2
$\begingroup$

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 $H_2(X) \approx \ker f \oplus \Bbb{Z}\approx \Bbb{Z}^2$.

$\endgroup$
  • $\begingroup$ What are $P^i$? $\endgroup$ – Rasmus Jun 27 '15 at 14:12
  • $\begingroup$ $P^i= \Bbb{R}P^i$ $\endgroup$ – iwriteonbananas Jun 27 '15 at 14:14
1
$\begingroup$

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 $H_2(X) \cong \ker d_2/\text{im}\, d_3 \cong \Bbb Z$

$\endgroup$
  • $\begingroup$ Nice, thanks for the alternative methods $\endgroup$ – iwriteonbananas Jun 27 '15 at 17:42
  • $\begingroup$ @iwriteonbananas Glad to help. $\endgroup$ – Balarka Sen Jun 28 '15 at 12:45
  • 1
    $\begingroup$ 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 $H_{2}(X) \approx \mathbb{Z}$. $\endgroup$ – TuoTuo Aug 5 '18 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.