -1 • g ∈ β, and hence there exists x ∈ M such that H -1 • g(x) = H -1 • g(τ (x)). Therefore g(x) = g(τ (x)), and we conclude that H(β) has the Borsuk-Ulam property. The converse follows in a similar manner using H -1 . Proposition 2.2. Let τ : T 2 → T 2 be a free involution, and let β, β ′ ∈ [T 2 ; K 2 ] such that β # , β ′ # are both of Type 1, 2 or 3. Suppose that the second coordinates of β # (ω) and β ′ # (ω) are equal for all ω ∈ π 1 (T 2 , * ) and the integer i that defines the homomorphism β # (resp. β ′ # ) is equal to 0 (resp. 1). Then β has the Borsuk-Ulam property with respect to τ if and only if β ′ does.

Proof. Let h : Z ⋊ Z → Z ⋊ Z be the homomorphism defined on the generators of Z ⋊ Z by h(1, 0) = (1, 0) and h(0, 1) = (1, 1). Then h is well defined, and it is an isomorphism whose

inverse h -1 : Z ⋊ Z → Z ⋊ Z is given by h -1 (1, 0) = (1, 0) and h -1 (0, 1) = (-1, 1). By [12, Theorem 5.6.2], there exists a homeomorphism H : (K 2 , * ) → (K 2 , * ) such that H # = h and H -1 # = h -1 .
Suppose that β # and β ′ # are both of Type 1, and that they satisfy the hypothesis of the statement, and let f : (T 2 , * ) → (K 2 , * ) be a representative map of β. Without loss of generality, we may suppose that f # = β # . Assume that β has the Borsuk-Ulam property with respect to τ . Then:

(H • f ) # (1, 0) = h(β # (1, 0)) = h(0, 2s 1 + 1) = h(0, 1) 2s 1 +1 = ((1, 1)(1, 1)) s 1 (1, 1) = (0, 2) s 1 (1, 1) = (1, 2s 1 + 1), and (H • f ) # (0, 1) = h(β # (0, 1)) = h(0, 2s 2 ) = h(0, 1) 2s 2 = ((1, 1)(1, 1)) s 2 = (0, 2) s 2 = (0, 2s 2 ).
Then H • f is a representative of β ′ , and the conclusion follows from Lemma 2.1. The converse follows in a similar way using H -1 instead H. The arguments for the cases of homomorphisms of Types 2 and 3 are analogous, and the details are left to the reader. Remark 2.3. Using Lemma 2.1, one may show that Proposition 2.2 holds in more generality. However the above statement will be sufficient for our purposes.

3 The normal closure of σ 2 in P 2 (K 2 )

Let S be a compact, connected surface without boundary. The ordered 2-point configuration space of S is defined by F 2 (S) = {(x, y) ∈ S × S | x = y}, D 2 (S) is the orbit space of F 2 (S) by the free involution τ S : F 2 (S) → F 2 (S) defined by τ S (x, y) = (y, x), and P 2 (S) = π 1 (F 2 (S)) and B 2 (S) = π 1 (D 2 (S)) are the pure and full 2-string braid groups of S respectively [START_REF] Fadell | Configuration spaces[END_REF]. We have a short exact sequence:

1 → P 2 (S) → B 2 (S) π → Z 2 → 1, (2)
where π : B 2 (S) → Z 2 is the homomorphism that to an element of B 2 (S) associates its permutation. If p 1 : F 2 (S) → S is the projection onto the first coordinate, the map (p 1 ) # : P 2 (S) → π 1 (S) may be interpreted geometrically as the surjective homomorphism that forgets the second string. Let σ ∈ B 2 (S)\P 2 (S) be the Artin generator of B 2 (S) that geometrically swaps the two basepoints.

Then σ 2 ∈ P 2 (S), and the normal closure of σ 2 in P 2 (S), which we denote by σ 2 , is also the normal closure of σ 2 in B 2 (S) by [START_REF] Fadell | Configuration spaces[END_REF]. In this section, we shall take S to be the Klein bottle, and we will show that σ 2 is a free group of countably-infinite rank for which we shall exhibit a basis. We will also express certain elements of σ 2 in this basis.

The following proposition summarises some results of [START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF]Section 4] regarding the structure of P 2 (K 2 ) and the action by conjugation by σ on this group. Proposition 3.1. [START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF]Theorems 19 and 20] The group

P 2 (K 2 ) is isomorphic to the semi-direct product F (u, v) ⋊ θ (Z ⋊ Z)
, where F (u, v) is the free group of rank 2 on the set {u, v} and the action θ : Z ⋊ Z → Aut(F (u, v)) is defined as follows:

θ(m, n) :      u → B m-δn u εn B -m+δn v → B m vu -2m B -m+δn B → B εn , where δ n = 0 if n is even 1 if n is odd, ε n = (-1
) n and B = uvuv -1 . With respect to this description, the following properties hold:

• the element σ ∈ B 2 (K 2 ) satisfies σ 2 = (B; 0, 0).

• if l σ : P 2 (K 2 ) → P 2 (K 2
) is the homomorphism defined by l σ (b) = σbσ -1 for all b ∈ P 2 (K 2 ), then:

l σ (u r ; 0, 0) = ((Bu -1 ) r B -r ; r, 0) l σ (1; m, 0) = (1; m, 0) l σ (v s ; 0, 0) = ((uv) -s (uB) δs ; 0, s) l σ (1; 0, n) = (B δn ; 0, n) l σ (B; 0, 0) = (B; 0, 0)
for all m, n, r, s ∈ Z, where the symbol 1 denotes the trivial element of F (u, v).

• the homomorphism (p 1 ) # :

P 2 (K 2 ) → π 1 (K 2 ) = Z ⋊ Z satisfies (p 1 ) # (w; r, s) = (r, s).
From now on, we identify P 2 (K 2 ) with F (u, v) ⋊ θ (Z ⋊ Z) without further comment. Remark 3.2. It follows from Proposition 3.1 that for all m, n ∈ Z, the automorphism θ(m, n) : F (u, v) → F (u, v) depends only on m and the parity of n, in particular θ(m, n) = θ(m, δ n ).

Consider the following maps:

ι : F (u, v) -→ P 2 (K 2 )
w -→ (w; 0, 0) and p F :

P 2 (K 2 ) -→ F (u, v) (w; m, n) -→ w.
Note that ι is a homomorphism, but due to the action θ, p F is not. Consider the map ρ : F (u, v) → F (u, v) defined by:

ρ = p F • l σ • ι, (3)
and the homomorphism g : F (u, v) → Z ⋊ Z defined on the basis {u, v} by:

g(u) = (1, 0) g(v) = (0, 1). (4)
Using Proposition 3.1 and (3), we obtain the following commutative diagram:

F (u, v) F (u, v) ι / / g , , Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y Y P 2 (K 2 ) lσ / / P 2 (K 2 ) = F (u, v) ⋊ θ (Z ⋊ Z) (p 1 ) # p F O O Z ⋊ Z = π 1 (K 2 ), (5)
from which it follows that: l σ (w; 0, 0) = (ρ(w); g(w)) [START_REF] Gonçalves | Coincidence properties for maps from the torus to the Klein bottle[END_REF] for all w ∈ F (u, v). Further if w, z ∈ F (u, v) then: ρ(wz) = (p F • l σ )(wz; 0, 0) = p F (l σ (w; 0, 0). l σ (z; 0, 0)) = p F (ρ(w); g(w)). (ρ(z); g(z))

= p F ρ(w)θ(g(w))(ρ(z)); g(w)g(z) = ρ(w)θ(g(w))(ρ(z)).

Thus the map ρ : F (u, v) → F (u, v) is not a homomorphism, but if w ∈ Ker (g) then:

ρ(wz) = ρ(w)ρ(z). (7)
In Theorem A.3 of the Appendix, we prove that:

Ker (g) = B k,l := v k u l Bu -l v -k , k, l ∈ Z |-. ( 8
)
Let w ∈ F (u, v) and (m, n) ∈ Z ⋊ Z. Using Proposition 3.1, we see that B ∈ Ker (g), and then that:

θ(m, n)(wBw -1 ) = θ(m, n)(w)B εn θ(m, n)(w) -1 ∈ Ker (g) . ( 9
)
Further, l σ (wBw -1 ; 0, 0) = l σ (w; 0, 0)l σ (B; 0, 0)l σ (w; 0, 0) -1 (5) = (ρ(w); g(w))(B; 0, 0)(ρ(w); g(w)

) -1 = (ρ(w)θ(g(w))(B); g(w))(θ(g(w) -1 )(ρ(w) -1 ); g(w) -1 ) = (ρ(w)θ(g(w))(B)ρ(w) -1 ; 0, 0). (10)
Composing [START_REF] Murasugi | A study of braids[END_REF] by p F , it follows from ( 5) and ( 9) that:

ρ(wBw -1 ) = ρ(w)θ(g(w))(B)ρ(w) -1 ∈ Ker (g) . ( 11
)
In particular, ρ(B k,l ) ∈ Ker (g) for all k, l ∈ Z, and using ( 7) and ( 8), the restriction of ρ to Ker (g) yields an endomorphism of Ker (g), which we also denote by ρ. Further, θ(m, n)(B k,l ) ∈ Ker (g) for all k, l, m, n ∈ Z using ( 9), and thus θ determines a homomorphism from Z ⋊ Z to Aut(Ker (g)), which we also denote by θ. Note that for all m, m ∈ Z, the endomorphism θ(m, n) : Ker (g) → Ker (g) is indeed surjective. To see this, let k, l ∈ Z, and let

ξ ∈ F (u, v) be such that θ(m, n)(ξ) = v k u l . Then θ(m, n)(ξB εn ξ -1 ) = B k,l
, and ξB εn ξ -1 ∈ Ker (g) because B ∈ Ker (g). Hence the image of θ(m, n) : Ker (g) → Ker (g) contains the basis {B k,l } k,l∈Z of Ker (g), and so this homomorphism is surjective.

