Let $M_1$ and $M_2$ be two mobius strips with boundaries $S_1$ and $S_2$, respectively. Say we form a space $X$ by gluing $M_1$ and $M_2$ along their boundaries by a 4-fold covering map, $f: S_1 \rightarrow S_2$. Compute $H_k(X)$.
So, $H_2(M_i)=0$ because a mobius band is a non-orientable surface. Also $H_2(S^1)=0$ where $S^1$ refers to the intersection of $M_1$ and $M_2$ after gluing, aka their now shared boundary.
So, if we consider reduced homlogy, then our L.E.S. looks like:
$0 \rightarrow H_2(X) \rightarrow^a H_1(S^1) \rightarrow^b H_1(M_1) \oplus H_1(M_2) \rightarrow^c H_1(X) \rightarrow^d 0 \rightarrow ...$
Where I believe $b(1) = (2,8)$.
Is this all correct? How do i calculate $H_2(X)$ and $H_1(X)$? Thanks!