Relation between the fundamental group and the first homology group

Question

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.

Answer 1

$p\circ f^{-1}$ (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.

Answer 2

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.