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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 $\partial_n \otimes 1$??

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    $\begingroup$ Which explicit examples have you tried? $\endgroup$ Commented Apr 16, 2013 at 1:32

2 Answers 2

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I know this is an old post, but came across it when I was asking myself exactly the same question. And I would like to share my thoughts on that.

So, I believe that the point is that the maps $\partial_n \otimes 1: C_{n} \otimes G \rightarrow B_{n-1} \otimes G $ and the boundary $ \partial_n \otimes 1: C_{n} \otimes G \rightarrow C_{n-1} \otimes G $ are not the same maps.

Indeed, let $(\partial_n)^r : C_{n} \rightarrow B_{n-1}$ be the restriction of $ \partial_n: C_{n} \rightarrow C_{n-1} $ to its image, and let $j: B_{n-1} \rightarrow C_{n-1}$ be the inclusion. Then, notice that the map between $C_n \otimes G$ and $B_{n-1} \otimes G$ appearing in the question actually is $ (\partial_n)^r \otimes 1.$ Moreover, it is clear that $\partial_n = j \circ (\partial_n)^r.$ From this, and from the fact the tensor functor may not preserve injectivity, one can conclude that $\Phi: H_n(C, G) \rightarrow B_{n-1}\otimes G,$ which is induced by $(\partial_n)^r,$ is not necessarily zero.

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The group $H_{n}(C,G)$ is defined as the homology of the sequence $$...C_{n+1}\otimes G\rightarrow C_{n}\otimes G\rightarrow C_{n-1}\otimes G...$$

If we denote the map $C_{i}\rightarrow C_{i-1}$ by $d_{i}$, then this is $\ker d_{n}\otimes i/Im d_{i+1}\otimes i$. Here $B_{n-1}$ lives in $C_{n-1}$, and $Z_{n}$ lives in $C_{n}$. So by taking the kernel of $Im d_{i+1}\otimes i$ you can only quotient out $B_{n+1}$, not $B_{n}$. I suspect this is the source of your confusion. Some pencil and paper may be helpful for the purpose if you are confused with the notations.

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