The following result describes the subgroup σ 2 .

Proposition 3.3. The injective homomorphism ι : F (u, v) → P 2 (K 2 ) defined by ι(w) = (w; 0, 0) for all w ∈ F (u, v), restricts to an isomorphism between Ker (g) and σ 2 , which we also denote by ι. In particular, σ 2 is a free group of infinite rank for which a basis is given by {(B k,l ; 0, 0)} k,l∈Z , and l σ ( σ 2 ) ⊂ Ker ((p 1 ) # ). Up to this isomorphism, the homomorphisms θ : Z⋊Z → Aut(Ker (g)) and ρ : Ker (g) → Ker (g) induce homomorphisms Z ⋊ Z → Aut( σ 2 ) and σ 2 → σ 2 , which we also denote by θ and ρ respectively. Further, the following diagram is commutative:

σ 2 ρ / / _ σ 2 _ P 2 (K 2 ) lσ / / P 2 (K 2 ).
Proof. Let H = ι(Ker (g)) ⊂ P 2 (K 2 ). We start by showing that σ 2 = H. To see this, first note that σ 2 = (B; 0, 0) and ι(B k,l ) ∈ σ 2 for all k, l ∈ Z, hence H ⊂ σ 2 by (8). Conversely, for all w ∈ F (u, v), q ∈ Z ⋊ Z and k, l ∈ Z, we have:

(w; q)(B k,l ; 0, 0)(w; q) -1 =(wθ(q)(B k,l ); q)(θ(q -1 )(w -1 ); q -1 ) = (wθ(q)(B k,l )w -1 ; 0, 0).

Since θ(q)(B k,l ) ∈ Ker (g) and Ker (g) is normal in F (u, v), it follows that wθ(q)(B k,l )w -1 ∈ Ker (g), and so (w; q)(B k,l ; 0, 0)(w; q) -1 ∈ ι(Ker (g)) = H by [START_REF] Zieschang | Surfaces and planar discontinuous groups[END_REF]. Hence H is a normal subgroup of P 2 (K 2 ) by ( 8), and since σ 2 ∈ H, we conclude that σ 2 ⊂ H. Thus σ 2 = H as required. Since ι is injective, the induced homomorphism ι : Ker (g) → σ 2 is an isomorphism, and the image of the basis {B k,l } k,l∈Z by ι yields a basis of the free group σ 2 . The commutativity of the given diagram follows by considering the images of the elements of this basis {(B k,l ; 0, 0)} k,l∈Z and using [START_REF] Murasugi | A study of braids[END_REF].

Remark 3.4. Although Ker (g) and σ 2 are isomorphic by Proposition 3.3, our results will be stated in terms of σ 2 , since we will formulate most of our equations in this group.

The following result provides a normal form for elements of F (u, v) in terms of g and σ 2 .

Proposition 3.5. Let w ∈ F (u, v), and let g(w) = (r, s). Then there exists a unique element x ∈ σ 2 such that w = u r v s x.

Proof. Let w be as in the statement, and let x = v -s u -r w. Then w = u r v s x, and:

g(x) = g(v -s u -r w) = g(v) -s g(u) -r g(w) = (0, -s)(-r, 0)(r, s) = (0, -s)(0, s) = (0, 0). So x ∈ σ 2 . Clearly x is unique. Let p, q ∈ Z. Since σ 2 is a normal subgroup of F (u, v), the following homomorphism is well defined: c p,q : σ 2 -→ σ 2 x -→ v p u q xu -q v -p . ( 13
)
Lemma 3.6. Let k, l, p, q ∈ Z, and let (m, n) ∈ Z ⋊ Z. With the notation of Proposition 3.1, there exist γ, λ, η ∈ σ 2 such that:

(a) θ(m, n)(B k,l ) = γB εn k,εnl-2δ k m γ -1 . (b) ρ(B k,l ) = λB ε k -k,ε (k+1) l λ -1 . (c) c p,q (B k,l ) = ηB k+p,l+ε k q η -1 .
Proof. During the proof, we will make use freely of Proposition 3.1. First, we have:

θ(m, n)(B k,l ) = θ(m, n)(v k u l Bu -l v -k ) = θ(m, n)(v k u l )B εn θ(m, n)(v k u l ) -1 = γB εn k,εnl-2δ k m γ -1 , where γ = θ(m, n)(v k u l )u ε n+1 l+2δ k m v -k ∈ F (u, v).
To complete the proof of item (a), it remains to show that γ ∈ Ker (g). Since:

γ = (B m vu -2m B -m+δn ) k (B m-δn u εn B -m+δn ) l u ε n+1 l+2δ k m v -k ,
and B ∈ Ker (g), it follows that:

g(γ) = ((0, 1)(-2m, 0)) k (ε n l, 0)(ε n+1 l + 2δ k m, 0)(0, -k) = (2m, 1) k (ε n l + ε n+1 l + 2δ k m, -k) = (2δ k m, k)(2δ k m, -k) = (2δ k m + 2δ k ε k m, 0) = (0, 0), using the fact that ε k = -1 if k is odd, and δ k = 0 if k is even. Hence γ ∈ Ker (g).
To prove item (b), first note that:

θ(g(v k u l ))(B) = θ((0, k)(l, 0))(B) = θ(ε k l, k)(B) = B ε k . ( 14
)
By [START_REF] Whitehead | Elements of homotopy theory[END_REF] and ( 14), we have:

ρ(B k,l ) = ρ(v k u l Bu -l v -k ) = ρ(v k u l )θ(g(v k u l ))(B)ρ(v k u l ) -1 = λB ε k -k,ε (k+1) l λ -1 , where λ = ρ(v k u l )u ε k l v k ∈ F (u, v).
It remains to show that λ ∈ Ker (g). We have:

g(λ) = g(ρ(v k u l )u ε k l v k ) = g((p F • l σ • i)(v k u l )). (ε k l, k) = g(p F (((uv) -k (uB) δ k ; 0, k)((Bu -1 ) l B -l ; l, 0))). (ε k l, k) = g((uv) -k (uB) δ k θ(0, k)((Bu -1 ) l B -l )). (ε k l, k) = (1, 1) -k (δ k , 0). g((B ε k (B -δ k u ε k B δ k ) -1 ) l B -ε k l ). (ε k l, k) = (δ -k , -k)(δ k , 0)(-ε k l, 0)(ε k l, k) = (δ -k + ε -k δ k , 0) = (0, 0), since δ k = 0 if k is even, and ε k = -1 if k is odd. Hence λ ∈ Ker (g).
Finally we prove item (c).

We have:

c p,q (B k,l ) = v p u q B k,l u -q v -p = v p u q v k u l Bu -l v -k u -q v -p = ηB k+p,l+ε k q η -1 , where η = v p u q v k u ε k+1 q v -k-p ∈ F (u, v).
It remains to show that η ∈ Ker (g). This is indeed the case because:

g(η) = (0, p)(q, k)(ε k+1 q, -k -p) = (0, p)(q + ε k ε k+1 q, -p) = (0, p)(0, -p) = (0,
:= v k u l Bu -l v -k | k, l ∈ Z} is a basis, namely: σ 2 Ab = k,l∈Z Z [B k,l ] .
For all (m, n) ∈ Z ⋊ Z and p, q ∈ Z, the endomorphisms θ(m, n), ρ and (c p,q ) of σ 2 induce endomorphisms θ(m, n) Ab , ρ Ab and (c p,q ) Ab of σ 2 Ab respectively, and by Lemma 3.6, for all k, l ∈ Z, they satisfy:

θ(m, n) Ab (B k,l ) = ε n B k,εnl-2δ k m (15) ρ Ab (B k.l ) = ε k B -k,ε (k+1) l , and (16)
(c p,q ) Ab (B k,l ) = B k+p,l+ε k q . ( 17
)
Let k, l ∈ Z and r ∈ {0, 1}. If x and y are elements of a group, let [x, y] = xyx -1 y -1 denote their commutator. In the rest of the paper, we will be particularly interested in the following elements of F (u, v):

(I) T k,r = u k (B εr u -εr ) kεr . (II) I k = v k (vB) -k . (III) O k,l = v 2k , u l . (IV) J k,l = v 2k (vu l ) -2k .
As we shall now see, T k,r , I k , O k,l and J k,l are elements of σ 2 , and their projections into σ 2 Ab will be denoted by T k,r , I k , O k,l and J k,l respectively. The following result describes the decomposition of these Abelianisations in terms of the basis {B k,l } k,l . If l ∈ Z, let σ l denote its sign, i.e σ l = 1 if l > 0, σ l = -1 if l < 0, and σ l = 0 if l = 0.

