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

  • 0
    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.2012-12-19
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    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](http://i.stack.imgur.com/G7pZu.png) ([original image from Wikipedia](http://en.wikipedia.org/wiki/Pair_of_pants)) with the edges oriented so that the "waist" is the outer edge of the disk, and the "pants cuffs" are the punctures.2012-12-19
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    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?2012-12-19
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    You should not identify $\delta$ with $\beta$, only $\gamma$ and $\epsilon$. The result cannot be drawn as something planar without identifications.2012-12-19
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    @user17786: I thought it wouldn't be a problem, since after all couldn't we have (from the start) just not used either $\beta$ or $\delta$, and taken our punctures so that $x=y=z$?2012-12-19
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    @zev: I don't think you can do that, because $y$ and $z$ lie in disjoint circles, so they cannot be equal from the beginning.2012-12-19
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    @user17786: But the problem just says that the open disks we're removing have to be disjoint, not their boundaries.2012-12-19
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    @zev: I think the good identification picture would be the as first one but writing $\gamma$ instead of $\epsilon$ and $y$ instead of $z$.2012-12-19
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    @zev: I've assumed that the adherence of each of the balls is disjoint from the adherence of the other (it is true that that is not assumed by the OP). Nevertheless, if you don't make that assumption, you may as well consider the case when the adherence of the balls meet in more that one point. That gives infinitely many cases with different homology.2012-12-19
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6790/discussion-between-user17786-and-zev-chonoles)2012-12-19
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    @ZevChonoles You possibly misunderstood the comment by Steve D. I think about the space in question exactly as Steve D does namely as a punctured Klein-bottle. I computed the homology with this picture in mind (by the same method you used) and my result was $\mathbb{Z}\times \mathbb{Z}$ for the first homology. (Of course it's possible that I made a mistake at some point in my calculation...)2012-12-19
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    @user17786: I've done the calculation as you suggested, and I agree now that the first homology should be $\mathbb{Z}^2$. Does my answer look right now?2012-12-19
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    @Dave: Thanks for your comment - user17786's suggestion of how to draw the diagram was what made me realize my error. And indeed, a punctured Klein bottle and punctured torus should both be homotopy equivalent to $\mathbb{S}^1\vee\mathbb{S}^1$, so this homology makes sense.2012-12-19
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    @ZevChonoles: I think so.2012-12-19
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    @zevchonoles: The puctured klein bottle has $H_1 = Z^2 + Z/2$, and corresponds to make the opposite identification ($\gamma$ should be oppositely oriented in one of the circles)2012-12-19
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    @user17786 : If $\gamma$ was oppositely oriented in one of the circles we would end up with a punctured torus.2012-12-19
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    @dave: You're totally right. The mistake is at the end, when it says $\langle \alpha,\gamma,\beta-\delta\rangle/\langle \alpha + 2\gamma\rangle \cong Z^2$. It should say $\cong Z + Z + Z/2$.2012-12-19
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    Thanks to you guys again for catching my mistake!2012-12-19
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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.2012-12-19
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    @ZevChonoles: Your computation of the kernel and image are correct, I think you are mistakenly adding torsion when there isn't any. The relation $\alpha+2\gamma=0$ simply says $\alpha=2\gamma$, which just means you can remove the generator $\alpha$, given a group (freely) generated by $\gamma$ and $\beta-\delta$.2012-12-19
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    @SteveD: You're right. the quotient given by Zev at the beginning was right. I'm sorry about this.2012-12-19
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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 $H_1(X) = \mathbb{Z} \oplus \mathbb{Z}$.