Universal Coefficients Theorem For Homology - Tor map 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?
 A: The homology classes in $H_n(C;G)$ are represented by elements of the kernel of $\partial\otimes id\colon C_n\otimes G\rightarrow C_{n-1}\otimes G$ (the differential of the chain complex $C_{\bullet}\otimes G$). The notational abuse is a detriment here, but this is not the same thing as the kernel of $\partial_B\otimes id\colon C_n\otimes G\rightarrow B_{n-1}\otimes G$ (I've put an index on this map to differentiate it from the other one). Why, you ask? Consider that we have a commutative triangle (I'm only drawing a square, because this site doesn't have proper support for drawing commtuative diagrams)
$\require{AMScd}$
\begin{CD}
C_n @>\partial_B>> B_{n-1}\\
@V\partial  V V @VVV\\
C_{n-1} @= C_{n-1}.
\end{CD}
Tensoring this diagram with $G$, we obtain the diagram of interest
\begin{CD}
C_n\otimes G @>\partial_B\otimes id>> B_{n-1}\otimes G\\
@V\partial\otimes id  V V @VVV\\
C_{n-1}\otimes G @= C_{n-1}\otimes G.
\end{CD}
The commutativity of this diagram implies that $\ker(\partial_B\otimes id)\subseteq\ker(\partial\otimes id)$, but the catch is that whilst $B_{n-1}\rightarrow C_{n-1}$ is an inclusion, tensoring with $G$ can destroy its injectivity and so $B_{n-1}\otimes G\rightarrow C_{n-1}\otimes G$ is not injective anymore in general, which makes the kernel of $\partial\otimes id$ larger. You are correct to observe that $\phi$ vanishes on $\ker(\partial_B\otimes id)$, so, in a sense, what the non-triviality of $\phi$ measures is precisely the failure of $-\otimes G$ to preserve the injectivity of $B_{n-1}\rightarrow C_{n-l}$.
To see an explicit example, consider the chain complex
$$\dotsc\rightarrow0\rightarrow\mathbb{Z}\stackrel{2\cdot}{\rightarrow}\mathbb{Z}\rightarrow0\rightarrow\dotsc.$$
Here, the non-trivial terms are in degree $1$ and $0$. The inclusion $B_0\rightarrow C_0$ then is the inclusion $2\mathbb{Z}\rightarrow\mathbb{Z}$. If you tensor with $G=\mathbb{Z}/2\mathbb{Z}$, you obtain the zero map $\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}/2\mathbb{Z}$, which is not injective. In fact, $\phi$ in this case is an isomorphism $H_1(C;\mathbb{Z}/2\mathbb{Z})\rightarrow\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}$.
