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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 $X$ : $1$ $0$-cell,$1-$ $1-cell$ and $1$ $2$-cell. Attaching the $2$-cell to the $1$-skeleton by first diving the $S^{1}$ into $3$ parts and map 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!

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2 Answers 2

up vote 12 down vote accepted

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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Shouldn't the first homology be $\mathbb{Z}\times\mathbb{Z}$? I'm thinking about this by taking a cylinder, and gluing the boundary as decribed in the question, then puncturing the result. –  user641 Dec 19 '12 at 16:38
    
When we construct the space in the way you describe, we just get a punctured torus, which would indeed have $H_1\cong\mathbb{Z}^2$, but I think the difference is explained by the outer edge of the disk having the opposite orientation of the inner circles, so that it ends up mattering which pair of circles you connect. Here is a pair of pants (original image from Wikipedia) with the edges oriented so that the "waist" is the outer edge of the disk, and the "pants cuffs" are the punctures. –  Zev Chonoles Dec 19 '12 at 16:58
    
Your construction attaches a pant cuff to the waist, and my construction attaches the pants cuffs together. However, I'm no longer sure if my construction is the one intended in the problem, or possibly if what I'm saying is making sense. What do you think? –  Zev Chonoles Dec 19 '12 at 17:00
    
You should not identify $\delta$ with $\beta$, only $\gamma$ and $\epsilon$. The result cannot be drawn as something planar without identifications. –  user17786 Dec 19 '12 at 17:01
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The punctured Klein bottle has no torsion in its homology. An easy way to see this is to compute the fundamental group, which is free on two generators, then abelianize. –  user641 Dec 19 '12 at 17:56
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$X$ as mentioned earlier is a punctured Klein Bottle, hence deformation retracts onto wedge of two circles. So $H_1(X) = \mathbb{Z} \oplus \mathbb{Z}$.

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