Proposition 3.7. Let k, l ∈ Z and r ∈ {0, 1}. (a) The elements T k,r , I k , O k,l and J k,l belong to σ 2 . (b) If k = 0 then T 0,r = I 0 = 0, and if k = 0 or l = 0 then O k,l = J k,l = 0.
(c) For all k, l = 0:

T k,r = σ k σ k k i=1 B 0,σ k (i+(σ k (1-2r)-1)/2) I k = -σ k σ k k i=1 B σ k i+(1-σ k )/2,0 J k,l = -σ k σ l σ k k i=1 σ l l j=1 B σ k (2i-1),σ l (j-(1+σ l )/2) O k,l = σ k σ l σ k k i=1 σ l l j=1 B σ k (2i-1),-σ l j+(σ l -1)/2 -B σ k (2i-1)-1,σ l j-(1+σ l )/2 .
The proof of Proposition 3.7, which is divided into the following four lemmas, consists essentially in manipulating each of the elements to obtain recurrence relations, and using induction to obtain expressions for T k,r , I k , O k,l and J k,l in σ 2 . Part (c) of Proposition 3.7 is obtained by Abelianising these expressions.

Lemma 3.8. If k ∈ Z and r ∈ {0, 1} then T k,r = σ k k i=1 B σ k 0,k-σ k i-r+(σ k +1)/2 .
Proof. Let r ∈ {0, 1}. Then clearly T 0,r = 1 and T 1,r = u(B εr u -εr ) εr = B 0,1-r , so the result holds for k = 1. Suppose that the formula is valid for some k ≥ 1. Then:

T k+1,r = u k+1 (B εr u -εr ) (k+1)εr = u. u k (B εr u -εr ) kεr . (B εr u -εr ) εr = uT k,r u -1 . u(B εr u -εr ) εr = u k i=1 B 0,k-i+1-r u -1 . B 0,1-r = k+1 i=1 B 0,k+1-i+1-r ,
and so the result holds for all k ≥ 0 by induction. Suppose that k ≥ 1. Then:

T -k,r = u -k (B εr u -εr ) -kεr = u -k (u k (B εr u -εr ) kεr ) -1 u k = u -k T -1 k,r u k = k i=1 B 0,-i+1-r -1 = k i=1 B -1 0,-k+i-r
using the result for T k,r in the case k ≥ 1, which proves the formula for all k ∈ Z.

I k = σ k k i=1 B i+k(1-σ k )/2,0 -σ k . Proof. If k = 0, then clearly I 0 = 1. If k = 1, I 1 = v(vB) -1 = vB -1 v -1 = B -1 1,0
, and result holds in this case. Suppose that the formula for I k holds for some k ≥ 1, and let us prove the formula for k + 1. We have:

I k+1 = v k+1 (vB) -k-1 = vv k (vB) -k v -1 v(vB) -1 = vI k v -1 I 1 = v k i=1 B i,0 -1 v -1 B -1 1,0 = k i=1 B i+1,0 -1 B -1 1,0 = k+1 i=1 B i,0 -1
, so by induction, the formula for I k holds for all k ≥ 0. If k < 0 then -k > 0 and so:

I k =v k (vB) -k = v k (vB) -k v k v -k = v k v -k (vB) k -1 v -k = v k I -1 -k v -k =v k -k i=1 B i,0 v -k = -k i=1 B k+i,0 .
It follows that the formula given in the statement holds for all k = 0.

Lemma 3.10. Let k, l ∈ Z, let ε ∈ {-1, 1}, and let ω = 0 if ε = -1 1 if ε = 1.
If k = 0 or l = 0, then J k,l = 1, and if k, l > 0, then:

J k,εl = k i=1 l j=1 B -ε 2k-2i+1,εj-ω and J -k,εl = k i=1 l j=1 B -ε -2i+1,εj-ω -1
.

Proof. If k = 0 or l = 0, then clearly J k,l = 1. Now let k = 1 and ε ∈ {1, -1}. Then for all l ≥ 0, we have:

J 1,ε(l+1) = v 2 (vu ε(l+1) ) -2 = v 2 u -ε u -εl v -1 u -ε u -εl v -1 = v 2 (u -εl v -1 u -εl v -1 )vu εl vu -ε v -1 u -ε u -εl v -1 = J 1,εl vu εl+(ε-1)/2 (u (1-ε)/2 vu -ε v -1 u -ε+(ε-1)/2 )u -εl+(1-ε)/2 v -1 = J 1,εl vu εl+(ε-1)/2 B -ε u -εl+(1-ε)/2 v -1 = J 1,εl B -ε 1,εl+(ε-1)/2 . ( 18
)
Hence J 1,1 = B -1 1,0 and J 1,-1 = B 1,-1 , which are in agreement with the expressions of the statement of the lemma for k = l = 1. Suppose now that these expressions hold for k = 1 and some l ≥ 1. Now εl + (ε -1)/2 = ε(l + 1) -(ε + 1)/2 = ε(l + 1) -ω, so by (18), we have:

J 1,ε(l+1) = J k,εl = l j=1 B -ε 1,εj-ω B -ε 1,εl+(ε-1)/2 = l+1 j=1 B -ε 1,εj-ω ,
and thus the expressions of the statement hold for all k, l ≥ 1 by induction. From this, for all k, l > 0 and ε ∈ {1, -1}, it follows that:

J -k,εl = v -2k (vu εl ) 2k = v -2k v 2k (vu εl ) -2k -1 v 2k = v -2k J -1 k,εl v 2k = v -2k k i=1 l j=1 B -ε 2k-2i+1,εj-ω -1 v 2k = k i=1 l j=1 B -ε -2i+1,εj-ω -1
, which completes the proof of the lemma.

Lemma 3.11. Let k, l ∈ Z, let ε ∈ {-1, 1}, and let ω = 0 if ε = -1 1 if ε = 1. If k = 0 or l = 0, then O k,l = 1. If k, l > 0, then for all 1 ≤ i ≤ k and 1 ≤ j ≤ l, there exist η i,j,ε , ζ i,j,ε , µ i,j,ε , ν i,j,ε ∈ σ 2 such that: (a) O εk,l = l j=1 k i=1 η i,j,ε B 2ωk-2i+1,-j η -1 i,j,ε ζ i,j,ε B -1 2ωk-2i,j-1 ζ -1 i,j,ε ε . (b) O εk,-l = l j=1 k i=1 µ i,j,ε B 2ωk-2i+1,l-j µ -1 i,j,ε ν i,j,ε B -1 2ωk-2i,-l+j-1 ν -1 i,j,ε ε -1
.

By taking ε = 1 or -1 in parts (a)
O 1,1 = v 2 uv -2 u -1 = vu -1 uvuv -1 uv -1 vu -1 v -1 u -1 = vu -1 Buv -1 B -1 = B 1,-1 B -1 0,0 ,
and so the expression for O 1,1 given in (a) is correct by taking η 1,1,1 = ζ 1,1,1 = 1. We now suppose that the expression for O k,1 given in (a) holds for some k ≥ 1, where

η i,1,1 = ζ i,1,1 = 1 for all 1 ≤ i ≤ k. Then: O k+1,1 = v 2(k+1) uv -2(k+1) u -1 = v 2 v 2k uv -2k u -1 v -2 v 2 uv -2 u -1 = v 2 O k,1 v -2 O 1,1 = v 2 k i=1 B 2k-2i+1,-1 B -1 2k-2i,0 v -2 B 1,-1 B -1 0,0 = k i=1 B 2(k+1)-2i+1,-1 B -1 2(k+1)-2i,0 B 1,-1 B -1 0,0 = k+1 i=1 B 2(k+1)-2i+1,-1 B -1 2(k+1)-2i,0 .
By induction, it follows that that the expression for O k,1 given in (a) is correct for all k ≥ 1, where

η i,1,1 = ζ i,1,1 = 1 for all 1 ≤ i ≤ k.
Using this, for all k ≥ 1, we obtain:

O -k,1 = v -2k uv 2k u -1 = v -2k (v 2k uv -2k u -1 ) -1 v 2k = v -2k O -1 k,1 v 2k = v -2k k i=1 B 2k-2i+1,-1 B -1 2k-2i,0 -1 v 2k = k i=1 B -2i+1,-1 B -1 -2i,0 -1 .
Hence the expression for O -k,1 given in (a) is correct for all k ≥ 1, where η i,1,-1 = ζ i,1,-1 = 1 for all 1 ≤ i ≤ k. We now suppose that the expression for O εk,l given in (a) is correct for some l ≥ 1 and all k ≥ 1 and ε ∈ {1, -1}, where η i,j,ε , ζ i,j,ε ∈ σ 2 for all 1 ≤ i ≤ k and 1 ≤ j ≤ l. We just saw that this is the case if l = 1. We now study the case l + 1. By Lemma 3.6(c), for all

