For a group $G$, its abelianization is given by $\tilde G := G/[G,G]$ where $[G,G]$ is the commutator subgroup of $G$. I am almost done proving the Hurewicz theorem (statement follows), and I need help with a final step. Note that $[f]$ represents a homotopy class, and $[[f]]$ represents a homology class.

Hurewicz Theorem: If $(X,x_0)$ is a path-connected topological space, then $h: \pi_1(X,x_0) \to H_1(X)$ induces an isomorphism $h': \widetilde{\pi}_1(X,x_0) \to H_1(X)$. Here, $h$ takes homotopy classes to homology classes, i.e. $h: [f]\mapsto [[f]]$.

What I have accomplished so far:

  1. I have shown that $h: \pi_1(X,x_0) \to H_1(X)$ given by $h: [f]\mapsto [[f]]$ is a well-defined homomorphism.
  2. By the universal property of abelianization (which follows from the universal property of quotient groups), I have shown that $h$ induces a homomorphism $\widetilde{\pi}_1(X,x_0) \to H_1(X)$ such that $h = h' \circ q$ where $q : \pi_1(X,x_0) \to \widetilde\pi_1(X,x_0)$ is the canonical quotient map.
  3. To prove that $h'$ is an isomorphism, it suffices to show two things, namely (i) $h$ is surjective and (ii) $\ker h = [\pi_1(X,x_0), \pi_1(X,x_0)]$. I have shown that $h$ is surjective and $[\pi_1(X,x_0), \pi_1(X,x_0)] \subset\ker h$.

What I need help with:

To complete the proof, we need $$\color{blue}{\ker h \subset [\pi_1(X,x_0), \pi_1(X,x_0)]}$$ Suppose $[f] \in \ker h$, i.e. $h[f] = [[f]] = 0 \in H_1(X)$. It suffices to show that $[f]$ is trivial in $\widetilde\pi_1(X,x_0)$. As $f$ is a loop, $f(0)= f(1)$, so $f$ is a $1$-cycle. $[[f]] = 0$ means that $f$ is homologous to zero. So, $f$ must be the boundary of some $2$-chain, $\sigma = \sum_i n_i \sigma_i$ where $\sigma_i: \triangle^2\to X$ are singular $2$-simplices. Allowing $\sigma_i = \sigma_j$ for $i\ne j$, we can take $n_i = \pm 1$ for simplicity. Thus, $\sigma = \sum_i \sigma_i$. For each singular $2$-simplex $\sigma_i$, $\partial \sigma_i = \lambda_{i_0} - \lambda_{i_1} + \lambda_{i_2}$ for three $1$-chains $\lambda_{i_j}$ where $0\le j\le 2$. We have $$\gamma = \partial\left(\sum_i n_i \sigma_i \right) = \sum_i n_i \partial \sigma_i = \sum_{i}\sum_{j=0}^2 (-1)^j n_i \lambda_{i_j}$$

How do I proceed? I'm stuck here. I'd appreciate any help in completing my attempt. Thank you!

Several questions about the Hurewicz theorem have been asked on this site earlier, but this is not a duplicate - since I have requested help with a specific part of my own proof attempt.

  • $\begingroup$ I think the main observation here is that $\sigma_i$ defines a homotopy between $\lambda_{i_1}$ and $\lambda_{i_0}+\lambda_{i_1}$, so the right hand side is going to be null homotopic in some sense. $\endgroup$
    – Berci
    Nov 13, 2021 at 19:04

1 Answer 1


See pages 4-5 of this document.


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