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...
Reducing the nlab-theorem to tom Dieck's theorem breaks down when one tries to show that the interiors of $\tilde X_0, \tilde X_1$ cover $\tilde X$. At least there is no simple proof - but nevertheless it could be true. Anyway, we do not need it. In fact, tom Dieck's theorem relies on two ingredients:
Theorem (2.6.1) which states a pushout property for fundamental groupoids under the assumption that $X_0$ and $X_1$ are subspaces of $X$ such that the
interiors cover $X$.
The existence of a retraction functor $r : \Pi(Z) \to \Pi(Z,z)$ which tom Dieck only defines for path connected $Z$. This works as follows: For each object $x$ of $\Pi(Z)$ (i.e. each point $x \in Z$) we define $r(x) = z$. For the morphisms we proceed as follows: We choose any morphism $u_x : x \to z$ if $x \ne z$ and take $u_z = id_z$= path homotopy class of the constant path at $z$. Given a morphism $\alpha : x \to y$ in $\Pi(Z)$, we define $r(\alpha) = u_y \alpha u_x^{-1}$.
We shall see that 2. can be generalized so that we can prove
Theorem (Seifert - van Kampen). Let $X$ be a topological space and $X_0,X_1\subset X$ be subsets whose interiors cover $X$ such that $X_{01}=X_0\cap X_1$ is path connected. Then for any choice of base point $\ast\in X_{01}$
\begin{matrix}\pi_1(X_{01},\ast) & \to & \pi_1(X_0,\ast)\\
\downarrow&&\downarrow \\
\pi_1(X_1,\ast) & \to & \pi_1(X,\ast)
\end{matrix}
is a pushout in the category of groups.
Proof. As Tyrone suggested in his comment, for a pointed space $(Z,z)$ let us denote by $\tilde Z$ the path-component of $Z$ containing the basepoint $z$. From Jackson's and your answers we know that for $X_{01}$ path connected and $* \in X_{01}$ we have $\tilde X_{01} := \tilde X_0 \cap \tilde X_1 = X_{01}$ and $\tilde X = \tilde X_0 \cup \tilde X_1$. Note that $X_{01}$ path connected is essential for both equations.
We apply tom Dieck's retraction contruction to $Z = \tilde X$ and $z = * \in X_{01} = \tilde X_{01}$ by first chosing $u_x$ in $\Pi(X_{01})$ for all $x \in X_{01}$ (where of course $u_* = id_*$), then $u_x$ in $\Pi(\tilde X_0)$ for all $x \in \tilde X_0 \setminus X_{01}$ and finally $u_x$ in $\Pi(\tilde X_1)$ for all $x \in \tilde X_1 \setminus X_{01}$. Since $\Pi(X_{01}),\Pi(\tilde X_0), \Pi(\tilde X_1)$ are subcategories of $\Pi(\tilde X)$, this gives us a choice of $u_x$ in $\Pi(\tilde X)$ for all $x \in \tilde X$ providing a retraction $\tilde r : \Pi(\tilde X) \to \Pi(\tilde X,*) = \Pi(X,*)$. We extend it to a retraction $r : \Pi(X) \to \Pi(X,*)$ as follows: Given a morphisms $\alpha : x \to y$ in $\Pi(x)$, then $x,y$ belong to same path component $P$ of $X$. If $P = \tilde X$, we define $r(\alpha) = \tilde r(\alpha)$. If $P \ne \tilde X$, we define $r(\alpha) =id_*$. Consider the restriction $r_{01}: \Pi(X_{01}) \to \Pi(X,*)$. The category $\Pi(X_{01})$ is a subcategory of $\Pi(\tilde X,*)$ and by construction $r_{01}(\Pi(X_{01})) = \tilde r(\Pi(X_{01})) \subset \Pi(X_{01},*)$, i.e. we may regard $r_{01}$ as a map $r_{01} : \Pi(X_{01}) \to \Pi(X_{01},*)$. Next consider the restriction $r_0 : \Pi(X_0) \to \Pi(X,*)$. For the subcategory $\Pi(\tilde X_0) \subset \Pi(X_0)$ we have by construction $r_0(\Pi(\tilde X_0)) = \tilde r(\Pi(\tilde X_0)) \subset \Pi(X_0,*)$. Let $\tilde P_0$ be a path component of $X_0$ different from $\tilde X_0$, i.e. $\tilde P_0 \cap \tilde X_0 = \emptyset$. Then $\tilde P_0 \cap \tilde X = \tilde P_0 \cap \tilde X_1 = \tilde P_0 \cap X_0 \cap \tilde X_1 \subset \tilde P_0 \cap X_0 \cap X_1 = \tilde P_0 \cap \tilde X_0 \cap \tilde X_1 \subset \tilde P_0 \cap \tilde X_0 = \emptyset$. Thus $\tilde P_0 \cap \tilde X = \emptyset$ and therefore by construction $r_0(\Pi(\tilde P_0)) = r(\Pi(\tilde P_0) = \{id_*\} \subset \Pi(X_0,*)$. We conclude $r_0(\Pi(X_0)) \subset \Pi(X_0,*)$, i.e. we may regard $r_0$ as a map $r_0 : \Pi(X_0) \to \Pi(X_0,*)$. Similarly $r$ restricts to $r_1 : \Pi(X_1) \to \Pi(X_1,*)$. Therefore we get a commutative diagram
\begin{matrix}\Pi(X_0) & \hookleftarrow & \Pi_1(X_{01}) & \hookrightarrow & \Pi(X_1)\\
\downarrow r_0 && \downarrow r_{01} &&\downarrow r_1 \\
\Pi(X_0,*) & \hookleftarrow & \Pi(X_{01},*) & \hookrightarrow & \Pi(X_1,*)
\end{matrix}
Now the same argument as in the proof of tom Dieck's Theorem (2.6.2) applies.
Best Answer
I think you are slightly misreading the statement: there is no assertion of injectivity of the homomorphism $$\widetilde H_1(A \cap B) \xrightarrow{\Phi} \widetilde H_1(A) \oplus \widetilde H_1(B) $$ What he writes is that $\widetilde H_1(X)$ is the quotient of $\widetilde H_1(A) \oplus \widetilde H_1(B)$ by the image of $\Phi$, which is true by exactness if you had left off the $0 \to$ at the beginning of your exact sequence.