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If $f:X \rightarrow Y$ is a homotopy equivalence with homotopy inverse $g$ and I want to prove that $f_*$ is an isomorphism then Hatcher (on page 37) uses the following:

$$ \pi_1 (X, x_0) \xrightarrow{f_*} \pi_1 (Y, f(x_0)) \xrightarrow{g_*} \pi_1 (X, g(f(x_0))) \xrightarrow{f_*} \pi_1 (X, f(g(f(x_0))))$$

My question is: why is it useful or necessary to apply the functions several times? If I have $gf \simeq id_X$ then I know $(gf)_* = \beta_h$ (preceding lemma) where $\beta_h$ is an isomorphism so $f_*$ is injective. Doing the same for $fg \simeq id_Y$ yields $f_*$ is surjective, so I'm done.

What am I missing? Thanks for your help!

Rudy the Reindeer
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2 Answers2

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You're missing nothing. And Hatcher is doing the same thing, but just draws both $(fg)_*$ and $(gf)_*$ on the same diagram.

Grigory M
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I think it is necessary because the first $f_*$ is not identical with the third $f_*$ in the diagram (Actually they differ by two isomorphisms).
The first $f_*$ is induced by $f:(X,x_0)\rightarrow (Y,f(x_0))$ but the third is induced by $f:(X,g(f(x_0)))\rightarrow (Y, f(g(f(x_0))))$. Notice that these two morphisms in $Top_*$ are not identical.
So the proper way to prove this proposition is to prove that $g_*$ is both injective and surjective thus isomorphism. The first $f_*$ is now isomorphism from the fact $(gf)_*=\beta_h$.

qwertywelch
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