1 ≤ i ≤ k, there exist η i,l+1,ε , ζ i,l+1,ε ∈ σ 2 such that u l B 2ωk-2i+1,-1 u -l = η i,l+1,ε B 2ωk-2i+1,-l-1 η -1 i,l+1,ε and u l B 2ωk-2i,0 u -l = ζ i,l+1,ε B 2ωk-2i,l ζ -1 i,l+1,ε . Thus: O εk,l+1 =v 2εk u l+1 v -2εk u -l-1 = v 2εk u l v -2εk u -l u l v 2εk uv -2εk u -1 u -l =O εk,l u l O εk,1 u -l = l j=1 k i=1 η i,j,ε B 2ωk-2i+1,-j η -1 i,j,ε ζ i,j,ε B -1 2ωk-2i,j-1 ζ -1 i,j,ε ε u l k i=1 B 2ωk-2i+1,-1 B -1 2ωk-2i,0 ε u -l = l j=1 k i=1 η i,j,ε B 2ωk-2i+1,-j η -1 i,j,ε ζ i,j,ε B -1 2ωk-2i,j-1 ζ -1 i,j,ε ε . k i=1 η i,l+1,ε B 2ωk-2i+1,-l-1 η -1 i,l+1,ε ζ i,l+1,ε B 2ωk-2i,l ζ -1 i,l+1,ε ε = l+1 j=1 k i=1 η i,j,ε B 2ωk-2i+1,-j η -1 i,j,ε ζ i,j,ε B -1 2ωk-2i,j-1 ζ -1 i,j,ε ε .
By induction, it follows that (a) holds for all k, l ≥ 1 and ε ∈ {1, -1}. We now use part (a) to prove part (b). Let k, l ≥ 1 and ε ∈ {1, -1}. By Lemma 3.6(c), for all

1 ≤ i ≤ k and 1 ≤ j ≤ l, there exist µ i,j,ε , ν i,j,ε ∈ σ 2 such that u -l η i,j,ε B 2wk-2i+1,-j η -1 i,j,ε u l = µ i,j,ε B 2ωk-2i+1,l-j µ -1 i,j,ε and u -l ζ i,j,ε B 2wk-2i,j-1 ζ -1 i,j,ε u l = ν i,j,ε B 2ωk-2i,-l+j-1 ν -1 i,j,ε . O εk,-l = v 2εk u -l v -2εk u l = u -l (v 2εk u l v -2εk u -l ) -1 u l = u -l O -1 2εk,l u l = u -l l j=1 k i=1 η i,j,ε B 2ωk-2i+1,-j η -1 i,j,ε ζ i,j,ε B -1 2ωk-2i,j-1 ζ -1 i,j,ε ε -1 u l = l j=1 k i=1 µ i,j,ε B 2ωk-2i+1,l-j µ -1 i,j,ε ν i,j,ε B -1 2ωk-2i,-l+j-1 ν -1 i,j,ε ε -1 , which proves part (b).
(i) al σ (b) = ba. (ii) (p 1 ) # (al σ (a)) = α # (1, 0). (iii) (p 1 ) # (b) = α # (0, 1).
Proof. The result may be obtained in a manner similar to that of [ Proof. Since the statement is symmetric with respect to α and α ′ , it suffices to show that if α does not have the Borsuk-Ulam property then neither does α ′ . If α does not have the Borsuk-Ulam property, there exist a, b ∈ P 2 (K 2 ) satisfying (i)-(iii) of Lemma 4.1. By hypothesis, there exist

α # : (1, 0) → (r 1 , s 1 ) (0, 1) → (r 2 , s 2 ) and α ′ # : (1, 0) → (r 1 , s ′ 1 ) (0, 1) → (r 2 , s ′ 2 ) for some r 1 , r 2 , s 1 , s ′ 1 , s 2 , s ′ 2 ∈ Z. If s 1 ≡ s ′ 1 mod
k 1 , k 2 ∈ Z such that s ′ 1 = s 1 + 4k 1 and s ′ 2 = s 2 + 2k 2 .
Let a ′ = a(1; 0, 2k 1 ) and b ′ = b(1; 0, 2k 2 ) in P 2 (K 2 ). It suffices to show that a ′ and b ′ satisfy (i)-(iii) of Lemma 4.1.Using Proposition 3.1, one may check that the centre of B 2 (K 2 ) is the subgroup (1; 0, 2) . Thus: (i) α ′ # (1, 0) = (0, 2s 1 + 1 mod 4) and α ′ # (0, 1) = (0, j) if α # is of Type 1 (in which case j = 0), or is of Type 2 (in which case j = 1).

a ′ l σ (b ′ ) = a(1; 0, 2k 1 )l σ (b(1; 0, 2k 2 )) = al σ (b)(1; 0, 2k 1 + 2k 2 ) (i) = ba(1; 0, 2k 1 + 2k 2 ) = b ′ a ′ , (p 1 ) # (a ′ l σ (a ′ )) = (p 1 ) # (a(1; 0, 2k 1 )l σ (a(1; 0, 2k 1 ))) = (p 1 ) # (al σ (a)(1; 0, 4k 1 )) (ii) = (r 1 , s 1 )(0, 4k 1 ) = (r 1 , s ′ 1 ) = α ′ # (1, 0), and (p 1 ) # (b ′ ) = (p 1 ) # (b(1; 0, 2k 2 )) (iii) = (r 2 , s 2 )(0, 2k 2 ) = (r 2 , s ′ 2 ) = α ′ # (0,
(ii) α ′ # (1, 0) = (0, 2s 1 mod 4) and α ′ # (0, 1) = (0, 1) if α # is of Type 3. So for each of Types 1, 2 and 3, there are two cases to consider, s 1 = 0, and s 1 = 1.

(b) Suppose that α # is of Type 4, and let r 1 , r 2 , s 1 and s 2 be the integers that appear in the description of α # in Proposition 1.1, where r 1 ≥ 0. By Proposition 2.2 and Corollary 4.2, α has the Borsuk-Ulam property with respect to τ 1 if and only if α ′ does, where α ′ ∈ [T 2 , * ; K 2 , * ] satisfies α ′ # (1, 0) = (r 1 , 2s 1 mod 4) and α ′ # (0, 1) = (r 2 , 0). So for each pair of integers (r 1 , r 2 ), where r 1 ≥ 0, there are two cases to consider, s 1 = 0, and s 1 = 1.

To prove Theorem 1. Then α has the Borsuk-Ulam property with respect to τ 1 .

Σ = {(0, 2s + 1, 0, j) | j, s ∈ {0, 1}} ∪ {(r 1 , 0, r 2 , 0) | r 1 , r 2 ∈ Z, r 1 ≥ 0} ∪ {(0, 2, 0, 0)}. Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class such that α # : (1, 0) → (ρ, γ) (0, 1) → (ξ,
The following lemma will be used in the proofs of some of these propositions. Lemma 4.8. Let a, b ∈ P 2 (K 2 ). Then there exist x, y ∈ σ 2 and a i , b i , m i , n i ∈ Z, where i ∈ {1, 2}, such that:

a = (u a 1 v a 2 x; m 1 , n 1 ) and b = (u b 1 v b 2 y; m 2 , n 2 ). ( 19
)
Suppose further that a and b satisfy the relation of Lemma 4.1(i). Then:

b 2 = 0 and (1 + (-1) δn 1 +1 )m 2 = (1 + (-1) δn 2 +1 )m 1 + (-1) δn 1 b 1 , ( 20
)
so b 1 is even, and:

u b 1 yB m 2 -δn 2 u a 1 εn 2 (B δn 2 vu -2m 2 ) a 2 B δn 2 -m 2 θ(m 2 , δ n 2 )(x) = u a 1 v a 2 xB m 1 -δn 1 (B εn 1 u -εn 1 ) b 1 B -εn 1 b 1 +δn 1 -m 1 θ(m 1 + ε n 1 b 1 , δ n 1 )(ρ(y))B δn 2 εn 1 . (21)
Proof. Let a, b ∈ P 2 (K 2 ). Proposition 3.5 implies that there exist x, y ∈ σ 2 and a i , b i , m i , n i ∈ Z, where i ∈ {1, 2}, for which (19) holds. First, we have:

ba = (u b 1 v b 2 y; m 2 , n 2 )(u a 1 v a 2 x; m 1 , n 1 ) = (u b 1 v b 2 yθ(m 2 , δ n 2 )(u a 1 v a 2 x); m 2 + (-1) δn 2 m 1 , n 2 + n 1 ). ( 22
)
Now:

l σ (b) =((Bu -1 ) b 1 B -b 1 ; b 1 , 0)((uv) -b 2 (uB) δ b 2 ; 0, b 2 )(ρ(y); 0, 0)(1; m 2 , 0)(B δn 2 ; 0, n 2 ) =((Bu -1 ) b 1 B -b 1 ; b 1 , 0)((uv) -b 2 (uB) δ b 2 ; 0, b 2 )(ρ(y)B δn 2 ; m 2 , n 2 ) =((Bu -1 ) b 1 B -b 1 θ(b 1 , 0)((uv) -b 2 (uB) δ b 2 ); b 1 , b 2 )(ρ(y)B δn 2 ; m 2 , n 2 ) =((Bu -1 ) b 1 B -b 1 θ(b 1 , 0)((uv) -b 2 (uB) δ b 2 )θ(b 1 , δ b 2 )(ρ(y)B δn 2 ); b 1 + (-1) δ b 2 m 2 , b 2 + n 2 ). (23)
Thus:

(p 1 ) # (al σ (b)) = (m 1 , n 1 )(b 1 + (-1) δ b 2 m 2 , b 2 + n 2 ) = (m 1 + (-1) δn 1 b 1 + (-1) δn 1 +δ b 2 m 2 , n 1 + b 2 + n 2 ). (24)
Suppose that a and b satisfy the relation of Lemma 4.1(i). It follows from ( 22) and (24) that b 2 = 0, and then that m 2 + (-1) δn 2 m 1 = m 1 + (-1) δn 1 b 1 + (-1) δn 1 m 2 , which yields (20). This implies that b 1 is even. We now expand and simplify the remaining parts of ( 22) and (23):

p F (ba) = u b 1 yθ(m 2 , δ n 2 )(u a 1 v a 2 x) = u b 1 yB m 2 -δn 2 u a 1 εn 2 (B δn 2 vu -2m 2 ) a 2 B δn 2 -m 2 θ(m 2 , δ n 2 )(x), (25)
and

p F (l σ (b)) = (Bu -1 ) b 1 B -b 1 θ(b 1 , 0)(ρ(y)B δn 2 ) = (Bu -1 ) b 1 B -b 1 θ(b 1 , 0)(ρ(y))B δn 2 .
Hence:

p F (al σ (b)) =u a 1 v a 2 xθ(m 1 , δ n 1 )((Bu -1 ) b 1 B -b 1 θ(b 1 , 0)(ρ(y))B δn 2 ) =u a 1 v a 2 xB m 1 -δn 1 (B εn 1 u -εn 1 ) b 1 B -εn 1 b 1 +δn 1 -m 1 θ(m 1 + ε n 1 b 1 , δ n 1 )(ρ(y))B δn 2 εn 1 . (26)
Equation ( 21) then follows from the hypothesis, and equations ( 25) and (26).

