$X=S^2/{\sim}$ where any point on the equator is identified with its antipodal point. Compute $\pi_1(X)$ and $H_\ast(X)$

Question

My instructor gave me an idea. He used Van Kampen's Theorem to calculate the fundamental group, with $A=B=\mathbb{R}P^2$ and $A \cap B=S^1$ (the equator). I know $\mathbb{R}P^2$ is the quotient space of $S^2$ where each point is identified with its antipodal point. How can $A$ and $B$ can be $\mathbb{R}P^2$ if we only identify points on the equator of $S^2$ with their antipodal points?

By Van Kampen's Theorem, $\pi_1(X)$ is the pushout of $\pi_1(\mathbb{R}P^2) \leftarrow \pi_1(S^1) \rightarrow \pi_1(\mathbb{R}P^2)$ or $\pi_1(X)\cong \pi_1(\mathbb{R}P^2) \ast \pi_1(\mathbb{R}P^2)/N$. How exactly do I find the normal subgroup $N$? I know that $\pi_1(\mathbb{R}P^2)\cong \mathbb{Z}/2$ and $\pi_1(S^1) \cong \mathbb{Z}$.

I think I can use Mayer Vietoris sequence to calculate homology, with the same $A$ and $B$. If $n>1$, then $H_n(S^1)=0$ so $H_n(\mathbb{R}P^2) \oplus H_n(\mathbb{R}P^2) \cong H_n(X)$.

However, I'm stuck on calculating $H_0(X)$ and $H_1(X)$. I believe the Mayer Vietoris sequence looks like this $\rightarrow H_2(X)\rightarrow H_1(S^1) \rightarrow H_1(\mathbb{R}P^2) \oplus H_1(\mathbb{R}P^2)\rightarrow H_1(X) \rightarrow H_0(S^1) \rightarrow H_0(\mathbb{R}P^2)\oplus H_0(\mathbb{R}P^2)\rightarrow H_0(X)$. This reduces to $\rightarrow \mathbb{Z} \oplus \mathbb{Z}\rightarrow \mathbb{Z} \rightarrow \mathbb{Z}/2 \oplus \mathbb{Z}/2\rightarrow H_1(X) \rightarrow \mathbb{Z} \rightarrow \mathbb{Z} \oplus \mathbb{Z}\rightarrow H_0(X)$.

I know the sequence is exact so the kernel of the map going out of $H_1(X)$ is the image of the map coming in, but how does that help or is there another way to calculate $H_1(X)$ and $H_0(X)$? I don't know what the image or kernel of any of these maps is.

Answer 1

How can $A$ and $B$ be $\mathbb{RP}^2$ if we only identify points on the equator of $S^2$ with their antipodal points?

Let $S^2$ be the sphere as usual. Now, cut $S^2$ along its equator gives us 2 $D^2$ (disks). Since $\sim$ is generated by $x\sim -x$ for all $x$ on the equator (i.e. $x$ is equivalent to its antipodal point iff its on the equator). We finally obtain two identical quotient space $A\approx D^2/{\sim}$ and $B\approx D^2/{\sim}$ where the same equivalence relation in the restricted situation identifies $x$ with $-x$ if $x\in\partial D^2\approx S^1$. Now, you can easily prove that $A\approx B\approx \Bbb{RP}^2$.

Remark: I think you confused the definition of $\Bbb{RP}^2$ with $A,B$. Indeed, $\Bbb{RP}^2=S^2/(x\sim-x)$. But, after identifying the interior of the upper hemisphere with the interior of the lower hemisphere, you'll get $D^2/(x\sim-x)$ where $x\in\partial D^2$ which is just what your instructor mentioned in the first paragraph.

How exactly do I find the normal subgroup $N$?

First, note that $X\not\approx\Bbb{RP}^2$ but $X$ is obtained by attaching $\partial(I^2/{\sim})$ where $(x,0)\sim (1-x,1)$ and $(0,y)\sim(1,1-y)$ to another identical copy of it (i.e. identifying the boundary of two $D^2/{\sim}$ together). Here is what it's like:

Let $U$ represent the blue half which is $\Bbb{RP}^2$ and $V$ be the other half. Clearly, $\pi_1(A\cap B)\cong \pi_1(S^1)=\langle a\rangle$. By Van-Kampen's Thm, $j_1:(U,x_0)\to(X,x_0)$ and $j_2:(V,x_0)\to(X,x_0)$ induce an epimorphism $j_*:\pi_1(U)*\pi_1(V)\to\pi_1(X)$. The amalgamated relation which is $N$ must be given by $i_{1*}:\pi_1(U\cap V)\to\pi_1(U)$ and $i_{2*}:\pi_1(U\cap V)\to\pi_1(V)$ so they have the form of $i_{1*}(a)^{-1}i_{2*}(a)=1$ where $a$ is a generator of $\pi_1(U\cap V)$. To see this, You can easily show that $N\subset \ker(j_*)$ using the commutativity of the diagram of this theorem, that is splitting $U\cap V\to X$ into two branches. Then, show the injectivity of $\pi_1(U)*\pi_1(V)/N\to\pi_1(X)$ which should be included in the proof of Seifert Van-Kampen's Thm.

We get $\pi_1(X)\cong(\pi_1(\Bbb{RP}^2)*\pi_1(\Bbb{RP}^2))/(aa=1,bb=1)$, but since the generators of the two parts are the same (by the equivalence relation), we simplify the result to be $\pi_1(X)=\langle a,b\mid a=b,aa=1,bb=1\rangle=\langle a\mid aa=1\rangle\cong\Bbb{Z}/2$. Here $aa=a^2=1$ because in the picture $a^2\simeq$ (the outer boundary of the square and is a trivial loop).

Compute $H_*(X)$:

I prefer Cellular homology in this case because $X$ is just $\Bbb{RP}^1$ with 2 $e^2$ attached by gluing their boundaries with $\Bbb{RP}^1$.

1. $H_0(X)\cong\Bbb{Z}$ because $X$ is connected. $H_p(X)=0$ if $p>2$.

For other cases, consider the cellular chain complex: $$0\to\Bbb{Z}\oplus\Bbb{Z}\overset{\partial_2}{\to}\Bbb{Z}\overset{0}{\to}\Bbb{Z}\to0$$

2. For $H_2(X)$, Let's assign a counterclockwise orientation to 2-cells $\gamma_1,\gamma_2$, then we can see that $\partial_2(n\gamma_1+k\gamma_2)=0$ iff $n=k\implies Z_2(X)=\ker(\partial_2)=\Bbb{Z}$ because $\partial(\gamma_1)=2e^1$, $\partial(\gamma_2)=-2e^1$. So, $H_2(X)\cong\Bbb{Z}$

3. For $H_1(X)$, the image of the boundary map $D_1(X)\to D_0(X)$ is $0$ because the only 1-cell under the boundary map gives us $v_0-v_0=0$ which implies $Z_1(X)=\Bbb{Z}$ and from number 2, we know that $B_1(X)=im(\partial_2)=2\Bbb{Z}$ because $\partial_2(\gamma_1)$ wraps the equator two times and so does $\gamma_2$ (different directions). So, $H_1(X)=Z_1(X)/B_1(X)=\Bbb{Z}/2$.

To sum up, $$ H_p(X;\Bbb{Z})= \begin{cases} \Bbb{Z} & p=0,2\\ \Bbb{Z}/2 & p=1\\ 0 & \text{otherwise } \end{cases} $$

I think that Using MV sequence is a little bit complicated in this case, but you can check out this which is a similar situation and that person uses MV sequence which I think it's more difficult than Cellular homology.

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