Homology of some quotient of $S^2$ 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.

 A: 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,
$$ H_0(X)=C_0(X)\cong \mathbb{Z}.$$