Proof of Proposition 4.4. Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class such that α # (1, 0) = (ρ, γ) and α # (0, 1) = (ξ, τ ), where (ρ, γ, ξ, τ ) ∈ Σ. Note that either ρ = ξ = 0 or γ = τ = 0. In what follows, we will use the identities δ δq = δ q and ε δq = ε q for all q ∈ Z. Let a, b ∈ P 2 (K 2 ) be such that a = (u δρ v δγ B δγ (γ-δγ )/2 ; (ρ -δ ρ )/2, (γ -δ γ )/2) and b = (B -δρξ ; ξ, τ ). With respect to the notation of (19), x = B δγ (γ-δγ )/2 and y = B -δρξ . To prove the result, we show that conditions (i)-(iii) of Lemma 4.1 are satisfied. Clearly, (p 1 ) # (b) = (ξ, τ ) = α # (0, 1). Further, by taking b = a in (24), we have:

(p 1 ) # (al σ (a)) = 1 + (-1) δ (γ-δγ )/2 +δγ (ρ -δ ρ ) 2 + ε (γ-δγ )/2 δ ρ , γ . ( 27
)
If ρ = ξ = 0 then the first coordinate of the right-hand side of ( 27) is equal to zero, while if γ = τ = 0, this coordinate is equal to ρ. In both cases, we conclude that

(p 1 ) # (al σ (a)) = (ρ, γ) = α # (1, 0).
Hence conditions (ii) and (iii) of Lemma 4.1 are satisfied. It remains to check condition (i). Note that in the proof of Lemma 4.8, the only condition that we have applied to obtain equations ( 25) and ( 26) is that b 2 = 0. But this coefficient is zero in our case, and so these equations are also satisfied here. So using ( 22), (25) and Proposition 3.1, we see that:

ba = B -δρξ θ(ξ, δ τ )(u δρ v δγ B δγ (γ-δγ )/2 ); ξ + ε τ (ρ -δ ρ ) 2 , τ + (γ -δ γ ) 2 = B -δρξ B ξ-δτ u δρετ (B δτ vu -2ξ ) δγ B δτ -ξ B ετ δγ (γ-δγ )/2 ; ξ + ε τ (ρ -δ ρ ) 2 , τ + (γ -δ γ ) 2 . ( 28
)
In a similar manner, by ( 24) and ( 26), we see that:

al σ (b) = u δρ v δγ B δγ (γ-δγ )/2+(δτ -δρξ)ε (γ-δγ )/2 ; (ρ -δ ρ ) 2 + ε (γ-δγ )/2 ξ, τ + (γ -δ γ ) 2 . ( 29
) If ρ = ξ = 0 (resp. γ = τ = 0) then (p 1 ) # (ba) = (0, τ + (γ -δ γ )/2) = (p 1 ) # (al σ (b)) (resp. (p 1 ) # (ba) = (ξ + (ρ -δ ρ )/2, 0) = (p 1 ) # (al σ (b))). So it remains to show that p F (ba) = p F (al σ (b)). (a) If ρ = ξ = 0, p F (ba) = B -δτ (B δτ v) δγ B δτ B ετ δγ (γ-δγ )/2 and p F (al σ (b)) = v δγ B δγ (γ-δγ )/2+δτ ε (γ-δγ )/2 .
(i) Suppose that δ γ = 0. Since (ρ, γ, ξ, τ ) ∈ Σ, it follows that δ τ = 0, and thus p F (ba

) = p F (al σ (b)) = 1. (ii) Suppose that δ γ = 1. Then p F (ba) = vB δτ +ετ (γ-1)/2 and p F (al σ (b)) = vB (γ-1)/2+δτ ε (γ-1)/2 . If δ τ = 0 then p F (ba) = vB (γ-1)/2 = p F (al σ (b)). So suppose that δ τ = 1. Since (ρ, γ, ξ, τ ) ∈ Σ, it follows that (γ -1)/2 ∈ {0, 1}, and one can check easily that 1 -(γ - 1)/2 = (γ -1)/2 + ε (γ-1)/2 . It follows that p F (ba) = p F (al σ (b)). (b) If γ = τ = 0 then p F (ba) = B (1-δρ)ξ u δρ B -ξ and p F (al σ (b)) = u δρ B -δρξ , and one sees that p F (ba) = 1 = p F (al σ (b)) if δ ρ = 0, and p F (ba) = uB -ξ = p F (al σ (b)) if δ ρ = 1.
In order to prove Proposition 4.5, we will make use of the following lemma.

Lemma 4.9. Let α ∈ [T 2 , * ; K 2 , * ] be such that α # :

(1, 0) → (0, 2s) (0, 1) → (0, 1) for some s ∈ {0, 1}. If α does not have the Borsuk-Ulam property with respect to τ 1 then there exist x, y ∈ σ 2 Ab and m 1 , n 1 ∈ Z that satisfy the following equation in σ 2 Ab :

µ 1 (x) + µ 2 (y) = I 2n 1 -2s -T -2m 1 ,δn 1 -O n 1 -s,-2m 1 -(m 1 + δ n 1 + ε n 1 )B 0,0 -(m 1 -δ n 1 )B 0,-2m 1 -B 2n 1 -2s,-2δ n 1 +1 m 1 + B 2n 1 -2s,0 , (30)
where µ 1 , µ 2 : σ 2 Ab → σ 2 Ab are the homomorphisms defined on the basis elements of σ 2 Ab by:

µ 1 (B k,l ) =B k,l-2ε k m 1 + B k,-l and µ 2 (B k,l ) =ε k ε δn 1 B -k,ε δn 1 ε k+1 l+2δ k m 1 -B k+2n 1 -2s,l-2ε k δ n 1 +1 m 1 .
Proof. By hypothesis, there exist a, b ∈ P 2 (K 2 ) satisfying (i)-(iii) of Lemma 4.1. Proposition 3.5 implies that a and b may be written in the form (19). Lemma 4.1(iii) implies that (p 1 ) # (b) = (0, 1), so (m 2 , n 2 ) = (0, 1). Using Lemma 4.1(ii) and taking b = a in (23), we obtain:

(0, 2s) = (p 1 ) # (al σ (a)) = (m 1 , n 1 )(a 1 + (-1) a 2 m 1 , n 1 + a 2 ) = (m 1 + (-1) n 1 a 1 + (-1) n 1 +a 2 m 1 , 2n 1 + a 2 ).
It follows that a 2 = 2s -2n 1 is even and a 1 = -2δ n 1 +1 m 1 . Lemmas 4.1(i) and 4.8 then imply that b 2 = 0 and b 1 = -2ε n 1 m 1 . We now analyse (21), which holds because Lemma 4.1(i) does. The left-hand side of ( 21) is equal to p F (ba), and may be rewritten as:

p F (ba) =u b 1 yB -1 u -a 1 (Bv) 2s-2n 1 Bθ(0, 1)(x) = u b 1 yB -1 u -a 1 B(vB) 2s-2n 1 θ(0, 1)(x) =u b 1 -a 1 v 2s-2n 1 . v 2n 1 -2s u a 1 yu -a 1 v 2s-2n 1 . v 2n 1 -2s u a 1 B -1 u -a 1 v 2s-2n 1 . v 2n 1 -2s Bv 2s-2n 1 . v 2n 1 -2s (vB) 2s-2n 1 θ(0, 1)(x) =u b 1 -a 1 v 2s-2n 1 c 2n 1 -2s,a 1 (y)B -1 2n 1 -2s,a 1 B 2n 1 -2s,0 I 2n 1 -2s θ(0, 1)(x), (31)
and the right-hand side of ( 21) is equal to p F (al σ (b)), and may be rewritten as:

p F (al σ (b)) =u a 1 v 2s-2n 1 xB m 1 -δn 1 (B εn 1 u -εn 1 ) b 1 B m 1 +δn 1 θ(-m 1 , δ n 1 )(ρ(y))B εn 1 =u 2m 1 +a 1 v 2s-2n 1 [v 2n 1 -2s , u -2m 1 ]c 0,-2m 1 (x)B m 1 -δn 1 0,-2m 1 . u -2m 1 (B εn 1 u -εn 1 ) -2εn 1 m 1 B m 1 +δn 1 0,0 θ(-m 1 , δ n 1 )(ρ(y))B εn 1 0,0 =u 2m 1 +a 1 v 2s-2n 1 O n 1 -s,-2m 1 c 0,-2m 1 (x)B m 1 -δn 1 0,-2m 1 T -2m 1 ,δn 1 B m 1 +δn 1 0,0 . θ(-m 1 , δ n 1 )(ρ(y))B εn 1 0,0 , (32)
Since p F (ba) = p F (al σ (b)), and using the fact that b 1 -a 1 = 2m 1 +a 1 , it follows by Abelianising (31) and (32) that:

