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

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

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 4
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.