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I have been given the following result without proof, so I would like to show it is true:

Let $I=[0,1]$, then: $$H^\bullet(I,\partial I;R)\cong H^\bullet(I/\partial I,*;R)\cong H^\bullet(S^1,*;R)=R\cdot x$$ where $x$ is a generator of degree $1$. So let $H^\bullet(I,\partial I;R)=R\cdot\mu$.

Claim: Let $Y$ be any topological space and $n\ge0$. Then the map: $$H^n(Y;R)\to H^{n+1}(Y\times I,Y\times\partial I;R)$$ given by $$a\mapsto a\times\mu$$ is an isomorphism of $R$-modules.

Now here's my idea about how to proceed. I would really much like some opinions about whether it can be correct or not.

Proof: A version of the Künneth formula says that the following short sequence is exact: $$0 \to (H^\bullet(Y;R)\otimes_R H^\bullet(I,\partial I;R))^{n+1}\to H^{n+1}(Y,Y\times\partial I; R)\to \big(\operatorname{Tor}_R(H^\bullet(Y;R),H^\bullet(I,\partial I;R))\big)^{n+1}\to 0$$ where the map on the left is the cross product (see here). Since $H^\bullet(I,\partial I;R)$ is free, the term on the right should vanish, and since it is $0$ at every degree except $1$, we have the isomorphism: $$H^n(Y;R)\otimes_RH^1(I,\partial I;R)\stackrel{\cong}{\longrightarrow}H^{n+1}(Y,Y\times\partial I; R)$$ $$a\otimes_R\mu\longmapsto a\times\mu$$ Moreover, since $H^1(I,\partial I;R)=R\cdot\mu$ we have $H^n(Y;R)\otimes_RH^1(I,\partial I;R)\cong H^n(Y;R)$ via $\sum_{k=1}^m(r_ka_k\otimes_R\mu)=\left(\sum_{k=1}^mr_ka_k\right)\otimes_R\mu\mapsto\sum_{k=1}^mr_ka_k$, and thus our original map is an isomorphism.

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Daniel Robert-Nicoud
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1 Answers1

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This seems right. Conjugating by your isomorphism $\tilde H^\bullet(S^1) \cong H^\bullet(I,\partial I)$, and $\tilde H^\bullet(Y \times S^1) \cong H^\bullet(Y \times I,Y \times \partial I)$ it's the same to demand that if $\sigma \in H^1(S^1)$ is the fundamental class, then $$H^n(Y) \overset{\mathrm{id} \otimes \sigma}\longrightarrow H^n(Y) \otimes H^1(S^1) \overset\times\to H^{n+1}(Y \times S^1)$$

is an isomorphism, which also follows from Künneth.

jdc
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