(c 0,-2m 1 ) Ab (x) -θ(0, 1) Ab (x) + θ(-m 1 , δ n 1 ) Ab • ρ Ab (y) -(c 2n 1 -2s,a 1 ) Ab (y) = I 2n 1 -2s -T -2m 1 ,δn 1 -O n 1 -s,-2m 1 -(m 1 + δ n 1 + ε n 1 )B 0,0 -(m 1 -δ n 1 )B 0,-2m 1 -B 2n 1 -2s,a 1 + B 2n 1 -2s,0 in σ 2
Ab , where the projection of x (resp. y) in σ 2 Ab is also denoted by x (resp. y). By ( 15)-( 17), for all k, l ∈ Z, we have:

(c 0,-2m 1 ) Ab (B k,l ) -θ(0, 1) Ab (B k,l ) = B k,l-2ε k m 1 + B k,-l = µ 1 (B k,l ) and θ(-m 1 , δ n 1 ) Ab • ρ Ab (B k,l ) -(c 2n 1 -2s,a 1 ) Ab (B k,l ) = θ(-m 1 , δ n 1 ) Ab (ε k B -k,ε k+1 l ) -B k+2n 1 -2s,l+ε k a 1 = ε k ε δn 1 B -k,ε δn 1 ε k+1 l+2δ k m 1 -B k+2n 1 -2s,l+ε k a 1 = µ 2 (B k,l ).
The result follows by noting that a 1 = -2δ n 1 +1 m 1 .

We are now able to complete the proof of Proposition 4.5.

Proof of Proposition 4.5. We argue by contradiction. Suppose that α does not have the Borsuk-Ulam property with respect to τ 1 . Then there exist x, y ∈ σ 2 Ab that satisfy equation (30) given in the statement of Lemma 4.9. Let ξ : σ 2 Ab → Z 2 be the homomorphism defined on the basis {B k,l } k,l∈Z of σ 2

Ab by ξ(B k,l ) = 1 for all k, l ∈ Z. From the definition of the maps µ 1 and µ 2 , it follows that the left-hand side is sent to 0. By Proposition 3.7, ξ( I 2n 1 -2s ) = ξ( T -2m 1 ,δn 1 ) = ξ( O n 1 -s,-2m 1 ) = 0, and it follows that the right-hand side is sent to ε n 1 , which is different from 0. This yields a contradiction. We conclude that α has the Borsuk-Ulam property with respect to τ 1 .

Proof of Proposition 4.6. Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class such that α # (1, 0) = (r 1 , 2) and α # (0, 1) = (r 2 , 0), where r 1 , r 2 ∈ Z are such that r 1 > 0, and either r 2 = 0, or r 2 = 0 and e(r 1 ) ≤ e(r 2 ). Let o(r 1 ) = r 1 /2 e(r 1 ) , and let m = r 2 /2 e(r 1 ) . Then o(r 1 ) > 0 is odd, and m ∈ Z by hypothesis. Let α ′ ∈ [T 2 , * ; K 2 , * ] be the homotopy class for which α ′ # (1, 0) = (r 1 , 2o(r 1 )) and α ′ # (0, 1) = (r 2 , 2m). By Corollary 4.2, to prove the result, it suffices to exhibit a, b ∈ P 2 (K 2 ) that satisfy conditions (i)-(iii) of Lemma 4.1 for α ′ . Let c = (u 2 e(r 1 ) v 2 ; 0, 0) ∈ P 2 (K 2 ), and let a = (cσ) o(r 1 ) σ -1 and b = (cσ) 2m . Then a, b ∈ P 2 (K 2 ), and by Proposition 3.1, we see that Ab and (m 1 , n 1 ) ∈ Z ⋊ Z such that:

(p 1 ) # (l σ (c)) = (2 e(r 1 ) , 2). Now al σ (b) = (cσ) o(r 1 ) σ -1 σ(cσ) 2m σ -1 = (cσ) 2m (cσ) o(r 1 ) σ -1 = ba, so condition (i) of Lemma 4.1 is satisfied. Next, a = ((cσ) 2 ) (o(r 1 )-1)/2 c = (cl σ (c)σ 2 ) (o(r 1 )-1)/2 c, hence l σ (a) = (l σ (c)σ 2 c) (o(r 1 )-1)/2 l σ (c), and since (p 1 ) # (c) = (p 1 ) # (σ 2 ) = (0, 0), it follows that (p 1 ) # (al σ (a)) = ((p 1 ) # (l σ (c))) o(r 1 ) = (2 e(r 1 ) , 2) o(r 1 ) = (2 e(r 1 ) o(r 1 ), 2o(r 1 )) = (r 1 , 2o(r 1 )) = α # (1, 0). So condition (ii) of Lemma 4.1 holds. Finally, b = (cσcσ) m = (cl σ (c)σ 2 ) m , so (p 1 ) # (b) = (p 1 ) # (l σ (c)) m = (2 e(r 1 ) , 2) m = (2 e(r 1 ) m, 2m) = (r 2 , 2m) = α ′ # (0, 1
µ(x) + ν(y) = J n 1 -1,-2r 2 -O n 1 -1,2δn 1 r 2 -T 2δn 1 r 2 ,δn 1 + (33) r 2 B 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 -(m 1 -δ n 1 )B 0,2δn 1 r 2 -(δ n 1 (1 -2r 2 ) -m 1 + r 2 )B 0,0 in σ 2 Ab , where µ, ν : σ 2 Ab → σ 2
Ab are the homomorphisms defined on the elements of the basis {B k,l } k,l∈Z of σ 2

Ab by:

µ(B k,l ) =B k,l+2ε k δn 1 r 2 -B k,l-2δ k r 2 (34) ν(B k,l ) =ε k ε n 1 B -k,εn 1 ε k+1 l-2δ k (m 1 +2δn 1 r 2 ) -B k+2(n 1 -1),l+ε k (2δ n 1 +1 m 1 -εn 1 r 1 ) . ( 35
)
Proof. Let α ∈ [T 2 , * ; K 2 , * ] be a homotopy class that does not have the Borsuk-Ulam property, and that is represented by the homomorphism α # : π 1 (T 2 ) → π 1 (K 2 ) given by α # (1, 0) = (r 1 , 2) and α # (0, 1) = (r 2 , 0), where r 1 , r 2 ∈ Z. Then there exist a, b ∈ P 2 (K 2 ) satisfying conditions (i)-(iii) of Lemma 4.1. We write these elements in the form of equation ( 19). Condition (iii) implies that (m 2 , n 2 ) = (r 2 , 0), and taking b = a in (24), condition (ii) implies that (r

1 , 2) = (m 1 (1 + (-1) n 1 +a 2 ) + (-1) n 1 a 1 , 2n 1 + a 2 ). It follows that a 2 = 2(1 -n 1 ) and that a 1 = ε n 1 r 1 -2δ n 1 +1 m 1 . By (20), we have b 2 = 0 and (1 + (-1) δn 1 +1 )m 2 = (1 + (-1) δn 2 +1
)m 1 + (-1) δn 1 b 1 , from which we see that b 1 = -2δ n 1 r 2 using the fact that n 2 = 0. Substituting this information into (25) and (26), and using that fact that ε n 1 δ n 1 = -δ n 1 , we obtain:

p F (ba) =u -2δn 1 r 2 yB r 2 u εn 1 r 1 -2δ n 1 +1 m 1 (vu -2r 2 ) 2(1-n 1 ) B -r 2 θ(r 2 , 0)(x) =u εn 1 r 1 -2δn 1 r 2 -2δ n 1 +1 m 1 v 2(1-n 1 ) . v 2(n 1 -1) u 2δ n 1 +1 m 1 -εn 1 r 1 yB r 2 . u εn 1 r 1 -2δ n 1 +1 m 1 v 2(1-n 1 ) . v 2(n 1 -1) (vu -2r 2 ) 2(1-n 1 ) B -r 2 0,0 θ(r 2 , 0)(x) =u εn 1 r 1 -2δn 1 r 2 -2δ n 1 +1 m 1 v 2(1-n 1 ) c 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 (y). B r 2 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 J n 1 -1,-2r 2 B -r 2 0,0 θ(r 2 , 0)(x), p F (al σ (b)) =u εn 1 r 1 -2δ n 1 +1 m 1 v 2(1-n 1 ) xB m 1 -δn 1 (B εn 1 u -εn 1 ) 2εn 1 δn 1 r 2 . B δn 1 (1-2r 2 )-m 1 θ(m 1 + 2δ n 1 r 2 , δ n 1 )(ρ(y)) =u εn 1 r 1 -2δ n 1 +1 m 1 -2δn 1 r 2 v 2(1-n 1 ) . v 2(n 1 -1) u 2δn 1 r 2 v 2(1-n 1 ) u -2δn 1 r 2 . u 2δn 1 r 2 xu -2δn 1 r 2 . u 2δn 1 r 2 B m 1 -δn 1 u -2δn 1 r 2 . u 2δn 1 r 2 . (B εn 1 u -εn 1 ) 2εn 1 δn 1 r 2 B δn 1 (1-2r 2 )-m 1 0,0 . θ(m 1 + 2δ n 1 r 2 , δ n 1 )(ρ(y)) =u εn 1 r 1 -2δ n 1 +1 m 1 -2δn 1 r 2 v 2(1-n 1 ) O n 1 -1,2δn 1 r 2 c 0,2δn 1 r 2 (x)B m 1 -δn 1 0,2δn 1 r 2 . T 2δn 1 r 2 ,δn 1 B δn 1 (1-2r 2 )-m 1 0,0 θ(m 1 + 2δ n 1 r 2 , δ n 1 )(ρ(y)).
Applying condition (i) of Lemma 4.1, we see that: In what follows, we suppose that the hypotheses of Proposition 4.7 hold, namely r 1 ≥ 0, r 2 = 0, and either r 1 = 0, or r 1 > 0 and e(r 1 ) > e(r 2 ). With the notation of Lemma 4.10, we define the homomorphism ξ n 1 ,r 2 : σ 2 Ab → Z 2 on the basis {B k,l } k,l∈Z as follows: We now take the image of (33) by ξ n 1 ,r 2 . Using (47) and Lemma 4.10, it follows that:

