Cellular Homology of the quotient of $S^1\times D^2$ by its boundary

Question

I was trying to figure out the homology of the space $(S^1\times D^2)/( S^1\times S^1)$ using a cell complex decomposition. First of all, we know that:

• $S^1$ has one $0$-cell, one $1$-cell and $D^2$ has one $0$-cell, one $1$-cell and one $2$-cell.

So,

• $S^1\times D^2$ has one $0$-cell, two $1$-cells, two $2$- cells, one $3$-cell, and
• $S^1\times S^1$ has one $0$-cell, two $1$-cells and one $2$-cell.

So by taking the quotient, $(S^1\times D^2)/(S^1\times S^1)$ we are left with one $0$-cell, no $1$-cells, one $2$-cell and one $3$-cell.

So we have a sequence, $\mathbb Z\stackrel{d}{\to}\mathbb{Z}\to 0\to \mathbb{Z}$. How can I compute $d$? I feel like it should be $0$ but I cant prove it.

Answer 1

Short answer: in the cell decomposition of $S^1\times D^2$, the boundary of the $3$-cell is exactly made of the $2$-cell that you are going to quotient by. Therefore in the quotient CW structure, the boundary of the $3$-cell is $0$ as you expected.

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Long answer:

Concretely how do you build the quotient using these cells:

• First, you have a point ($0$-cell), which you can represent as a the (equivalence class of the) torus $S^1\times S^1$ that you quotiented by and which is therefore now a $0$-cell.
• Then, you add a $2$-cell $\{\text{pt}\}\times D^2$ for some point $\text{pt}\in S^1$: the boundary of this $2$-cell is entirely mapped to the $0$-cell as it should.
• Finally, fill in the solid torus with a $3$-cell: all the boundary of your $3$-ball goes to the $0$-cell (the boundary of the original solid torus), except for the $2$-cell which is covered twice, with different orientations.

This final sentence is the reason why your map is $0$ as you expected.