# Gluing two Möbius strips together along their boundary

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

• You need to use reduced homology to assert that all the $0$th homologies are $0$, which is fine. Jul 18 '19 at 20:01