A Construction of a map $h:H^n(C;G)\rightarrow{\rm Hom}(H_n(C),G)$ I'm trying to understand the map $h:H^n(C;G)\rightarrow {\rm Hom}(H_n(C),G)$ in the following extract from Hatcher. I undestand that $\varphi$ is zero on $B_n$. And if $\psi:G\rightarrow H$ is a group homomorphism we have and induced map $\overline{\psi}:G/\ker(\psi)\rightarrow H$. But $B_n$ is not nesscailry the kernel of $\varphi$?

Why is does $\varphi_0$ induce a quoitent homomorphism?

It doesn't seem to be true in general that: if $\psi:G\rightarrow H$ is a group homomorphism and $N$ is a subset of $\ker(\psi)$ then there is an induced map $G/N \rightarrow H$. Trivially if $N$ is not a subgroup then $G/N$ does not even make sense.

 A: I turned my comments into an answer, so that the question doesn't appear as unanswered.
In the passage from the book, $\varphi_0:Z_n \to G$ is a group homomorphism vanishing on the subgroup $B_n = \mathrm{Im}(\partial_{n+1}) \subseteq Z_n$ (notice that $B_n$ is really a subgroup, since it is the image of the homomorphism $\partial_{n+1}$). Since all the groups involved are abelian, $B_n$ is normal in $Z_n$, so the quotient $Z_n/B_n$ is a group.
Now, since $B_n \subseteq \ker \varphi_0$, we have an induced homomorphism $\overline{\varphi_0}:Z_n/B_n \to G$ defined in the following way: given a class $z_n+B_n \in Z_n/B_n$, we define $\overline{\varphi_0}(z_n+B_n) = \varphi_0(z_n)$. The fact that $B_n \subseteq \ker \varphi_0$ ensures that $\overline{\varphi_0}$ is well defined, and using the definiton of the group operation on the quotient $Z_n/B_n$ and the fact that $\varphi_0$ is a group homomorphism, it is straightforward to show that $\overline{\varphi_0}$ is also a group homomorphism.
