I can see that the paths $(\cos(\pi s), \sin(\pi s))$ and $(\cos(\pi s), -\sin(\pi s))$ in $\mathbb{R}^2 \setminus \{0\}$ are 'homotopic' But can't construct an explicit homotopy between them.

Could anyone suggest me an explicit homotoy function?

Here I mean just homotopy, not path-homotopy

  • $\begingroup$ What do you mean when you say "just homotopy"? Not endpoint-preserving? Also, what's the range of $s$? $[0,1]$, $[0,2]$...? $\endgroup$ – Najib Idrissi Oct 7 '14 at 13:58
  • $\begingroup$ (Please look at how I edited your post to get an idea of how to properly format your questions. You can also read this.) $\endgroup$ – Najib Idrissi Oct 7 '14 at 14:02
  • $\begingroup$ oh sorry I have not been specific enough. I mean not endpoint-preserving and the range of s is [0,1] $\endgroup$ – Keith Oct 7 '14 at 16:17

You can explicitly construct a "retraction" of the first path to the constant path based at $(1,0)$ ($s=0$), by putting $\gamma_t(s) = (\cos(\pi s t), \sin(\pi s t))$. You can do the same for the second path. Now do the first "retraction", then the second one in reverse direction.

  • $\begingroup$ oh I see. Thank you :) $\endgroup$ – Keith Oct 7 '14 at 16:51

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