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Let $X$ be space obtained by first removing the the interior of two disjoint closed disks from the unit closed disk in $\mathbb R^{2}$ and then identifying their boundaries clockwise. Compute the homology of this space.

My idea is to do this using cellular homology: We can have cell complex structure on $X$: one $0$-cell, one $1$-cell and one $2$-cell. Attaching the $2$-cell to the $1$-skeleton by first diving the $S^{1}$ into $3$ parts, then mapping these parts to the $1$-skeleton in the same direction.

Thus the cellular boundary map $d_2$ will be multiplication by $3$ and we have the homology groups $H_{0}(X)=\mathbb Z$ and $H_{1}(X)=\mathbb Z_{3}$ and $H_{i}(X)=0$, otherwise.

Please check the calculations and share some ideas for such questions. Thanks in advance!

2 Answers 2

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Many thanks to Steve D, user17786, and Dave Hartman for their helpful corrections.

First, I put a cell structure on the twice-punctured disk with 3 0-cells, 5 1-cells, and 1 2-cell: enter image description here

Note that the boundary of the 2-cell $D$ is $d_2D=\alpha+\beta+\gamma-\beta+\delta+\epsilon-\delta=\alpha+\gamma+\epsilon,$ and that the boundaries of the 1-cells are $\begin{align} d_1\alpha&=0\\ d_1\beta&=y-x\\ d_1\gamma&=0\\ d_1\delta&=z-x\\ d_1\epsilon&=0 \end{align} $ Now, we identify $y$ with $z$, and $\gamma$ with $\epsilon$, to produce a cell structure on $X$:

enter image description here

For $X$, the chain groups are $\begin{align} C_0(X)&=\langle x,y\rangle\\ C_1(X)&=\langle \alpha,\beta,\gamma,\delta\rangle\\ C_2(X)&=\langle D\rangle \end{align}$ where $D$ is our 2-cell, and we have $\begin{align} d_1\alpha&=0\\ d_1\beta&=y-x\\ d_1\gamma&=0\\ d_1\delta&=y-x \end{align} $

$d_2D=\alpha+\beta+\gamma-\beta+\delta+\gamma-\delta=\alpha+2\gamma.$

Thus, $H_0(X)=\ker(d_0)/\mathrm{im}(d_1)=\langle x,y\rangle/\langle y-x\rangle=\left\langle\overline{x}\right\rangle\cong\mathbb{Z}$ $H_1(X)=\ker(d_1)/\mathrm{im}(d_2)=\langle \alpha,\gamma,\beta-\delta\rangle/\langle \alpha+2\gamma\rangle=\left\langle\overline{\gamma},\overline{\beta-\delta}\right\rangle\cong\mathbb{Z}^2$ $H_2(X)=\ker(d_2)/\mathrm{im}(d_3)=0/0\cong 0.$

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    @ZevChonoles I was stuck on this problem too while preparing for my AT exam last semester. My lecturer then gave me the hint of adding the two lines to connect the two circles to the boundary and then I solved soon after that.2012-12-20
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$X$ as mentioned earlier is a punctured Klein Bottle, hence deformation retracts onto wedge of two circles. So $H_1(X) = \mathbb{Z} \oplus \mathbb{Z}$.