When Hatcher discusses the universal coefficient theorem for homology (section 3.A, pg. 261), he first takes the exact sequence of chain complexes
$$0 \rightarrow Z_n \xrightarrow{i_n} C_n \xrightarrow{\partial_n} B_{n-1} \rightarrow 0,$$
where $i_n$ is the inclusion and $\partial_n$ is the boundary. Here, the complexes $(Z_n)$ and $(B_n)$ have trivial boundary maps.
Then, he shows that we can tensor it with a given abelian group $G$ to obtain a new short exact sequence of chain complexes
$$0 \rightarrow Z_n \otimes G \xrightarrow{i_n \otimes~1} C_n \otimes G \xrightarrow{\partial_n \otimes~1} B_{n-1} \otimes G \rightarrow 0.$$
From this new short exact sequence of chain complexes, he constructs a long exact sequence of homology, as usual. Because of the triviality of the boundary maps in $(B_n)$, $(Z_n)$, this long exact sequence of homology looks like this:
$$\dots \rightarrow B_n \otimes G \rightarrow Z_n \otimes G \rightarrow H_n(C; G) \xrightarrow{\Large \Phi} B_{n-1} \otimes G \rightarrow \dots$$
What I can't understand is: why is the map $H_n(C ; G) \xrightarrow{\Large \Phi} B_{n-1} \otimes G$ not always zero? After all, isn't $H_n(C ; G)$ just a quotient of the kernel of $\partial_n \otimes 1$ to begin with? And shouldn't this map, $\Phi$, be essentially induced by $\partial_n \otimes 1$??