c 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 (y)B r 2 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 J n 1 -1,-2r 2 B -r 2 0,0 θ(r 2 , 0)(x) = O n 1 -1,2δn 1 r 2 c 0,2δn 1 r 2 (x)B
ξ n 1 ,r 2 (B k,l ) = 0 if k = n 1 -1, or if k = n 1 -
ξ n 1 ,r 2 ( O n 1 -1,2δn 1 r 2 ) = δ n 1 1. (44)
0 = ξ n 1 ,r 2 (µ(x) + ν(y)) = ξ n 1 ,r 2 ( J n 1 -1,-2r 2 -O n 1 -1,2δn 1 r 2 -T 2δn 1 r 2 ,δn 1 ). (48)
From ( 42), ( 44) and (45), we obtain the following conclusions:

  ⊕ Z, Z ⋊ Z) whose elements are described as follows:Type 1:

Theorem 1 . 3 .

 13 Given a non-zero integer t, let e(t) be its 2-adic evaluation. With the notation of Proposition 1.1, let β ∈ [T 2 , K 2 ] and β # ∈ Hom(Z ⊕ Z, Z ⋊ Z). Then β has the Borsuk-Ulam property with respect to τ 1 if and only if one of the following conditions is satisfied: (a) β # is a homomorphism of Type 3. (b) β # is a homomorphism of Type 4, where s 1 is odd and r 2 = 0, and additionally e(r 1 ) > e(r 2 )

Lemma 3 . 9 .

 39 Let k ∈ Z. If k = 0 then I k = 1, and if k = 0 then:

  and (b) of Lemma 3.11 and Abelianising, one may check that the formula for O k,l given in Proposition 3.7 is correct. Proof of Lemma 3.11. If k = 0 or l = 0, then clearly O k,l = 1. So suppose that k, l > 0, and let ε ∈ {-1, 1}. We start by considering the case l = 1. If k = ε = 1 then:

  3, it suffices to study the cases described by Remark 4.3. This will be carried out in Propositions 4.4-4.7 below. Part of Proposition 4.4 (resp. Proposition 4.5) treats the cases of Remark 4.3(a)(i) (resp. Remark 4.3(a)(ii)), and part of Proposition 4.4 and Propositions 4.6 and 4.7 deal with the cases of Remark 4.3(b).In each case, we will make use of Proposition 3.1 and its notation, as well as the commutative diagram[START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF].

Proposition 4 . 4 .

 44 Let:

1 . 4 . 5 .Proposition 4 . 6 . 1 . 4 . 7 .

 14546147 τ ), where (ρ, γ, ξ, τ ) ∈ Σ. Then α does not have the Borsuk-Ulam property with respect to τ Proposition Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class such that α # :(1, 0) → (0, 2s) (0, 1) → (0, 1), where s ∈ {0, 1}. Then α has the Borsuk-Ulam property with respect to τ 1 . If α # :(1, 0) → (r 1 , 2) (0, 1) → (r 2 , 0), where r 1 , r 2 ∈ Z, r 1 > 0, and one of the following conditions holds:(a) r 2 = 0.(b) r 2 = 0 and e(r 1 ) ≤ e(r 2 ).Then α does not have the Borsuk-Ulam property with respect to τProposition Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class such that α # :(1, 0) → (r 1 , 2) (0, 1) → (r 2 , 0), where r 1 , r 2 ∈ Z, r 1 ≥ 0, r 2 = 0, and one of the following conditions holds: (a) r 1 = 0.(b) r 1 > 0, and e(r 1 ) > e(r 2 ).

7 . 4 . 10 .

 7410 ), and condition (iii) of Lemma 4.1 is satisfied, which proves the proposition. The rest of this section is devoted to proving Proposition 4.Lemma Let α ∈ [T 2 , * ; K 2 , * ] be a homotopy class such that α # : π 1 (T 2 ) → π 1 (K 2 ) satisfies α # (1, 0) = (r 1 , 2) and α # (0, 1) = (r 2 , 0), where r 1 , r 2 ∈ Z. With the notation of Proposition 3.7, if α does not have the Borsuk-Ulam property, then there exist x, y ∈ σ 2

m 1 -δn 1 0,2δn 1 r 2 T(c 0 ,2δn 1 r 2 ) 1 -

 12021 2δn 1 r 2 ,δn 1 B δn 1 (1-2r 2 )-m 1 0,0 θ(m 1 + 2δ n 1 r 2 , δ n 1 )(ρ(y)),in σ 2 , and by Abelianising this equation, we obtain the following equality in σ 2Ab : Ab (x) -θ(r 2 , 0)Ab (x) + θ(m 1 + 2δ n 1 r 2 , δ n 1 ) Ab • ρ ab (y) -(c 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 ) Ab (y) = J n 1 -1,-2r 2 -O n 1 -1,2δn 1 r 2 -T 2δn 1 r 2 ,δn 1 + r 2 B 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r (m 1 -δ n 1 )B 0,2δn 1 r 2 -(δ n 1 (1 -2r 2 ) -m 1 + r 2 )B 0,0 . (36)Now by (15)-(17), for all k, l ∈ Z, we may check that:µ(B k,l ) =(c 0,2δn 1 r 2 ) Ab (B k,l ) -θ(r 2 , 0) Ab (B k,l) and (37)ν(B k,l ) =θ(m 1 + 2δ n 1 r 2 , δ n 1 ) Ab • ρ ab (B k,l ) -(c 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 ) Ab (B k,l ).(38)Equation (33) then follows from (36), (37) and (38).

( c )T 2δn 1 r 2 ,δn 1 = T 2r 2 , 1 = σ r 2 σr 2 2r 2 i=1B 0 , 2 -(δ n 1 ( 1 -

 c2121220211 Consider the term T 2δn 1 r 2 ,δn 1 . If n 1 is even then T 2δn 1 r 2 ,δn 1 = 0 by Proposition 3.7(b), and thus ξ n 1 ,r 2 ( T 2δn 1 r 2 ,δn 1 ) = 0. So assume that n 1 is odd. By Proposition 3.7(c), we have: σr 2 (i-(σr 2 +1)/2) .(45)If n 1 = 1 then ξ n 1 ,r 2 ( T 2r 2 ,1 ) = 0 by (39). So suppose that n 1 = 1. Then the set {σ r 2 (i -(σ r 2 + 1)/2) | i = 1, . . . ,σ r 2 2r 2 } of indices consists of 2 |r 2 | consecutive integers, and thus contains o(r 2 ) elements divisible by 2 e(r 2 )+1 by Remark 4.12. It follows from (39) that ξ n 1 ,r 2 ( T 2r 2 ,1 ) = 1.Hence:ξ n 1 ,r 2 ( T 2δn 1 r 2 ,δn 1 ) = 1 if n 1 = 1 0 otherwise. (46) (d) Let χ = r 2 B 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 -(m 1 -δ n 1 )B 0,2δn 1 r 2 -(δ n 1 (1 -2r 2 ) -m 1 + r 2 )B 0,0 . If n 1 = 1 then it follows from (39) that ξ n 1 ,r 2 (χ) = 0. So suppose that n 1 = 1. Then χ = r 2 B 0,r 1 -(m 1 -1)B 0,2r 2 -(1 -m 1 -r 2 )B 0,0 .By hypothesis, e(r 1 ) > e(r 2 ), and we see from (39) that ξ n 1 ,r 2 (χ) = r 2 + m 1 -1 + 1 -m 1 -r 2 = 0. Thus: ξ n 1 ,r 2r 2 B 2(n 1 -1),2δ n 1 +1 m 1 -εn 1 r 1 -(m 1 -δ n 1 )B 0,2δn 1 r 2r 2 ) -m 1 + r 2 )B 0,0 = 0. (47)

  homotopy class, then β has the Borsuk-Ulam property with respect to τ if and only if H(β) has the Borsuk-Ulam property with respect to τ .

