Relative homology group of the solid torus relative to hollow torus

algebraic-topologyhomology-cohomologysolution-verification

I try to compute the relative homology groups $H_\bullet(X,\partial X)$ where $X=D^2\times S^1$ and $\partial X=S^1\times S^1$. I have a partial answer here.

Here is my attempt.

Since $X$ is homotopy equivalent to $S^1$ (since $D^2$ is contractible) we can easily deduce that

$$H_0(X)\cong\mathbb{Z},\quad H_1(X)\cong \mathbb{Z},\quad H_i(X)=0\quad\text{for $i\geq 2$}.$$

The homology groups of $\partial X$ is also well-known;

$$H_0(\partial X)\cong\mathbb{Z},\quad H_1(\partial X)\cong\mathbb{Z}\oplus\mathbb{Z},\quad H_2(\partial X)\cong\mathbb{Z},\quad H_i(X)=0\quad\text{otherwise}.$$

From these fact I can compute the relative homology groups as follows.

  • $H_i(X,\partial X)=0$ for $i>3$. This follows from the exact sequence

    $$H_i(X)\longrightarrow H_i(X,\partial X)\longrightarrow H_{i-1}(\partial X)$$

    where $H_i(X)=H_{i-1}(\partial X)=0$.

  • $H_3(X,\partial X)\cong\mathbb{Z}$. This follows from the exact sequence

    $$H_3(X)\longrightarrow H_3(X,\partial X)\longrightarrow H_2(\partial X)\longrightarrow H_2(X)$$

    and since $H_3(X)=H_2(X)=0$, $H_2(\partial X)\cong H_3(X,\partial X)$ must hold.

  • $H_2(X,\partial X)\cong\mathbb{Z}$ since from

    $$H_2(X)\longrightarrow H_2(X,\partial X)\longrightarrow H_1(\partial X)\overset{f_\ast}{\longrightarrow} H_1(X)$$

    we have $H_2(X,\partial X)\cong\ker f_\ast$ but $f_\ast$ maps $\mathbb{Z}\oplus\mathbb{Z}$ by projection (by filling the hollow torus, the vertical loop on the surface of $S^1\times S^1$ will be killed) so $H_2(X,\partial X)\cong\mathbb{Z}$.

  • Finally $H_1(X,\partial X)=0$ since $H_0(\partial X)\longrightarrow H_0(X)$ is the identity map (injective) and $H_1(X,\partial X)\longrightarrow H_0(\partial X)\longrightarrow H_0(X)$ exact. The same reasoning applied to the right (using surjectivity) of $H_0(\partial X)\longrightarrow H_0(X)$ gives $H_0(X,\partial X)=0$.

So here is my question.

  1. Is my computation correct?

  2. By viewing the solid torus as CW complex, with one $0$-cell, two $1$-cell, two $2$-cell (one from the exterior of the torus, and the other inside the torus) and one $3$-cell (see the figure in page 3 of this paper) I suspect that the above computation can be explained by saying,

    “$X$ can be obtained by attaching one $2$-cell, one $3$-cell to $\partial X$ so the relative homology group is $\mathbb{Z}$ for $n=2,3$ and $0$ otherwise.”

    This sounds correct for me, but I cannot find the proposition supporting the reasoning. The closest one is Lemma 2.34 of Hatcher, but it is slightly different, so I wonder whether the assertion is valid or not.

Best Answer

Your computation is correct. One thing to note, which you probably understand, is that $H_1(X,\partial X)=0$ follows not just because $H_0(\partial X)\longrightarrow H_0(X)$ is an isomorphism, but because also the map coming to $H_1(X,\partial X)$ is zero by the surjectivity of $f_\ast$ that you showed above.

Homology is not as simple as counting cells, unfortunately. After all, any CW-complex can be seen as built by attaching cells to a point. However, amounts of cells don't determine the ("relative") homology groups $H_{\bullet}(X, pt) \simeq \tilde{H}_{\bullet}(X)$. For example, the homology of the projective plane, which is made from a $1$-cell and a $2$-cell, is not $\mathbb{Z}$ but $\mathbb{Z}/2$ in dimension $1$.

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