Let's suppose $$\phi: \pi_1(X,x_0)\to H_1(X,\mathbb{Z})$$ $$[f]\to [[f]]$$

I already prove that this function is well defined and is a group homomorphism.

I have to show that this $\phi$ function induces a group homomorphism $$\Phi: \pi_1(X,x_0)^{ab}\to H_1(X,\mathbb{Z})$$ where $\pi_1(X,x_0)^{ab}$ is the abelianization of the group $\pi_1(X,x_0)$.

I have thought to show that $[\pi_1(X,x_0),\pi_1(X,x_0)]\subset ker(\phi)$ to conclude that $\Phi$ is a homomorphism. Let $[f], [g]\in \pi_1(X,x_0)$, notice that $[[f],[g]]=[f]*[g]*[f]^{-1}*[g]^{-1}$ and thus $\phi([[f],[g]])=\phi([f]*[g]*[f]^{-1}*[g]^{-1})=[[f]]+[[g]]-[[f]]-[[g]]=0$ (This assuming that $\phi([h]^{-1})=-[[h]]\in H_1(X)$, how can I prove this?). With this, we can conclude that $[[f],[g]]\in ker(\phi)$ and so $[\pi_1(X,x_0),\pi_1(X,x_0)]\subset ker(\phi)$ because $[\pi_1(X,x_0),\pi_1(X,x_0)]$ is generated by all the $[[f],[g]]$. Is this test OK? Thank you.


This assuming that $\phi([h]^{-1})=-[[h]]\in H_1(X)$, how can I prove this?

This is simply because $\phi$ is a group homomorphism. Generally if $f:G\to H$ is a group homomorphism then you have


and so $f(g^{-1})$ is the inverse of $f(g)$. In symbolic notation $f(g)^{-1}=f(g^{-1})$.

It's just that $H_1(X)$ is abelian and so inverses are denoted with "$-$" symbol.

Your proof is correct. You also may wanna have a look at abelianization because what you've shown is a special case of more general fact. When you have a group homomorphism $f:G\to H$ with $H$ abelian and $\pi:G\to G^{ab}$ is the standard quotient map then there exists a unique group homomorphism $g:G^{ab}\to H$ such that $f=g\circ\pi$. Or in other words $f$ uniquely factorizes through $G^{ab}$ (and in fact $G^{ab}$ can be defined by an universal variation of this property, see wiki). The proof is pretty much the same but it is good to catch essentials. Because in this case homotopy and homology is just a distraction.


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