# Homology of $S^2/x\sim -x$ for $x$ on the equator

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 $H_i(X)$.

I wrote my solution/attempt below and I would like to know if it it correct and, more importantly, if there is a simpler solution. I tried using cellular homology but I'm not very experienced with it and kept getting nonsense.

Here is my attempt:

$X$ can be written as the following pushout: \begin{align*} \require{AMScd} \begin{CD} P^1 @>{i}>> P^2\\ @V{i}VV @VVV \\ P^2 @>>> X \end{CD} \end{align*} where $i:P^1\hookrightarrow P^2$ is the inclusion. So by a Mayer Vietoris argument, we have $H_i(X,P^2) \approx H_i(P^2, P^1)$. This can be simplified to $\tilde{H}_i(P^2/P^1) \approx \tilde{H}_i(S^2)$. Now we consider the LES of the pair $(X, P^2)$. $$0\to H_2(P^2)\to H_2(X)\to H_2(X, P^2)\to H_1(P^2)\to H_1(X) \to 0$$ We know all homology groups involved except those of $X$, so the LES reads: $$0\to \Bbb{Z}\to H_2(X) \to \Bbb{Z}\to \Bbb{Z}_2\to H_1(X)\to 0$$

The problem lies in identifying the map $\Bbb{Z}\to \Bbb{Z}_2$, call it $f$. The LES tells us that $H_1(X)\approx \Bbb{Z}_2/\text{im }f$. On the other hand, by the Van Kampen theorem for attaching 2-cells, $X$ can be viewed as $P^2$ with a 2-disk attached along the path $aa$, hence $\pi_1(X)\approx \Bbb{Z}_2$. This is abelian, so it coincides with $H_1(X)$. Then $f$ must be the $0$ map. Using that, from the LES we can also compute $H_2(X) \approx \ker f \oplus \Bbb{Z}\approx \Bbb{Z}^2$.

• What are $P^i$? – Rasmus Jun 27 '15 at 14:12
• $P^i= \Bbb{R}P^i$ – iwriteonbananas Jun 27 '15 at 14:14

Your method is almost correct : you seem to have assumed somewhere that $H_2(\Bbb RP^2) = \Bbb Z$, which is wrong. $H_2(\Bbb RP^2)$ is actually trivial. That said, $H_2(X)$ should be $\Bbb Z$ instead of $\Bbb Z^2$.

Here's another way of doing this : $\{U, V\}$ forms an open cover of $X$ where $U$ is the image of the open set consisting of a $\epsilon$-thickened north hemisphere of $S^2$ by the quotient map and $V$ is the same thing done with the south hemisphere.

Using Mayer-Vietoris on this open cover, we have the exact sequence

$$0 \to H_2(U \cap V) \to H_2(U) \oplus H_2(V) \to H_2(X) \to H_1(U \cap V) \to 0$$

$U \cap V$ is just $\Bbb RP^1 \cong S^1$, so $H_2(U \cap V) = 0$ and $H_1(U \cap V) = \Bbb Z$. $U$ and $V$ are both homeomorphic to $\Bbb RP^2$, so $H_2(U) = H_2(V) = 0$. Thus, the above sequence reduces to the short exact sequence

$$0 \to H_2(X) \to \Bbb Z \to 0$$

This implies $H_2(X) = \Bbb Z$. Using van Kampen, $H_1(X) = \pi_1^{ab}(X) = \Bbb Z/2\Bbb Z$, so you're done.

Yet another way to do it would be cellular homology : $X$ has the cell structure given by a $0$-cell, a $1$-cell and two $2$-cells attached by degree $2$ maps. Thus, the cellular chain complex is

$$0 \stackrel{d_3}{\to} \Bbb Z^2 \stackrel{d_2}{\to} \Bbb Z \stackrel{d_1}{\to} \Bbb Z \to 0$$

$d_1 = 0$ as there is only a single $0$-cell. By the cellular boundary formula, $d_2(e_1^2) = 2e^1$ and $d_2(e_2^2) = -2e^1$ where $e^n_i$ is the $i$-th $n$-dimensional cell. Note that $\text{ker} \, d_2$ is the infinite cyclic group generated by $e_1^2 + e_2^2$

Thus, $H_1(X) \cong \ker d_1/\text{im} \, d_2 \cong \Bbb Z/2\Bbb Z$ and $H_2(X) \cong \ker d_2/\text{im}\, d_3 \cong \Bbb Z$

• Nice, thanks for the alternative methods – iwriteonbananas Jun 27 '15 at 17:42
• @iwriteonbananas Glad to help. – Balarka Sen Jun 28 '15 at 12:45
• There is an error in your first solution using Mayer-Vietoris, namely $H_1(U) \oplus H_1(V)= \mathbb{Z}_2 \oplus \mathbb{Z}_2$ rather than $0$. So the sequence actually reduces to $0 \rightarrow H_2(X) \xrightarrow[]{\alpha} \mathbb{Z} \xrightarrow[]{\beta} \mathbb{Z}_{2} \oplus \mathbb{Z}_{2}$. Howver, there is a simple algebraic fix. By exactness, $\alpha$ is injective so $H_2(X) \approx \mathbb{Z}$ or $H_{2}(X)=0$ (since all subgroups of $\mathbb{Z}$ are infinite cyclic or trivial). But $\beta$ cannot be injective, so ker$\beta =$ im$\alpha$ is not trivial, so $H_{2}(X) \approx \mathbb{Z}$. – TuoTuo Aug 5 '18 at 1:15
• Sorry for late comment, but why wouldn't $H_1(X) = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$? – David Oct 22 '19 at 3:39
• Even later comment: why is the abelianisation of $\pi_1$ given as $\mathbb{Z}/2\mathbb{Z}$? Just saying 'by van Kampen' seems a very big leap – bounceback May 8 at 20:33