I am currently learning algebraic topology (using Munkres/Hatcher)
So far I've seen that the fundamental group of the "figure 8 space" is $\langle a, b \rangle$ or simply the free product on 2 generators.
Then, we can also calculate the fundamental group of the Klein bottle which was $\langle a, b : aba = b \rangle$ (an application of Van-Kampen's theorem).
I was wondering if this is enough information to distinguish the two spaces. Can I claim that these 2 free products are non-isomorphic? (the presentation with no relations seems to be different to one with relations). I know generators must be mapped to generators by an isomorphism but there may be different generating letters if I'm not mistaken (rather than just $a, b$, they could be a combination of them).
Do we have enough information from the fundamental groups to distinguish these spaces?
Best Answer
Let $G_1 := \langle a,b \rangle$ and $G_2 :=\langle a,b : aba=b \rangle$. If $G_1 \cong G_2$, then for every abelian group $A$ also $\mathrm{Hom}(G_1,A) \cong \mathrm{Hom}(G_2,A)$ as abelian groups*. But the universal property of the free group modulo relations implies that $\mathrm{Hom}(G_1,A) \cong A^2$, whereas $\mathrm{Hom}(G_2,A)$ is isomorphic to the subgroup of those $(x,y) \in A^2$ such that $x+y+x=y$ in $A$, i.e. $2x=0$ and $y$ is arbitrary. If $2 : A \to A$ is injective, we get $\mathrm{Hom}(G_2,A) \cong A$. So it suffices to find a single abelian group $A$ such that $2 : A \to A$ is injective and $A \not\cong A^2$. Well, any non-trivial finite abelian group of odd order is an example. But also $\mathbb{Z}$ is an example.
This proof is, of course, essentially equivalent to the one using abelianizations. But the thing is: you don't have to find or use the universal abelian quotient. The homomorphisms into concrete abelian groups are sufficient.
*Conceptually, this is because $\mathrm{Hom}(-,A) : \mathbf{Grp}^{\mathrm{op}} \to \mathbf{Ab}$ is a functor.