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In Hatcher's Algebraic Topology textbook, we are given some tools to calculate relative homology groups, the prime example of which, as I have seen, are finding the relative homology group of the solid torus relative to the hollow torus (see here and here). I would like to generalize the result to an arbitrary topological space $X$.

More precisely, If $D^2$ denotes the $2$-dimensional disk, I would like to express the relative homology groups $H_i(X \times D^2, X \times \partial D^2)$ in terms of homology groups of $X$.

Here is my work so far:

We know that $H_i(X \times \partial D^2) \cong H_i(X) \oplus H_{i-1}(X)$ for all $i$ (by convention, $H_{-1}(X) := 0$). Thus, from the short exact sequence $0 \to \partial X \times D^2 \hookrightarrow X \times D^2 \to (X \times D^2) / (X \times \partial D^2) \to 0$, we get the following long exact sequence

$$ \cdots \to H_i(X \times \partial D^2) \cong H_i(X) \oplus H_{i-1}(X) \to H_i(X \times D^2) \cong H_i(X) \to H_i(X \times D^2, X \times \partial D^2) \xrightarrow{\partial} H_{i-1}(X \times \partial D^2) \cong H_{i-1}(X) \oplus H_{i-2}(X) \to \cdots. $$

The difficulty I am having is, since we don't have nice $H_i($something$) = 0$ as we have in some problems, it's difficult to express $H_i(X \times D^2, X \times \partial D^2)$ in terms of $H_j(X)$ in this case. As $H_i$ and $\tilde{H_i}$ doesn't differ so much (and in fact are the same for relative homology groups), considering the reduced homology groups also didn't lead to much insight either.

I also was thinking of applying the relative homological version of the Mayer-Vietoris sequence, but I ran into a similar problem of having none of the factors I desired in a nice form.

Ted Shifrin
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Cordon Smith
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1 Answers1

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When this happens, unfortunately you have to dig into at least one of the induced maps or use fancy technology like a Künneth theorem.

Consider the map $i:X\times \partial D^2\hookrightarrow X\times D^2$. In homology, $i_*:H_i(X\times \partial D^2)\rightarrow H_i(X\times D^2)$ is basically an isomorphism on the homology of $X$ and sends the cycles in $S^1$ to zero since they all become boundaries under $i$. The isomorphism $H_i(X\times S^1)\cong H_i(X)\oplus H_{i-1}(X)$ tells you that the cycles in $X\times S^1$ come from pairing the cycles in $X$ with those in $S^1$. So we can say that $\ker(i_*)\cong H_{i-1}(X)$ and $\text{im}(i_*)\cong H_i(X)$. Essentially your $i_*$ is a projection map.

Let's call the next map $q_*$. We know that $\text{ker}(q_*) = \text{im}(i_*)\cong H_{i}(X)$. Also from $H_i(X\times D^2, X\times S^1)\xrightarrow{\partial} H_{i-1}(X\times \partial D^2)\xrightarrow{i_*}H_{i-1}(X)$ we get that $\text{im}(\partial) \cong H_{i-2}(X)$. So we can break the long exact sequence into shorter pieces.

$0\rightarrow \text{coker}(i_*)\xrightarrow{q_*}H_i(X\times D^2,X\times \partial D^2)\xrightarrow{\partial}\text{im}(\partial) \rightarrow 0$.

But $\text{coker}(i_*) = H_i(X)/\text{im}(i_*) = 0$, so $\partial$ is actually an isomorphism (in this SES).

Alternatively, you can try to use the relative Kunneth Formula, which I think should say something like $$H_i(X\times D^2,X\times S^1)\cong \bigoplus_{p+q = i}H_p(X)\otimes H_q(D^2,S^1)\oplus \bigoplus_{p+q=i} \text{Tor}(H_p(X),H_q(D^2,S^1))$$

The $H_q$ term is $\mathbb{Z}$ only when $q=2$, and thus the $\text{Tor}$ term contributes nothing and we have a single term when $p = i-2$. But you should assume I'm lying and check this out more carefully for yourself.

Wyatt Kuehster
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