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Hatcher has very minimal treatment of direct cohomology computation. For a greater depth of understanding I want to make sure that I can still compute cohomology straight from the definitions of cellular cohomology, the Mayer-vietoris sequence, etc...

I'm trying to compute the cohomology groups of the Klein bottle using cellular cohomology. We know that the Klein bottle has a CW complex consisting of one 0-cell, two 1-cells and one 2-cell. Hence we have a cochain complex given by $$0 \longleftarrow \mathbb{Z} \longleftarrow \mathbb{Z} \oplus \mathbb{Z} \longleftarrow \mathbb{Z} \longleftarrow 0.$$ My confusion is on how we compute the coboundary maps?

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You need to know what the maps are. Let $C_0,C_1$ and $C_2$ be the free groups on the cells. Then you have

$$0\rightarrow C_2\rightarrow C_1 \rightarrow C_1\rightarrow 0$$ $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0$$

and $$\partial_2 (n)=(2n,0)$$ $$\partial_1(n,m)=0$$ so for the homology groups we get $$H_2,H_1,H_0$$ $$0,\mathbb{Z}_2\oplus\mathbb{Z}, \mathbb{Z}$$

Now to get the cohomology we have to find the maps for

$$0\rightarrow \operatorname{Hom}(C_0, \mathbb{Z})\rightarrow \operatorname{Hom}(C_1, \mathbb{Z})\rightarrow \operatorname{Hom}(C_2, \mathbb{Z})\rightarrow 0$$ $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0$$

So $$\delta_0 f (a,b)=f(\partial_1(a,b))=0$$ thus $\delta_0=0$. $$\delta_1 g (n)=g(\partial_2(n))=g(2n,0)$$ By choosing an appropriate basis for $C^1$ we can think of this map as $(a,b)\mapsto 2a$.

Thus for cohomology we get

$$H^0,H^1,H^2$$ $$\mathbb{Z}, \mathbb{Z}, \mathbb{Z}_2$$

I hope this makes sense and that it helps !

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  • $\begingroup$ Could you please explain the part "By choosing an appropriate basis for $C^1$ we can think of this map as $(a,b)\mapsto 2a$"? Thanks in advance $\endgroup$ – perlman Apr 2 '17 at 21:24

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