Why is $H^*(S^1\times X,\{\text{pt}\}\times X;R)\cong H^*(D^1\times X,\partial D^1\times X;R)$?

Question

Let $X$ be a topological space, $R$ is a commutative, unital ring. In a proof from lecture there is claimed that $H^*(S^1\times X,\{\text{pt}\}\times X;R)\cong H^*(D^1\times X,\partial D^1\times X;R)$ in singular cohomology and I'm not sure how to justify it.

It is $\partial(D^1)=S^0=\{-1\}\coprod \{+1\}$ and maybe it's something like exicion and homotopy invariance, but I don't know how to apply excision here in detail. I would be happy if you explain me how to prove that $H^*(S^1\times X,\{\text{pt}\}\times X;R)\cong H^*(D^1\times X,\partial D^1\times X;R)$.

Answer 1

The inclusion $* \times X \subset S^1 \times X$ is a cofibration (the singleton $*$ is a sub-CW-complex of $S^1$, hence $* \subset S^1$ is a cofibration, and by this question the product of a cofibration with an identity map is a cofibration), thus: $$H^*(S^1 \times X, * \times X) \cong \tilde{H}^*\bigl( (S^1 \times X) / (* \times X) \bigr)$$ (this is a standard fact, see e.g. Hatcher, Proposition 2.22 which deals with the case of homology; the cohomological case is identical).

Similarly you have: $$H^*(D^1 \times X, \partial D^1 \times X) \cong \tilde{H}^*\bigl( (D^1 \times X) / (\partial D^1 \times X) \bigr).$$

But now the two spaces $(S^1 \times X) / (* \times X)$ and $(D^1 \times X) / (\partial D^1 \times X)$ are homeomorphic: in the first one the boundary of $D^1$ is already collapsed, while in the second one you collapse at the same time the boundary of $D^1$ and the subspace corresponding to $* \times X$ (drawing a picture might be helpful here). Thus they have isomorphic cohomology, and you get the result.