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Let $M_1$ and $M_2$ be Möbius strips with boundary circles $S_1$ and $S_2$. Suppose we construct a space $X$ by gluing $M_1$ to $M_2$ by identifying their boundary points using a 4-fold covering map $f: S_1 \rightarrow S_2$. Calculate $H_i(X)$.

After reviewing how gluing maps work: What does it mean to "Glue the boundary circle of a mobius strip to another circle in a 2:1 covering map"

I've come up with the following proof. Can somebody please help me polish it or correct it if need be? Thanks!

$proof$: Let $X = M_1 \cup_f M_2$ as according to the problem statement. We look at the Mayer-Vietoris sequence:

\begin{multline}0 \longrightarrow H_2(X) \xrightarrow{\enspace a\enspace} H_1(S) \xrightarrow{\enspace b\enspace} H_1(MB_1) \oplus H_1(MB_2) \xrightarrow{\enspace c\enspace} H_1(X) \\\xrightarrow{\enspace d \enspace} H_0(S) \xrightarrow{\enspace e\enspace} H_0(MB_1) \oplus H_0(MB_2) \xrightarrow{\enspace f\enspace} H_0(X) \rightarrow 0 \end{multline}

Where $H_1(S) \cong H_1(MB_i) \cong \mathbb{Z}$ and all the zeroth homologies are $\mathbb{Z}$ because all the spaces are path connected.

Since the circle that is formed by the intersection of the two spaces is wrapped around $\delta MB_2$ four times, and the deformation retraction that makes $MB \cong S^1$ is a 2:1 retraction, I believe that the map $b$ will be defined by:

$b(1) = (2,-8)$

Where $1$ is the generator for $H_1(S)$. Therefore $b$ is injective and so $H_2(X)=0$

Now the only tricky part left is trying to determine $H_1(X)$. We examine the following short exact sequence:

$0 \longrightarrow \operatorname{coker}(b) \xrightarrow{\enspace\tilde{c}\enspace} H_1(X) \xrightarrow{\enspace d\enspace} \operatorname{im}(d) \longrightarrow 0$

Where $\tilde{c}$ is the map induced by c, which is well defined because $\ker(c) = \operatorname{im}(b)$.

Now, $e(1) = (1,-1)$, where the first one is a generator of $H_0(S)$ and the $1$'s in the image are the generators of $MB_1$ and $MB_2$.

Thereforefore $e$ is injective, so $\operatorname{im}(d) = \ker(e) = 0$

Thus $\operatorname{coker}(b) \cong H_1(X)$

where $\operatorname{coker}(b) = \frac{\mathbb{Z} \oplus \mathbb{Z}}{\langle(2,-8)\rangle}$

Thanks to help gained in this thread: Is there a cleaner way to express the quotient $\frac{\langle(1,0),(0,1)\rangle}{\langle(2,-8)\rangle}$?

I can write $\frac{\mathbb{Z} \oplus \mathbb{Z}}{\langle(2,-8)\rangle} \cong \mathbb{Z} \oplus \mathbb{Z}_2$

Please help me make this better! Thanks!

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    $\begingroup$ You need to use reduced homology to assert that all the $0$th homologies are $0$, which is fine. $\endgroup$ Jul 18 '19 at 20:01
  • $\begingroup$ Good catch, edited it $\endgroup$
    – Nobel Cat
    Jul 18 '19 at 20:39

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