# Show that $\ker h =[\pi_1(X,x_0), \pi_1(X,x_0)]$ in the proof of Hurewicz Theorem

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.

• 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. Nov 13, 2021 at 19:04