# Relation between the fundamental group and the first homology group

Is it true that if the fundamental groups of two spaces are isomorphic, then their first homology groups are isomorphic? I think the answer is yes;

Let $f:\pi_1(X)\to\pi_1(Y)$ be an isomorphism and $p:\pi_1(X)\to \pi_1(X)^*, q:\pi_1(Y)\to \pi_1(Y)^*$ be the quotient maps, where $\pi_1(X)^*, \pi_1(Y)^*$ are the abelianizations of the fundamental groups. I need to show that $p\circ f$ and $q\circ f^{-1}$ are isomorphisms. But I don't how to do that.

• Shouldn't $p$ be a map $\pi_1(X) \to [\pi_1(X), \pi_1(X)]$ and likewise for $q$? – Michael Albanese Aug 28 '13 at 14:56
• Why? It is the quotient map $p:\pi_1(X_1)\to \pi_1(X_1)/[\pi_1(X), \pi_1(X)]$ – Xena Aug 28 '13 at 14:58
• Sorry, that's what I mean't. What you wrote in your comment is not what you wrote in the question. Also, what are $X_1$ and $X_2$? – Michael Albanese Aug 28 '13 at 14:59
• Isn't the abelianization the same thing that I wrote in my comment? $X_1, X_2$ are just topological spaces. – Xena Aug 28 '13 at 15:03
• A topological space doesn't have an abelianization, a group does. – Michael Albanese Aug 28 '13 at 15:04

$p\circ f^{-1}$ (this is the correct composition) is a homomorphism from $\pi_1(Y)$ to $\pi_1(X)^*$; it is not this map which is an isomorphism.

I will sketch the correct idea for you, avoiding the more sophisticated concepts from user87690's answer. Elements of $\pi_1(X)^*$ are equivalence classes of elements of $\pi_1(X)$; $g_1 \sim g_2$ if and only if $g_2 = g_1 h_1^{-1}h_2^{-1}h_1 h_2$ for appropriate $h_1, h_2 \in \pi_1(X)$. Given a homomorphism $f : \pi_1(X) \to \pi_1(Y)$, an obvious way to try to define a map from $\pi_1(X)^*$ to $\pi_1(Y)^*$ is $f^{\mathrm{ab}}: [g] \mapsto [f(g)]$, where $[]$ denotes equivalence class. You need to check that this is well-defined, i.e., independent of which choice of representative we take for the equivalence class, and is a homomorphism.

In the case that $f$ is an ismorphism, you should be able to check that $f^{\mathrm{ab}}$ is also an isomorphism.

• Great, thanks. It is all clear now. – Xena Aug 28 '13 at 16:00

Since $H_1(X)$ is isomorphic to abelianization of $π_1(X)$ for $X$ path-connected and isomorphic groups have isomorphic abelinizations, positive answer to your question follows.

Abelinization is a functor from groups to abelian groups. Let $Ab(G)$ mean the abelinization of $G$ and let $π_G: G \to Ab(G)$ be the canonical projection. Then for any group homomorphism $f: G \to H$ there is unique homomorphism $Ab(f): Ab(G) \to Ab(H)$ such that $Ab(f) π_G = π_H f$ (this follows from universal property of abelinization which follows from the homomorphism theorem for groups). Now $Ab$ is a functor which means that $Ab(fg) = Ab(f) Ab(g)$ and $Ab(id_G) = id_{Ab(G)}$. Functors always preserve isomorphisms since if $f: G \to H$ is isomorphism and $g$ is its inverse then $Ab(f) Ab(g) = Ab(fg) = Ab(id_G) = id_{Ab(G)}$ and similary for $Ab(g) Ab(f)$ so $Ab(f)$ is isomorphism between the abelinizations and $Ab(g)$ is its inverse.

Intuitivelly it's obvious that isomorphic groups have isomorphic abelinization since the abelinization depends only on group structure.

• Yes, this is exactly what I'm asking for: to prove that isomorphic groups have isomorphic abelianizations. – Xena Aug 28 '13 at 15:03
• @Fanni: Abelinization is a functor from groups to abelian groups since any group homomorphism factors to homomorphism between abelizations. – user87690 Aug 28 '13 at 15:05
• I didn't understand, what is a functor? What does it mean for a map to factor to something? – Xena Aug 28 '13 at 15:18
• @Fanni: I've added explanation to the answer. – user87690 Aug 28 '13 at 15:35