Homology of some quotient of $S^2$

Solution 1:

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