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Let X be the quotient space of $S^2$ under the identifications $x\sim-x$ for $x$ in the equator $S^1$. Compute the homology groups $H_i(X)$. Do the same for $S^3$ with antipodal points of the equator $S^2 \subset S^3$ identified.

This is probably related to cellular homology. Thanks.

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You can, indeed, do this via cellular homology (think of the CW structure with two 2-cells, one 1-cell, and one zero-cell --- this is what you get by crushing the standard CW structure on $S^2$ with two cells in each dimension) and then try to figure out the cellular boundary maps. –  Akhil Mathew Jan 10 '11 at 14:58
Hi Akhil. I did this and got $H_0(X)=Z$ and $H_i(X)=0$ for $i>0$. Does this make sense? Thanks! –  shay Jan 10 '11 at 15:42
I don't think that's right -- what CW decomposition are you using? –  Aaron Mazel-Gee Jan 10 '11 at 17:21
As Akhil said, I used one 0-cell, one 1-cell attached to the 0-cell in the obvious way and two 2-cells both attached to the previous with the antipodal map. So we get a sequence $0 \to^{d_3} Z^2 \to^{d_2} Z \to^{d_1} Z \to 0$ with $d_1 = 0$ and $d_2$ taking each generator of $Z^2$ to the generator of $Z$. That leaves us with $Ker d_1 = Z$, $Im d_2 = Z$, meaning $H_1=0$. Also, $Ker d_2 = 0$ and so $H_2 = 0$ –  shay Jan 10 '11 at 17:28

1 Answer 1

up vote 3 down vote accepted

Let $X=S^2/\mathord\sim$ and,letting $S^2_+\subseteq S^2$ be the upper closed hemisphere in the sphere, let $X_+=S^2_+/\mathord\sim$ be the quotient of $S^2_+$ by the restricted equivalence relation. Now consider the long exact sequence in reduced homology for the pair $(X,X_+)$.

Using excision &c, show that the relative homology of $(X,X_+)$ is the same as that of the result of collapsing $X_+$ to a point, so that you get a $2$-sphere. On the other hand, $X_+$ is a projective plane, so you also know its homology. Now use the long exact sequence.

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Hello Mariano! First of all, thanks for the comment. Following your advice, we get that $H_n(X,X_+) = \tilde{H}_n(X/X_+) = \tilde{H}_n(S^2)$ and then from the long exact sequence of $(X,X_+)$ I got that $H_0(X) = \mathbb{Z}$, $H_i(X)=0$ for i=1, i>2 and $H_2(X) = \mathbb{Z}^2$. Is this right? Thank you very much! –  shay Jan 11 '11 at 16:25
Oops, that should be $H_2(X) = \mathbb{Z} \times \mathbb{Z}_2$ there. –  shay Jan 11 '11 at 22:39
Is the answer Shay gave with $H_2(X)=\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$ correct? When I apply the argument given by Mariano, it seems to me the non-zero portion of the long exact sequence you get is $0\to H_2(X)\to\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to H_1(X)\to 0$. Now, I haven't worked out what the connecting map $\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ is, but it's either the quotient map (and then $H_2(X)=2\mathbb{Z}\cong\mathbb{Z}$ and$H_1(X)=0$), or the zero map (and then $H_2(X)=\mathbb{Z}$ and $H_1( –  user14780 Aug 19 '11 at 4:26
Could someone explain what the connecting map $\mathbb Z\rightarrow \mathbb Z/2\mathbb Z$ is? I am having trouble finding out what the non-trivial element in $H_1(\mathbb{RP^2})$ is. Thanks in advance. –  awllower Jun 23 '14 at 9:01

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