I'm trying to understand the Universal Coefficients Theorem for homology but I don't really get what the map $$ H_n(C; G) \to \mathrm{Tor}(H_{n-1}(C), G) $$ should be.
The construction should be the following: you take the short exact sequence $$ 0 \to Z_n \to C_n \xrightarrow{\partial} B_{n-1} \to 0 $$ which splits and hence exactness is preserved after tensoring with $G$: $$ 0 \to Z_n \otimes G \to C_n \otimes G \xrightarrow{\partial \otimes id} B_{n-1} \otimes G \to 0. $$
Hence, we can consider the long exact sequence in homology, which looks like $$ \dots \to Z_n \otimes G \to H_n(C; G) \xrightarrow{\phi} B_{n-1} \otimes G \to \dots $$ (here the outer chain complexes are endowed with the zero differentials).
What I don't understand is why $\phi$ is not the zero map. Every element of $H_n(C; G)$ is represented by an element of $\mathrm{ker}(\partial \otimes id)$, which is $Z_n \otimes G$ by exactness.
There is something weird going on with the tensor product but I can't get my head around it. I know this is a duplicate of a (really) old post but I don't think the answers there were satisfactory. Can anyone help me?