If 2 spaces are homotopy equivalent, then their fundamental group is the same

Is this true ?


$f:X\rightarrow Y$$\quad$$g:Y\rightarrow X$ s.t.

$f\circ g\simeq id_Y$$\quad$$g\circ f\simeq id_X$


$f_*:\pi(X,x_0)\rightarrow\pi(Y,y_0)$ and $\quad$$g_*:\pi(Y,y_0)\rightarrow\pi(X,x_0)$

Hence $f_*\circ g_*=(f\circ g)_*\simeq id_*$ and this implies what ?


If we consider $f_*\circ g_*$

$f_*\circ g_*:\pi(Y,y_0)\rightarrow\pi(Y,y_0)$

let $[\sigma]\in \pi(Y,y_0)$ then we have

$f_*\circ g_*([\sigma])=(f\circ g)_*([\sigma])=[f\circ g\circ\sigma]=[id\circ\sigma]=[\sigma]$

  • 2
    $\begingroup$ For those who are voting to close as "unclear what you are asking", here are the definitions of homotopy equivalence and fundamental group... $\endgroup$ – Julien Dec 24 '13 at 14:40
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    $\begingroup$ I’d like to point out an oversight: If $ f: X \to Y $ and $ g: Y \to X $ are homotopy inverses of each other, and if $ f(x_{0}) = y_{0} $, you can’t simply say that $ g(y_{0}) = x_{0} $. A homotopy inverse doesn’t have to be an inverse map. However, it’s still true that if $ f: X \to Y $ has a homotopy inverse and $ f(x_{0}) = y_{0} $, then $ f_{\ast}: {\pi_{1}}(X,x_{0}) \to {\pi_{1}}(Y,y_{0}) $ is a group isomorphism — you just have to invest more effort in proving it. $\endgroup$ – Transcendental Nov 25 '17 at 6:02

If two maps $\varphi, \psi : (X, x_0) \to (Y, y_0)$ are homotopic, then the induced maps $\varphi_*, \psi_* : \pi_1(X, x_0) \to \pi_1(Y, y_0)$ are equal. It doesn't make sense to say that the induced maps are homotopic (as your question suggests), since the induced maps are group homomorphisms, not continuous maps between topological spaces.

Once we know this, it's easy to prove the statement in your question. Since $f \circ g \simeq \operatorname{id}_Y$, we have $$ (f \circ g)_* = f_* \circ g_* = (\operatorname{id}_Y)_* = \operatorname{id}_{\pi_1(Y, y_0)}. $$

Similarly, since $g \circ f \simeq \operatorname{id}_X$, we have $$ (g \circ f)_* = g_* \circ f_* = (\operatorname{id}_X)_* = \operatorname{id}_{\pi_1(X, x_0)}. $$

It follows that $f_*$, $g_*$ are inverses of each other. Hence they are isomorphisms, and $\pi_1(X, x_0)$, $\pi_1(Y, y_0)$ are isomorphic.

  • $\begingroup$ That was nice. Is it necessary check that $f_*([\delta][\gamma])=f_*([\delta])f_*([\gamma])$ to conclude that the two fundamental groups are group isomorphic? $\endgroup$ – rgm Jan 20 '17 at 18:28
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    $\begingroup$ The check you mention is necessary to show that a continuous map induces a group homomorphism on fundamental groups. This should be proven before discussing the issue in question. $\endgroup$ – Ayman Hourieh Jan 21 '17 at 0:00
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    $\begingroup$ I’d like to point out an oversight: If $ f(x_{0}) = y_{0} $, then it isn’t necessarily true that $ g(y_{0}) = x_{0} $. Hence, $ (f \circ g)_{\ast} $ may not map $ {\pi_{1}}(Y,y_{0}) $ to itself and so may not be equal to $ \operatorname{Id}_{{\pi_{1}}(Y,y_{0})} $ at all. $\endgroup$ – Transcendental Nov 25 '17 at 6:10
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    $\begingroup$ Comment after six years ;-) Your proof only covers the case that pointed spaces are pointed homotopy equivalent. The OP allows free homotopy equivalences between unbased spaces. By the way, the question as it is has to be made precise. It seems that one has to restrict to path-connected spaces. $\endgroup$ – Paul Frost Sep 14 '19 at 17:41

$f, g$ induce homomorphisms of their respective fundamental groups, with $-$ $$g_* \circ f_* = Id_{\pi_1(X)}$$ and $$f_* \circ g_* = Id_{\pi_1(Y)}$$

We have to verify that $f_*, g_*$ are $1-1$ and onto. We can prove this for arbitrary groups actually $-$ Let $G, H$ be groups and let $p: G \rightarrow H, q: H \rightarrow G$ be group homomorphisms with $q \circ p = Id_G, p \circ q = Id_H$. Then $$p(g_1) = p(g_2) \implies g_1 = Id_G(g_1) = (q \circ p)(g_1) = q(p(g_1)) = q(p(g_2)) = (q \circ p)(g_2) = Id_G(g_2) = g_2$$ So $p$ is $1-1$. In addition, $$h \in H \implies q(h) \in G \implies p(q(h)) = (p \circ q)(h) = Id_H(h) = h$$ So that $p$ is onto as well, hence bijective. Likewise, $q$ is bijective. It follows that $p, q$ are in fact isomorphisms, with $p^{-1} = q$. Thus $G \cong H$.

  • $\begingroup$ What we need to prove are $g_* \circ f_* = Id_{\pi_1(X)}$ and $f_* \circ g_* = Id_{\pi_1(Y)}$. If we know these two identities, then they are the inverse of each other, and we are done. You just use these two identities without any proof to prove trivial stuff, which seems really funny to me. $\endgroup$ – Sam Wong Jan 28 at 5:09
  • $\begingroup$ You don't need to prove $g_* \circ f_* = Id_{\pi_1(X)}$, or $f_* \circ g_* = Id_{\pi_1(Y)}$. It's practically given to you. The whole exercise is trivial, so what? $\endgroup$ – Joe Shmo Jan 28 at 5:16
  • $\begingroup$ And by the way, if $X$ and $Y$ are not path-connected, then it makes no sense to write $\pi_1(X)$ and $\pi_1(Y)$. $\endgroup$ – Sam Wong Jan 28 at 5:17
  • $\begingroup$ We need to prove the two identities which are in the core step of the proof, and this exercise is not trivial. $\endgroup$ – Sam Wong Jan 28 at 5:18

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