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Let $X$ and $Y$ be topological spaces. If $f\colon X\to Y$ is a homeomorphism, then it induces an isomorphism $f_\sharp\colon\pi_1(X,x_0)\to \pi_1(Y,f(x_0))$. All good.

As far as I know, the result still holds if we weaken "homeomorphism" to "homotopy equivalence", but I am having trouble checking that. I know that $(f_2\circ f_1)_\sharp = (f_2)_\sharp\circ (f_1)_\sharp$, $({\rm id}_X)_\sharp = {\rm id}_{\pi_1(X,x_0)}$ and that $f_1\simeq f_2 ~ {\rm rel}\,\{x_0\}$ implies $(f_1)_\sharp = (f_2)_\sharp$.

If $f\colon X \to Y$ is a homotopy equivalence, then we have $g\colon Y\to X$ with $g\circ f \simeq {\rm id}_X$ and $f\circ g\simeq {\rm id}_Y$.

My problems are:

  • I can't guarantee that $g(f(x_0))=x_0$;
  • These homotopies don't need to be ${\rm rel}\,\{x_0\}$ and ${\rm rel}\,\{f(x_0)\}$, respectively.

If I have these two assumptions, I can make: $${\rm id}_{\pi_1(X,x_0)} = ({\rm id}_X)_\sharp = (g\circ f)_\sharp = g_\sharp\circ f_\sharp$$and repeat this for $f_\sharp\circ g_\sharp$.

How can I go around these problems?

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  • $\begingroup$ This is proved somewhere in chapter 1 of Hatcher. Search "homotopy equivalence" and you'll surely find it. $\endgroup$ – user98602 Aug 15 '15 at 21:31
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Let $H$ be a homotopy from $g\circ f$ to ${\rm id}_X$ and take a path $\gamma(t) = H(x_0, t)$, i.e. $\gamma(0) = g(f(x_o))$ and $\gamma(1) = x_0$. Consider

$\pi_1(X, x_0) \to \pi_1(Y, f(x_0)) \to \pi_1(X, g \circ f(x_0)) \to π_1(X, x_0)$,

where the first map is $f_{\sharp}$, the second $g_{\sharp}$ and the third is the one induced by $\gamma$. Hence $[\lambda] \in \pi_1(X, x_0)$ gets mapped to $[\gamma^{-1} g f \lambda \gamma]$. One immediately writes down (with the help of $H$) a homotopy from $\lambda$ to $g f \lambda$. Use it to write down a homotopy from $\lambda$ to $\gamma^{-1} g f \lambda \gamma$ concluding that the composition of those maps is indeed the identity.

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  • $\begingroup$ Thanks, I get the idea. Writing $\widetilde{H}(s,t) = H(\lambda(s),t)$ gives $g\circ f \circ \lambda \simeq \lambda$, but this $\simeq$ is not ${\rm rel}\,\{0,1\}$, shouldn't this be a problem? Or am I doing something wrong? +1 of course $\endgroup$ – Ivo Terek Aug 15 '15 at 22:12
  • $\begingroup$ The point is that you build a homotopy (without fixing endpoints) just to obtain that the composition of those maps is the identity, which is bijective. But if a composition of maps is bijective (i.e. injective + surjective) what can you say about the "compositioned" maps? $\endgroup$ – M.U. Aug 16 '15 at 11:39
  • $\begingroup$ Yes, sure, but that's not my problem. My problem is: to prove that the composition is the identity, I must get $[\gamma^{-1}gf\lambda\gamma]=[\lambda]$, that is, $\gamma^{-1}gf\lambda\gamma\simeq\lambda ~{\rm rel}\,\{0,1\}$, and the suggestion gives $gf\lambda \simeq \lambda $, but not ${\rm rel}\,\{0, 1\}$ $\endgroup$ – Ivo Terek Aug 16 '15 at 14:15
  • $\begingroup$ In my head, since the fundamental groups are classes from the relation $\simeq ~{\rm rel}\,\{0, 1\}$, we need to keep the endpoints fixed all the time $\endgroup$ – Ivo Terek Aug 16 '15 at 14:19
  • $\begingroup$ Oh, wait, I think I got it. You're saying that even if $\widetilde{H}$ doesn't keep endpoints fixed, when I do the next homotopy, the composition will keep them fixed alright? $\endgroup$ – Ivo Terek Aug 16 '15 at 14:21

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