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Show that if $\varphi: X \rightarrow Y$ is a homotopy equivalence, then the induced homomorphisms $\varphi_{*}:\pi_n(X,x_0) \rightarrow\pi_n(Y,\varphi(x_0))$ are isomorphisms, for all n$\in \mathbb{N}$.

I've already checked the proof given in 1.18 (case n=1) but I can't generalize to arbitrary dimension. Intuitively, I've tried to repeat the argument, in this case using the change-of-basepoint isomorphism given in page 341, but I can't write a correct proof. I'd appreciate if someone could help me with it. Thanks.

Here is one other attemp: If $\varphi: X \rightarrow Y$ is a homotopy equivalence, $\exists \psi:Y\rightarrow X$ s.t. $\varphi \circ \psi \simeq Id_Y$ and $\psi \circ \varphi \simeq Id_X$, so the induced applications statisfies:

$$(\varphi \circ \psi)_{*}\stackrel{Functor}{=}\varphi_{*} \circ \psi_{*}=Id_{\pi_n(Y)}$$ $$(\psi \circ \varphi)_{*}\stackrel{Functor}{=}\psi_{*} \circ \varphi_{*}=Id_{\pi_n(X)}$$ so $\varphi_{*}^{-1}=\psi_{*}$ and $\varphi_{*}$ is an isomorphism of groups.

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  • $\begingroup$ I am working on this question too. I am wondering exactly why can't we repeat the argument in Chapter 1.18 with $\pi_1$ replaced with $\pi_n$? $\endgroup$ – yoyostein Apr 14 '16 at 15:57
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Use $\psi:Y\to X$ such that $\psi\circ\varphi\simeq id_X$ and $\varphi\circ\psi\simeq id_Y$. What about $\psi_*\circ \varphi_*$ and $\varphi_*\circ\psi_*$

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  • $\begingroup$ I've written a proof with your idea @Hamou, was that what you were thinking? Thanks $\endgroup$ – user148502 Jul 29 '14 at 16:37
  • $\begingroup$ That is. it work. $\endgroup$ – Hamou Jul 29 '14 at 16:40

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