Rigorous application of Mayer-Vietoris to a quotient of $S^{2}\times I$ As part of exercise 3.3.24 in Hatcher's Algebraic Topology, I computed the homology groups of the space resulting from $S^{2}\times I$ by identifying $S^{2}\times\{0\}$ and $S^{2}\times\{1\}$ via a reflection. I could intuitively see the result, but I need help making my argument more rigorous.
I used Mayer-Vietoris to do this. I defined $A$ to be the image of $S^{2}\times(1/4, 3/4)$ and $B$ the image of $S^{2}\times(I-\{1/2\})$ in the quotient space. It's easy to see that $A$ and $B$ deformation retract onto $S^2$, $A\cap B$ deformation retracts onto $S^{2}\sqcup S^{2}$. The cases $n\not=2$ are straightforward. $n=2$ gives
$$H_{2}(A\cap B)\overset{\phi}{\rightarrow}H_2(A)\oplus H_2(B)\rightarrow H_2(X)\rightarrow 0$$
This reduces to
$$\mathbb{Z}\oplus\mathbb{Z}\overset{\phi}{\rightarrow}\mathbb{Z}\oplus\mathbb{Z}\rightarrow H_2(X)\rightarrow 0$$
Call the generators of the first group $x,y$ and those of the second group $a,b$. Intuitively, the "twist" resulting from the reflection changes the sign of one of the generators but not the other. If I put $\phi(x)=a+b,\ \phi(y)=a-b$ I get $H_2(X)=\mathbb{Z}_2$. 
This all makes sense. Many of the lecture notes and MSE answers I read online use MVS this way.
 However I don't feel comfortable relying on intuition, especially that I'm still developing it. How can I apply the same logic more rigorously? How can I demonstrate the effect of the "twist" mentioned above in a rigorous manner? 
Thanks
 A: Let $S^2$ be the unit sphere in $\mathbb{R}^3$. Let $X=S^2\times [0,1]/\sim$ where $\sim$ is the smallest equivalence relation containing $(x,0)\sim (r(x),1)$ where $r:S^2\rightarrow S^2$ is reflection in the $xy$-plane. Let $q:S^2\times [0,1]\rightarrow X$ be the quotient map.  Let $U=X-q(S^2\times \{1\})$, and let $V=X-q(S^2\times \{1/2\})$.  Using the product structure  as a guide we can deformation retract $U$ down onto $A=q(S^2\times \{1/2\})$, and $V$ down onto $B=q(S^2\times \{1/2\})$.  Similarly we can deformation retract $U\cap V$ down onto $C\cup D=q(S^2\times \{1/4\})\cup q(S^2\times \{3/4\})$.   Notice that in these two deformations, $A$ is carried to $C$ and $D$ by the identity map, using the natural identifications with $S^2$. However in the deformation of $A$ to $C$ you are passing through the identified end, so the restriction of the deformation, as a map from $B$ to $C$ is $r$, if you use the natural identifications of $B$ and $C$ from $S^2$. However the identification of $B$ with $D$ doesn't pass through the end so the defomation is as the identity.
Now you need to be careful naming the maps in your sequence, and you will get the answer you claimed.
A: To work this all out, you need to know what the maps in the MVS are. 
If $i:A\cap B \hookrightarrow A$ and $j: A\cap B \hookrightarrow B$ are the
inclusions, then the map
$H_*(A\cap B) \to H_*(A) \oplus H_*(B)$ is $i_* \oplus j_*.$
If $i': A\hookrightarrow X$ and $j': B\hookrightarrow X$ are the inclusions,
then the map
$H_*(A) \oplus H_*(B) \to H_*(X)$ is $i'_* - j'_*.$ (Actually, the signs on $j_*$ and $j'_*$ could be switched; I'm not sure what the standard convention is.)
Now if fix a generator of $H^2(A) = H^2(S^2)$, then we can choose generators 
of $H^2(A\cap B)$ so that the map $H^2(A \cap B) \to H^2(A)$ is given by $(a,b) \mapsto a + b$.  What we are doing here is identifying the two copies of $S^2$ in $A \cap B$ with the one one copy of $S^2$ in $A$ by homotoping them along (some subarc of) the arc $(1/4,3/4)$.  
If instead we were to identify the two copies of $S^2$ in $A \cap B$ with the one copy of $S^2$ in $B$ BY homotoping them along $(I \setminus \dfrac{1}{2} )/ 0 \sim 1,$ we would get the opposite identification (because when we pass from $0$ to $1$ we apply the reflection, which changes the sign of the generator
of $H^2(S^2)$).  
So, given that we fixed our basis so that the map $i_*$ is $(a,b) \mapsto a+b,$
the map $j_*$ is either $(a,b) \mapsto a -b$ or $(a,b) \mapsto -a+b$.  (If we interchange the two generators in $H^2(A\cap B)$, we can change the map from one
of these to the other, so there isn't any way to privilege one over the other.)
This gives what you want.
