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I know that homeomorhic spaces are homotopic, but am not sure if this applies, since I think they are not homeomorphic due to orientability.

I know $f$ and $g$ are homotopic if they represent: X$\rightarrow$Y, and there exists Homotopy map: $H: X \times [0,1] \rightarrow Y$, with: $H(x,0)=f(x)$ and $H(x,1)=g(x)$

So $X$ is a torus with 2-disc removed, $Y$ is the Klein bottle with 2-disc removed, but I am not sure how to apply the equation in practice.

It would be great if someone could help, and then I can practice more questions.

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  • $\begingroup$ Are you familiar with the fundamental polygons of those spaces? This should enable you to prove that both spaces are homeomorphic to a wedge sum of two circles. $\endgroup$ – Ayman Hourieh Feb 9 '16 at 21:22
  • $\begingroup$ Two "different" knots are homeomorphic but not homotopic. $\endgroup$ – ajotatxe Feb 9 '16 at 21:24
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    $\begingroup$ @ajotatxe, I cannot tell what the connection is between that and the question! $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '16 at 21:25
  • $\begingroup$ @thinker I've posted an answer, please have a look and let me know if you have doubts $\endgroup$ – Anubhav Mukherjee May 20 '16 at 13:42
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Look at the polygonal representation of two spaces. Now removing a disc from the middle, the rest of the space will be deformation retract into the boundary, which is nothing but wedge of two circles. (Just draw the picture of polygonal presentation, you can actually see what is happening.)

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Hint: show that both spaces deformation-retract to a wedge of two circles.

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  • $\begingroup$ hi I am not familar with the term 'wedge', could you elaborate? thanks $\endgroup$ – thinker Feb 9 '16 at 21:32
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    $\begingroup$ See en.wikipedia.org/wiki/Wedge_sum; it should be explained in any sensible textbook on algbraic topology! $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '16 at 21:39
  • $\begingroup$ @MarianoSuárez-Alvarez not exactly wedge of two circles, but take two circle and glue their north pole and south pole with each other because he is removing two discs $\endgroup$ – Anubhav Mukherjee Feb 9 '16 at 22:22
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    $\begingroup$ @Anubhav.K, the OP is not removing two discs, he is removing one $2$-disc... $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '16 at 23:40

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