Question about the proof of the universal coefficient theorem

Question

When deriving the universal coefficient theorem, in class we proceeded as follows:

We have the SES: $$0\to Z_\bullet\stackrel{i}{\longrightarrow}S_\bullet\stackrel{\partial}{\longrightarrow}B_{\bullet - 1}\to 0$$ where $S_n$ are the singular $n$-chains, $B_n$ the boundaries and $Z_n$ the cycles. The differential on $Z_\bullet$ and $B_\bullet$ is teh zero map, on $S_\bullet$ it is the usual differential $\partial$. This sequence is split exact at every level, and thus applying the functor $-\otimes_{\mathbb{Z}}G$, where $G$ is an abelian group, we have again a SES: $$0\to Z_\bullet\otimes_{\mathbb{Z}}G\stackrel{i\otimes_{\mathbb{Z}}1}{\longrightarrow}S_\bullet\otimes_{\mathbb{Z}}G\stackrel{\partial\otimes_{\mathbb{Z}}1}{\longrightarrow}B_{\bullet - 1}\otimes_{\mathbb{Z}}G\to 0$$ This induces a LES in homology: $$\ldots\stackrel{c}{\longrightarrow}Z_n\otimes_{\mathbb{Z}}G\stackrel{(i\otimes_{\mathbb{Z}}1)_*}{\longrightarrow}H_n(X;G)\stackrel{(\partial\otimes_{\mathbb{Z}}1)_*}{\longrightarrow}B_{n - 1}\otimes_{\mathbb{Z}}G\stackrel{c}{\longrightarrow}\ldots$$ where $c$ is the connectant. Then we proceded to analyze the various maps to conclude the proof.

My question is the following: the map $(\partial\otimes_{\mathbb{Z}}1)_*$ acts on any class $[\alpha]\in H_n(X;G)$ by: $$(\partial\otimes_{\mathbb{Z}}1)_*[\alpha\otimes_\mathbb{Z}g]=[(\partial\otimes_{\mathbb{Z}}1)(\alpha\otimes_{\mathbb{Z}}g)]=[\underbrace{\partial\alpha}_{=0}\otimes_{\mathbb{Z}}g]=0$$ or am I wrong? This looks wrong by looking at what we did later on, but there are a couple of steps that are not really clear...

Answer 1

First, a general element of $S\otimes G$ is of the form $\sum \alpha_i\otimes g_i$. But that is not the problem. The boundary of this is $\sum (\partial \alpha_i)\otimes g_i$ and you are assuming that its a cycle so it is zero in $S\otimes G$, however this does not imply that it is zero in $B\otimes G$.

Here is an example of this phenomena, which arises from calculating universal coefficients of the projective plane, which is a good thing to examine. Consider $2\mathbb{Z} \subseteq \mathbb{Z}$ Now $\mathbb{Z}\otimes \mathbb{Z}_2=\mathbb{Z}_2$ and note that $2\mathbb{Z}\cong \mathbb{Z}$, as groups, so you have also $2\mathbb{Z}\otimes \mathbb{Z}_2=\mathbb{Z}_2$. But the map from $2\mathbb{Z}\otimes \mathbb{Z}_2 \rightarrow \mathbb{Z}\otimes \mathbb{Z}_2$ is the zero map.