$X=S^2/{\sim}$ where any point on the equator is identified with its antipodal point. Compute $\pi_1(X)$ and $H_\ast(X)$ 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.
 A: 
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 $\Bbb{RP}^1$.

*

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

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


*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, $H_2(X)\cong\Bbb{Z}$


*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.

If this helps you solve the problem, please consider to click the tick to accept it because it took me some time to write it in Latex format...
