Homology of quotient of two copies of $\mathbb S^2$. I have been trying to solve the following exercise. My knowledge of homology reduces to Mayer-Vietorius and relative homology, so I would like to know if the problem can be solved using these tools.   

First of all to compute the homology groups of $X$, I used the Mayer-Vietorius long exact sequence. For that I considered the following sets $X_1 \simeq S^2, X_2 \simeq S^2$ enlarged versions of $A$ and $B$ so that their intersection is a set that can be deformated into the equator, that is $X_1 \cap X_2 \simeq S^1$. Computing the long exact sequence, the interesting part is the following one 
$$\cdots\rightarrow \underset{0} {\underbrace{H_{3}\big( X)}}  \rightarrow \underset{0} {\underbrace{H_{2}\big( \mathbb S^1)}} {\rightarrow } \underset{ \mathbb Z \oplus \mathbb Z} {\underbrace{H_{2}\big( \mathbb S^2) \oplus  {H_{2}\big( \mathbb S^2)}}}  \rightarrow H_2\big(X\big) \rightarrow \underset{\mathbb Z} {\underbrace{H_{1}\big( \mathbb S^1)}} {\rightarrow} 0 \cdots
$$
As $\mathbb Z$ is free this means that it splits so that $H_2(X) = H_1(\mathbb S^1) \oplus H_2(\mathbb S^2) \oplus H_2 (\mathbb S^2)$. So the homology groups of $X$ are the following.
$$H_n(X) = \left\{ \begin{array}{cl} \mathbb{Z} & \text{if} \ k=0 \\
\mathbb{Z^3} & \text{if} \ k=2 \\
 0 & \text{otherwise} \end{array}\right.$$
For the second part of the exercise I'm having a little bit more trouble. I tried using the relative homology exact sequence but it's not very inspiring. I couldn't quite understand what the inclusion $i_{*}:H_2(A) \to H_2(X)$ does. So I tried to use that the pair $(X,A)$ is such that $H_n(X,A) \simeq H_n(X/A)$ (I have to check that it is a strong deformation retract). This seems like a much better idea but I could not grasp what the space $X/A$ is. I'm thinking that it is $\mathbb S^2 \vee \mathbb S^2 $ as all the sphere $A$ is identified to the equator and then the equator is identified into a point. In that case it's the same as $\mathbb S^2$ with the equator identified which seems to be $\mathbb S^2 \vee \mathbb S^2 $.
Thanks in advance!
 A: Your Mayer-Vietoris computation for $H_n(X)$ is correct. Alternatively, one can deduce this using the following chain of homotopy equivalences:

The first space is your space $X$: two spheres glued along their equators, which we are drawing with one sitting inside the other. The second space is a sphere glued to a cylinder along one of its equators. The third one is a cylinder with two middle caps and the last one is a wedge sum of three spheres. It is clear that the last space has the homology groups you described.
For the second half, let us write down the relevant bits of the long exact sequence. One of them is
$0\to H_1(X,A)\to H_0(A)\to H_0(X)\to 0,$
since you know that $H_1(X)=0$ and $H_0(X,A)=0$. Now $H_0(A)\simeq H_0(X) \simeq \Bbb{Z}$, and the map between them is surjective and thus an isomorphism, so $H_1(X,A)=0$.
The other bit is
$0 \to H_3(X,A)\to H_2(A)\to H_2(X)\to H_2(X,A)\to 0$
since you know $H_3(X)\simeq H_1(A)=0$, and you also know that $H_2(A)\simeq \Bbb{Z}$ and $H_2(X)\simeq \Bbb{Z}^3$. One may describe the map $H_2(A)\to H_2(X)$ explicitly, since it is the one induced by the inclusion. The map $\sigma$ which radially expands a standard $2$-simplex to a $2$-sphere is a generator for $H_2(A)$. Similarly, if we suppose $A$ sits inside $X$ as in the drawing above, we may pick generators $x_1,x_2,x_3$ of $H_2(X)$ as in the following picture:

In this way, identifying $H_2(X)\simeq \Bbb{Z}^3$ by the map $x_i \mapsto e_i$ and $H_2(A)\simeq \Bbb{Z}$ by $\sigma\mapsto 1$, we see that the map $H_2(A)\to H_2(X)$ induces the map $\Bbb{Z}\to \Bbb{Z}^3$ given by $1\mapsto (0,1,0)$. Since this map is injective, $H_3(X,A)=0$, and moreover $H_1(X,A)\simeq \Bbb{Z}^3/\langle (0,1,0)\rangle\simeq \Bbb{Z}^2$.
Your argument using that $H_n(X,A)\simeq \tilde{H}_n(X/A)$ (notice it is the reduced homology groups) is correct as well, and you may check that the homology groups computed above agree with this. To see that $X/A$ is homeomorphic to a wedge of spheres, note that the map $X\to S^2\vee S^2$ taking the upper hemisphere of $B$ to all of one the spheres minus a single point, the lower hemisphere to the other sphere in a similar fashion and all of $A$ to the gluing point induces a continuous bijection $X/A\to S^2\vee S^2$. Since $X/A$ is compact and $S^2\vee S^2$ is Hausdorff, this is a homeomorphism.
