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Let $X$ be the quotient space of $S^2$ under the identifications $x\sim-x$ for $x$ in the equator $S^1$. I want to compute the homology groups $H_n(X)$. I've seen this but didn't really understand.

The quotient space $X$ will look like this, isn't this space homeomorphic to the wedge of two $S^2$'s? If this is the case, then it is easy to compute the homologies; they are $0$ for $n\not=2$ and $\Bbb{Z}\bigoplus\Bbb{Z}$ for $n=2$. But this shouldn't be that easy, there is something wrong I guess.

enter image description here

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Xena
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  • Your picture is wrong. You are supposed to identify antipodal points. Your picture doesn't do that at all. The only antipodal points identified are the points in the middle of the line you've drawn.

    What don't you understand about that previous post?

     – Devin Murray Oct 13 '13 at 15:28
  • If you don't like their use of excision you could also pretty easily use a Mayer-Vietoris sequence. What tools in homology theory do you know? – Devin Murray Oct 13 '13 at 15:36
  • It is still unclear why my picture is wrong, could you draw a correct picture? I didn't understand why should we use excision – Xena Oct 13 '13 at 15:36
  • I don't have Mayer-Vietoris yet but excision... But the thing I didn't understand seems to be more fundamental, like identifying points in a space – Xena Oct 13 '13 at 15:38
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    I sadly don't have a way of putting a picture on here so I'll try and describe your problem. Let's only consider the equatorial circle. If we identify antipodal points on a circle what do we get? It turns out that we get a circle again. But it's been wrapped around itself twice. Sit with this fact for a while and you should see why your picture is wrong. The equatorial circle should be again a circle under the identification, but with the top and bottom disks wrapped around it twice. (So we have two copies of $RP^2$ glued together in an interesting way. – Devin Murray Oct 13 '13 at 15:52
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    Mayer-Vietoris gives the following: $H_0=\mathbb{Z}, H_1=\mathbb{Z}/2\mathbb{Z}, H_2=\mathbb{Z}$. –  Oct 13 '13 at 17:34
  • @DevinMurray when you say " top and bottom disks wrapped around it twice." you mean each one twice? or both of them twice? – Brain Feb 19 '22 at 15:09

1 Answers1

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Let's do this directly from the definitions of cellular homology. We'll call your space $X$.

The chain groups are:

  • $C_2(X)$, generated by the two 2-cells in $X$, the northern hemisphere $n$ and the southern hemisphere $s$; we will write $C_2(X)=\langle n\rangle\oplus\langle s\rangle$.
  • $C_1(X)$, generated by the one 1-cell in $X$, coming from the equator of $S^2$, which we will denote $e$. So $C_1(X)=\langle e\rangle$.
  • $C_0(X)$, which is generated by some fixed point on that equator.

We have a sequence $$ C_2(X)\stackrel{\phi}{\rightarrow}C_1(X)\stackrel{\psi}{\rightarrow} C_0(X)\rightarrow 0$$

Let's work out what $\phi$ and $\psi$ do:

  • Each hemisphere is wound twice around that equatorial circle, but in opposite directions. In other words, $\phi(n)=2e$ while $\phi(s)=-2e$. Thus we have $$ H_2(X)=\ker(\phi)=\langle n+s\rangle\cong\mathbb{Z}.$$
  • The equatorial circle represented by $e$ meets our point twice (it's a loop), once with degree $+1$ and once with degree $-1$. So $\psi(e)=0$; that is, $\psi$ is the zero map. Hence $$ H_1(X)=\dfrac{\ker(\psi)}{im(\phi)}=\dfrac{\langle e\rangle}{\langle 2e\rangle}\cong \mathbb{Z}/2\mathbb{Z}.$$
  • Finally, since $\psi$ is the zero map, $$ H_0(X)=C_0(X)\cong \mathbb{Z}.$$