# How does this proof show $h_*$ (homomorphism induced by $h\colon (X,a)\to (Y,b)$) is an isomorphism?

Claim: if $h\colon(X,a)\to(Y,b)$ is a homeomorphism of $X$ with $Y$, then $h_*\colon \pi_1(X,a)\to \pi_1(Y,b)$ is an isomorphism.

where $\pi_1$ refers to the fundamental group and $h_*$ is the induced homomorphism defined by $h_*([f]) = [h(f)]$.

I already know $h_*$ is a homomorphism since $h(f\cdot g)=h(f) \cdot h(g)$. To show $h_*$ is an isomorphism, I thought it sufficed to show it's a bijection...

Munkres' Proof. Let $k: (Y,b)\to(X,a)$ be the inverse of $h$. Then $k_*\circ h_*=(k\circ h)_* = i_*$, where $i$ is the identity map of $(X,a)$. And $h_*\circ k_*= (h\circ k)_*=j_*$, where $j$ is the identity of $(Y,b)$. Since $i_*$ and $j_*$ are the identity homomorphisms of the groups $\pi_1(X,a)$ and $\pi_1(Y,b)$, respectively, $k_*$ is the inverse of $h_*$. $\Box$

How does this show $h_*$ is a bijection? Since $h$ is a homeomorphism, I know $h_*$ is injective.

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You just showed that $h^* \circ k^*$ is the identity on $\pi_1(Y,b)$ and $k^* \circ h^*$ is the identity on $\pi_1(X,a)$. This means $h^*$ and $k^*$ are two-sided inverses of each other. A function is bijective if and only if it has a two sided inverse. –  Bill Cook Oct 26 '11 at 18:40

Remember that a morphism is an isomorphism if and only if it has an inverse (that is also a morphism). For groups, a homomorphism is an isomorphism if and only if it has an inverse, if and only if it is bijective on underlying sets (that's why "it suffices" to show $h_*$ is bijective); this is not true for other settings (e.g., for topological spaces, a continuous map needs more than just being bijective in order to be an isomorphism, i.e. a homeomorphism), but for groups it is enough.
So, really, Munkres is showing that $h_*$ is an isomorphism directly, by producings its group-theoretic inverse.
But if you want to argue that $h_*$ is bijective, notice that $h_*$ and $k_*$ are both group homomorphisms, and they are inverses of each other as group homomorphisms. Any group homomorphism that has an inverse has to be bijective. (Because it is a function of the underlying set, and only bijective functions have inverses).