Homology of $\Sigma_{2}\times S^{1}$?

I'm quite at a loss with this...I want to use Mayer-Vietoris with open covers $A=\Sigma_{2}\times (S^{1}\setminus \{p\})$ and $B=\Sigma_{2}\times (S^{1}\setminus \{q\})$ so that $A$ and $B$ both deformation retract to $\Sigma_{2}$ and $A\cap B$ deformation retracts to $\Sigma_{2}\times\{0,1\}$, but I don't understand how to think about the inclusion maps $H_{n}(A\cap B) \hookrightarrow H_{n}(A)\bigoplus H_{n}(B)$.

$\Sigma_2$ denotes the orientable surface of genus two.

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What is $\Sigma_2$? – Alex Becker Jul 18 '12 at 2:49
@AlexBecker Probably the orientable surface of genus $2$. It would be nice to have this explicit in the question, I agree. – Dylan Moreland Jul 18 '12 at 2:50
@DylanMoreland That was my working hypothesis. I've drawn about 20 of those today though, so I want to be sure I'm not just seeing them everywhere. – Alex Becker Jul 18 '12 at 2:52
Yes, $\Sigma_{2}$ is the orientable surface of genus 2. – subfallen Jul 18 '12 at 17:37
@user36025, please add that information to the body of the question. – Mariano Suárez-Alvarez Jul 18 '12 at 21:27

Note that $A$ and $B$ are both homotopic to $\Sigma_2$, and so $H_n(A \cap B) \cong H_n(A) + H_n(B)$ for all $n$, if you choose your embeddings $i$ and $j$ wisely.
So by cellular homology I have $H_{0}(\Sigma_{2}) \simeq H_{2}(\Sigma_{2}) \simeq \mathbb{Z}$ and $H_{1}(\Sigma_{2}) \simeq \mathbb{Z}^{4}$ where $H_{0}$ is generated by the vertex $v$, $H_{2}$ by the 2-cell $F$, and $H_{1}$ by the 4 loops in the 1-skeleton $e_{1},\dots ,e_{4}$. So the map $H_{1}(A\cap B) \hookrightarrow H_{1}(A)\bigoplus H_{1}(B)$ takes a generator $(e_{i},e_{j})$ to $\dots$what? :( – subfallen Jul 18 '12 at 17:53
Well, $H_1(A)$ should be isomorphic to $\mathbb{Z}^2$, so should have two generators $(e_1^A, e_2^A)$; similarly for $H_1(B)$. And $H_1(A \cap B) \cong H_1(A) \oplus H_1(B)$, so what's the most obvious map you can think of between $H_1(A) \oplus H_1(B)$ and $H_1(A) \oplus H_1(B)$? – Kris Jul 18 '12 at 21:21
If $A$ is homotopic to $\Sigma_{2}$, then $H_{1}(A) \simeq \mathbb{Z}^{4}$ according to topospaces (topospaces.subwiki.org/wiki/… ). I'll try to think about this again tonight. – subfallen Jul 20 '12 at 19:59