# Cohomology groups for the following pair $(X,A)$

Let $X=S^1\times D^2$, and let $A=\{(z^k,z)\mid z\in S^1\}\subset X$. Calculate the groups and homomorphisms in the cohomology of the exact sequence of the pair $(X,A)$.

I know that theorically one has $$0\rightarrow C_n(A)\rightarrow C_n(X)\rightarrow C_n(X,A)\rightarrow 0$$ then apply Hom$(-,\mathbb{Z})$, and then apply the snake lemma to obtain the long exact sequence $$...\rightarrow H^n(X,A)\rightarrow H^n(X)\rightarrow H^n(A)\rightarrow H^{n+1}(X,A)\rightarrow ...$$

but I have never seen an example done to an actual space (I'm using Hatcher), so my idea was to try to compute the homology groups instead and using the universal coefficient obtain the cohomology groups, but even then I am not quite sure how I would obtain the maps.

If anyone could explain how to do this, or even give a link where they work out examples I would be very grateful :)

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One never ever EVER computes the (co)homology of a space by constructing the singular (co)chain complex and computing its (co)homology! That is NEVER done in practice. –  Mariano Suárez-Alvarez Dec 11 '13 at 5:48
@MarianoSuárez-Alvarez Yeah, for homology the way I compute it is using simplicial, cellular, or using nifty techniques like Mayer Viertoris sequences, but I have never computed LES for cohomology. So far the only way I know how to calculate cohomology is using cellular, simplicial, or using homology along with universal coefficient theorem. –  Daniel Montealegre Dec 11 '13 at 5:51
@MarianoSuárez-Alvarez True for the most part, except one can technically calculate with bare-hands the singular (co)homology of a single point. Beyond that, things get too messy to actually calculate. –  Daniel Rust Dec 12 '13 at 13:51

As both $X$ and $A$ are homotopic to $\mathbb S^1$, $H_1(X) = H_1(A) = \mathbb Z$ and all other homology groups vanish. The long exact sequence is

$$0 \to H_2(X, A) \to \mathbb Z \overset{f}{\to} \mathbb Z \to H_1(X, A)\to \mathbb Z \overset{g}{\to} \mathbb Z \to H_0(X, A)\to 0,$$

where the first two $\mathbb Z$'s corresponds to $H_1$ and the second two corresponds to $H_0$. Then $f = k$, as the generator $z$ of $H_1(A)$ is mapped to $z^k$ of $H_1(X)$. On the other hand, $g=1$ as $A$ and $X$ are both path connected and $g$ is the map induced by the inclusion $A\subset X$. Thus you have $H_2(X, A) = 0=H_0(X, A)$ and $H_1(X, A) = \mathbb Z/k\mathbb Z$. Thus you can use Universal coefficient theorem to find $H^i(X, A)$.

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How is $A$ homotopic to $S^1$? –  Daniel Montealegre Dec 11 '13 at 5:45
Why noy simply use the long exact sequence for cohomology? –  Mariano Suárez-Alvarez Dec 11 '13 at 5:46
$A$ is actually homeomorphic to $\mathbb S^1$ by projecting to the second component. –  John Dec 11 '13 at 5:47
@MarianoSuárez-Alvarez: Partly because the OP ask for such an approach, and partly because I do not want to say why $H^1(X) = Hom(H_1(X), \mathbb Z)$ (which somehow use again the universal coefficient theorem?). –  John Dec 11 '13 at 5:50
I don't understand your comment. –  Mariano Suárez-Alvarez Dec 11 '13 at 5:51

Given a pair $(X,A)$, there is a long exact sequence $$\cdots \to H^n(X,A)\to H^n(X) \to H^n(A) \to H^{n+1}(X,A) \to\cdots$$ Since $X$ and $A$ have the homotopy type of $S^1$, you know two out of every three of the abelian groups appearing here. Use that.

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I see that $H^i(X)=H^i(A)=\mathbb{Z}$ for $i=0,1$, but how can I find what the map is from $H^1(X)\rightarrow H^1(A)$? I see that the map in homology is multiplication by $k$, but how can I deduce the map in cohomology? –  Daniel Montealegre Dec 11 '13 at 6:00