Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

1 Answer 1

up vote 5 down vote accepted

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).

share|improve this answer
    
Ah yes, that's an important property about inverses. Thanx –  The Substitute Oct 26 '11 at 18:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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