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 🙂
Best Answer
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 $H^i(X, A)$.