# 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?

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 $$H_1(C;\mathbb{Z}/2\mathbb{Z})\rightarrow\mathbb{Z}\otimes\mathbb{Z}/2\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}$$.