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This is bad, but I was applying the UCT to a small complex and didn't seem to work. Namely the chain complex $0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \rightarrow 0$ where the nonzero map is, say, multiplication by 2. Then the homology groups should be (left to right) 0, 0, Z/2, 0 and the dual complex is essentially the backwards complex, with cohomology 0, Z/2, 0, 0. By the UCT then, Z/2 is isomorphic to 0+Ext(0). Sorry for the dumb question, but I can't see what I'm doing wrong and it's kind of stressing me out.

rob
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1 Answers1

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I'm going to assume that the factors of $\mathbb{Z}$ in this case are in dimensions 2 and 1, going down as you follow the arrows. Thus you have $H_2(C)=0$ and $H_1(C)\cong \mathbb{Z}/2$, where $C$ is the chain complex. The UCT in this case would say that $H^{2}(C)\cong Hom(0, \mathbb{Z}) \oplus Ext(\mathbb{Z}/2, \mathbb{Z})=Ext(\mathbb{Z}/2, \mathbb{Z})\cong \mathbb{Z}/2$, since the $Hom$ term is $Hom(H_2(C), \mathbb{Z})$ and the $Ext$ term is $Ext(H_{1}(C), \mathbb{Z})$, using the fact that $Ext(\mathbb{Z}/m\mathbb{Z}, G)\cong G/mG$ for any abelian group G. This is consistent with what you found by just computing with the chain complex directly. What the universal coefficient theorem tells you is that cohomology in a particular dimension is influenced by the structure of the homology in that dimension, as well as the homology in the dimension below.

Brian Klatt
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  • Thanks. Somehow I forgot that the dimensions decreased going from left to right, that was my problem. Maybe I should go get some air or something. – rob Oct 19 '13 at 19:28