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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!

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    I think you are correct that $b(1) = (2,8)$ (since the boundary circle of a mobius strip is twice a generator of $H_1$). To compute the groups, notice that in this case exactness implies $H_2(X) \cong ker(b)$ and $H_1(X) \cong coker(b)$. – William Jun 22 '19 at 23:30
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    Careful, most results involving orientation are about compact surfaces with no boundary. The homotopy type of a manifold with nonempty boundary is that of a noncompact surface. So the normal result about homology of compact surfaces doesn't hold. There might be some result still pertaining to this, but a safe bet is to just see that the mobius strip is homotopy equivalent to a circle. – Connor Malin Jun 22 '19 at 23:59
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    There are general statements you can make about all manifolds without boundary (a manifold with boundary is homotopy equivalent to it minus its boundary), but these require you to replace cohomology with cohomology with compact support. – Connor Malin Jun 23 '19 at 00:08
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    Every noncompact surface, and every surface with nonempty boundary, is homotopy equivalent to a 1-dimensional complex, hence its $H_2$ is trivial. – Lee Mosher Jun 23 '19 at 02:37

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