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Now, I am learning a proof that a homotopy equivalence induces an isomorphism. However, since I am a beginner in algebraic topology, I cannot fully understand the proof.

Suppose $\varphi:X\to Y$ is a homotopy equivalence. To prove $\varphi_\ast:\pi_1(X,x_0)\to\pi_1(Y,\varphi(x_0))$ is an isomorphism for every $x_0\in X$, consider the following diagram:

$$\pi_1 (X, x_0) \xrightarrow{\varphi_\ast} \pi_1 (Y, \varphi(x_0)) \xrightarrow{\psi_\ast} \pi_1 (X, \psi(\varphi(x_0))) \xrightarrow{\varphi_\ast} \pi_1 (X, \varphi(\psi(\varphi(x_0))))$$

where $\varphi\circ\psi\sim 1_Y$ and $\psi\circ\varphi\sim 1_X$.

Then there are three steps:

1) It follows that $(\varphi\circ\psi)_\ast$ is an isomorphism. Hence, $\varphi_\ast$ is surjective.

2) It follows that $(\psi\circ\varphi)_\ast$ is an isomorphism. Hence, $\varphi_\ast$ is injective.

3) Conclude that $\varphi_\ast$ is an isomorphism.

My Question is:

In 1), we actually prove that $\varphi_\ast:\pi_1 (X, x_0)\to\pi_1(Y,\varphi(x_0))$ is surjective.

In 2), we actually prove that $\varphi_\ast:\pi_1(X,\psi(\varphi(x_0)))\to \pi_1 (X, \varphi(\psi(\varphi(x_0))))$ is injective.

They are not the same function, although both are denoted by $\varphi_\ast$. So how to prove $\varphi_\ast:\pi_1(X,x_0)\to\pi_1(Y,\varphi(x_0))$ is injective?

Perhaps this question is really silly. I must miss something. Could anyone point it out? Thanks!

YYF
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2 Answers2

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Since $(\psi\circ\varphi)_*=ψ_*\circφ_*$ is an isomorphism, $ψ_*$ is surjective. And since $(φ∘ψ)_*$ is an isomorphism, $ψ_*$ must be injective. So $ψ_*$ is an isomorphism. It follows that $φ_*=ψ_*^{-1}∘(ψ∘φ)_*$ is an isomorphism, as well.

Stefan Hamcke
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This is a fair question. Here is how to complete the proof. Since $\varphi \circ \psi \sim 1_{Y}$ and $\psi \circ \varphi \sim 1_{X}$, the elements $x_0$ and $\psi(\varphi(x_0))$ are in the same path-component, using the path $\alpha$ induced by the homotopy. Similarly, $\varphi(\psi(\varphi(x_0)))$ and $\varphi(x_0)$ are in the same path-component, connected by a path $(\varphi \circ \alpha)^{-1}$. Thus, the corresponding fundamental groups are isomorphic, so if we know that

$$ \pi_1 (X, \psi(\varphi(x_0))) \xrightarrow{\varphi_\ast} \pi_1 (Y, \varphi(\psi(\varphi(x_0))))$$

is injective, we can define a map from $$\pi_1 (X, x_0) \xrightarrow{\sim} \pi_1 (X, \psi(\varphi(x_0))) \xrightarrow{\varphi_\ast} \pi_1 (Y, \varphi(\psi(\varphi(x_0)))) \xrightarrow{\sim} \pi_1 (Y, \varphi(x_0)) $$

this is the composition of isomorphisms and an injective map, hence it is also injective. Thus, it suffices to show that this composition map is the same map we started with $$\varphi_\ast:\pi_1 (X, x_0)\to\pi_1(Y,\varphi(x_0))$$

We know the change-of-basepoint isomorphisms above are given by conjugation by $[\alpha]$ and $[\varphi \circ \alpha]^{-1}$. Thus the full map sends a loop $\gamma$ to

$$[\gamma] \to [\alpha][\gamma][\alpha^{-1}] \to [\varphi \circ \alpha][\varphi\circ \gamma][\varphi \circ \alpha]^{-1} \to [\varphi \circ \alpha]^{-1}[\varphi \circ \alpha][\varphi\circ \gamma][\varphi \circ \alpha]^{-1}[\varphi \circ \alpha] = [\varphi \circ \gamma]$$

Everything cancels out, and we observe that the long composition above, which we know to be injective, is just $\varphi_{*}$ (the real one from the first direction of the proof).

Elchanan Solomon
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