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Consider the cell complex $X$ which is the circle with two discs attached along the attaching maps of degree $3$ and $5$. I would like to compute cohomology with integral coefficients using cellular cohomology and the universal coefficient theorem.

First, the cellular chain complex is as follows (from degree $2$ to degree $0$):

$$ 0 \to\mathbb Z\oplus\mathbb Z \to \mathbb Z \to \mathbb Z \to 0$$ where the first (non trivial, degree $2$) boundary map sends $(a,b) \to (3a+5b)$ while the degree $1$ map is simply $0$.

Computing the homology, we get $H_0(X) \cong H_2(X) \cong \mathbb Z$ and $H_1(X) \cong \mathbb Z/3\times\mathbb Z/5$.


Dualizing the complex to compute cohomology, we have (from degree $0$ to degree $2$): $$0 \to \mathbb Z \to \mathbb Z \to \mathbb Z\oplus \mathbb Z \to 0$$ where the final non trivial map is now $1 \to (3,5)$ and the other maps are trivial. Computing the cohomology, we now have: $$H^2(X) \cong H^0(X) \cong \mathbb Z$$ since $(3,5),(1,2)$ forms a basis for $\mathbb Z\oplus\mathbb Z$ while $H^1(X)$ is $0$.


Now, let me compute the cohomology rings using UCT:

We have the exact sequence: $$0 \to Ext^1(H_0(X),\mathbb Z)\to H^1(X) \to Hom(H_1(X),\mathbb Z)$$ and since both the $Ext$ and $Hom$ terms vanish, $H^1(X) \cong 0$ as expected.

On the other hand, we also have: $$0 \to Ext^1(H_1(X),\mathbb Z) \to H^2(X) \to Hom(H_2(X),\mathbb Z)\to 0.$$

Here however, the $Hom$ term is $\mathbb Z$ while the $Ext$ term is $\mathbb Z/3\mathbb Z \oplus \mathbb Z/5\mathbb Z$ so that $H^2 \cong \mathbb Z\oplus \mathbb Z/3\oplus\mathbb Z/5$ which does not match my earlier commputation.

Where did I go wrong?

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The image of $(a,b)\mapsto 3a+5b$ as a map from $\Bbb Z\oplus\Bbb Z$ to $\Bbb Z$ is all of $\Bbb Z$. So $H_1(X)=0$. There's now no problem with UCT: everything is free, so there's no Ext terms.

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  • $\begingroup$ Stupid mistake :( $\endgroup$ – Asvin Aug 3 '18 at 10:44

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