  Ab denote the Abelianisation of the group σ 2 . By abuse of notation, for all k, l ∈ Z, we denote the image of a generator B k,l in σ2 Ab by B k,l . By Proposition 3.3, σ 2 Ab is the free Abelian group for which {B k,l

	Let σ 2

[START_REF]1 is odd and n 1 = 1 then ξ n 1 ,r 2 ( J n 1 -1,-2r 2 ) = ξ n 1 ,r 2 ( T 2δn 1 r 2 ,δn 1 ) = 0 and ξ n 1 ,r 2 ( O n 1 -1,2δn 1 r 2 )[END_REF]

, which completes the proof.

4 Proof of Theorem 1.3

  Let α ∈ [T 2 , * ; K 2 , * ] and β ∈ [T 2 , K 2 ]. With the notation of[START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF] Theorem 4], suppose that α F = β. According to[START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF] Theorem 7], the pointed homotopy class α has the Borsuk-Ulam property with respect to the free involution τ 1 if and only if the homotopy class β has the Borsuk-Ulam property with respect to τ 1 . So to obtain Theorem 1.3, it suffices to prove the statement for pointed homotopy classes. Before doing so, we give an algebraic criterion, similar to that of[START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF] Lemma 22], to decide whether a pointed homotopy class has the Borsuk-Ulam property with respect to τ 1 .

Lemma 4.1. Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class. Then α does not have the Borsuk-Ulam property with respect to τ 1 if and only if there exist a, b ∈ P 2 (K 2 ) such that:

  Let α ∈ [T 2 , * ; K 2 , * ] be a pointed homotopy class, and let α # : π 1 (T 2 ) → π 1 (K 2 ) be the homomorphism described in[START_REF] Gonçalves | The Borsuk-Ulam property for homotopy classes of selfmaps of surfaces of Euler characteristic zero[END_REF] Theorem 4], and that is of one of the four types given in Proposition 1.1.

	1),
	which proves the corollary.
	Remark 4.3.

(a) Suppose that α # is of Type 1, 2 or 3, and let i ∈ {0, 1}, s 1 and s 2 be the integers that appear in the description of α # in Proposition 1.1. By Proposition 2.2 and Corollary 4.2, α has the Borsuk-Ulam property with respect to τ 1 if and only if α ′ does, where α ′ ∈ [T 2 , * ; K 2 , * ] satisfies:

  Ab → Z 2 and ξ n 1 ,r 2 • ν : σ 2 Ab → Z 2 are identically zero. (b) Consider the term O n 1 -1,2δn 1 r 2 . If n 1 is even or is equal to 1 then O n 1 -1,2δn 1 r 2 = 0 by Proposition 3.7(b), and ξ n 1 ,r 2 ( O n 1 -1,2δn 1 r 2 ) = 0. So assume that n 1 is odd and different from 1. By Proposition 3.7(c) and (39), we have:ξ n 1 ,r 2 ( O n 1 -1,2δn 1 r 2 ) =ξ n 1 ,r 2 ( O n 1 -1,2r 2 ) ,r 2 B σ n 1 -1 (2i-1),-σ 2r 2 j+(σ 2r 2 -1)/2 + ξ n 1 ,r 2 B σ n 1 -1 (2i-1)-1,σ 2r 2 j-(σ 2r 2 +1)/2 . (43)Observe that there is no integer i satisfying σ n 1 -1 (2i -1) = n 1 -1, and it follows thatξ n 1 ,r 2 B σ n 1 -1 (2i-1),-σ 2r 2 j+(σ 2r 2 -1)/2 = 0 for all i ∈ {1, . . . , σ n 1 -1 (n 1 -1)} and j ∈ {1, . . . , σ r 2 2r 2 }.For the second term of (43), note thatσ n 1 -1 (2i -1) -1 = n 1 -1 if and only if i = (σ n 1 -1 (n 1 -1)+σ n 1 -1 +1)/2,and in this case, i belongs to the allowed set {1, . . . , σ n 1 -1 (n 1 -1)} of indices. Now consider the terms of the form B n 1 -1,σ 2r 2 j-(σ 2r 2 -1)/2 , where j ∈ {1, . . . , σ r 2 2r 2 }. Then the set {σ 2r 2 j -(σ 2r 2 + 1)/2 | j = 1, . . . , σ r 2 2r 2 } consists of 2 |r 2 | consecutive integers, and thus contains o(r 2 ) elements divisible by 2 e(r 2 )+1 by Remark 4.12. It follows from (39) that ξ n 1 ,r 2 ( O n 1 -1,2r 2 ) = 1. Hence:

	σ n 1 -1 (n 1 -1)	σr 2 2r 2
	=	ξ n 1
	i=1	j=1

1 and 2 e(r 2 )+1 ∤ l 1 if k = n 1 -1 and 2 e(r 2 )+1 | l. (39) Lemma 4.11. With the notation of Lemma 4.10, the compositions ξ n 1 ,r 2 • µ : σ 2

Acknowledgements

This work is a continuation of part of the Ph.D. thesis [7] of the third author who was supported by CNPq project nº 140836 and Capes/COFECUB project n o 12693/13-8. The first and second authors wish to thank the 'Réseau Franco-Brésilien en Mathématiques' for financial support during their respective visits to the Laboratoire de Mathématiques Nicolas Oresme UMR CNRS 6139, Université de Caen Normandie, from the 9 th to the 24 th of November 2019, and to the Instituto de Matemática e Estatística, Universidade de São Paulo, from the 17 th to the 31 th of August 2019.

Proof. It suffices to prove that ξ n 1 ,r 2 • µ(B k,l ) = ξ n 1 ,r 2 • ν(B k,l ) = 0 for all k, l ∈ Z. We start with the case of ξ n 1 ,r 2 • µ. By (34), clearly

Using the fact that ε n 1 -1 δ n 1 = δ n 1 , we have:

Now (l + 2δ n 1 r 2 ) -(l -2δ n 1 -1 r 2 ) = 2r 2 (δ n 1 + δ n 1 -1 ) = 2r 2 , hence 2 e(r 2 )+1 | l + 2δ n 1 r 2 if and only if 2 e(r 2 )+1 | l -2δ n 1 -1 r 2 , and it follows from (39) and (40) that ξ n 1 ,r 2 • µ(B n 1 -1,l ) = 0 as required.

We now analyse the case of

, we obtain: 39) and (41) then imply that

We now complete the proof of Theorem 1.3. The following remark will be used in the proof of Proposition 4.7. and α # (0, 1) = (r 2 , 0), where r 2 = 0 and either r 1 = 0, or r 1 > 0 and e(r 1 ) > e(r 2 ). Suppose on the contrary that α does not have the Borsuk-Ulam property with respect to τ 1 . Then by Lemma 4.10, there exist x, y ∈ σ 2 Ab such that equation (33) holds. By Lemma 4.11, ξ n 1 ,r 2 (µ(x) + ν(y)) = 0. So to prove the result, it suffices to show that the image of the right-hand side of (33) by ξ n 1 ,r 2 is equal to 1. We analyse each of the terms in turn.

(a) We start by showing that:

If

By Proposition 3.7(c) and (39), we have:

If n 1 is odd, there is no integer i satisfying

A Appendix

Let g : F (u, v) → Z⋊Z be the homomorphism defined on the generators of F (u, v) by g(u) = (1, 0) and g(v) = (0, 1).

Proof. We use the Reidemeister-Schreier rewriting process that is described in detail in [8, Chapter 2, Theorem 2.8] and briefly in [10, Appendix I, Theorem 6.3]. We use the notation of [START_REF] Murasugi | A study of braids[END_REF]. Let S = {v k u l } k,l∈Z . We have g(v k u l ) = (0, k)(l, 0) = ((-1) k l, k). So, g| S : S → Z ⋊ Z is a bijection, and therefore S is a complete set of right coset representatives of Ker (g) in F (u, v). Moreover, S is a Schreier system of Ker (g). Let us compute the generators of Ker (g). We have

where for all w ∈ F (u, v), w is the unique element of S for which g(w) = g(w). Note that Γ k,l = 1 if and only if l = 0. Using the Reidemeister-Schreier rewriting process, we see that the group Ker (g) is freely generated by {Γ k,l } k,l∈Z, l̸ =0 .

The basis of Ker (g) given in Proposition A.1 is not well adapted to our calculations. We define a new basis that is more suitable.

, where B = uvuv -1 . Then we have the following relations in Ker (g):

, where Γ -1 k,0 = 1. Proof. We first prove part (a). We start by proving the result in the case k = 0. If l = 1, we have

B 0,l-i for some l ≥ 1, and let us show that the result holds for l + 1. We have:

By induction, the given formula is valid for k = 0 and all l ≥ 1. If l ≤ -1, the result holds for Γ 0,-l , and thus:

Hence the formula holds for k = 0 and all l ∈ Z \ {0}.

, and we obtain the formula for all k ∈ Z and l ∈ Z \ {0} using the results of the case k = 0. Part (b) then follows.

Theorem A.3. The set {B k,l = v k u l Bu -l v -k } k,l∈Z is a basis of Ker (g).

Proof. By Lemma A.2, the elements of the set {B k,l } k,l∈Z generate Ker (g). To show that this set is a basis, it suffices to prove that there are only trivial relations between these elements. Suppose on the contrary that there exists a word w ∈ Ker (g) for which:

Let S = {k 1 , k 2 , . . . , k n }. For each k ∈ S, we define the set R k consisting of those indices l j for which the element B k,l j appears in the word w. Let l m and l M be the minimal and maximal elements of R k respectively. We define the sets B k and Γ k as follows: