Gluing a mobius strip to a Torus with disc removed

Question

Let $D$ be a closed disk embedded in $S^1×S^1$. Then $X=S^1×S^1−\text{int}(D)$ has a circle $S$ for its boundary. Suppose we glue a Möbius strip to this space by identifying each point in the boundary circle of the Möbius strip with its image under a $2:1$ covering map of $S$

Compute the homology groups and fundamental group of this space.

Let's declare $R= X \cup_f MB=Z$ where $f$ is a $2:1$ covering map.

Here's what I've done so far:

After considering reduced homology, a portion of our Mayer-Vietoris sequence will look like this:

$0 \rightarrow H_2(R) \rightarrow^a H_1(S^1) \rightarrow^b H_1(MB) \oplus H_1(X) \rightarrow^c H_1(R) \rightarrow 0$

After substituting the following:

$H_1(X) \cong \mathbb{Z}^2$

$H_1(MB) \cong \mathbb{Z}$

$H_1(S^1) \cong \mathbb{Z}$

We get the following:

$0 \rightarrow H_2(R) \rightarrow^a \mathbb{Z} \rightarrow^b \mathbb{Z} \oplus \mathbb{Z}^2 \rightarrow^c H_1(R) \rightarrow 0$

Now, my main question is how $b$ acts on the generator of $\mathbb{Z}$, that is, what is $b(1)$?

$b(1)$ restricted to the summand representing $H_1(MB)$ will be equal to $2$, where $H_1(MB)$ is generated by $1$. $b$ here is induced by the inclusion map $\delta MB \rightarrow MB$, and the boundary of a moibus band wraps twice around the core circle, and thus $b(1)=2$. Note that this homomorphism is completely independent of the covering ratio between $\delta MB$ and $\delta X$, in our case a $2:1$ ratio.

Also, $b(1)$ restricted to the second two summands, the summands representing $H_1(X)$, I believe will have image $(0,0)$. I drew a picture of a square with opposite sides identified and removed a disc from it. Going once around the boundary of this disc would be homotopic to going around the perimter, and for each side when you go across the identified side you go reverse orientation, so it all cancels out.

So, I believe that $b(1)= (2,(0,0))$

Thus $b$ is injective and thus $H_2(R)=0$.

$H_1(R) \cong \frac{\mathbb{Z}^3}{\langle (2,0,0) \rangle} \cong \mathbb{Z}_2 \oplus \mathbb{Z} \oplus \mathbb{Z}$

How is this looking? Thanks in advance!! I also need help calculating the fundamental group of this space.

Answer 1

I would suggest to consider the fundamental group first. Here you can use the theorem of van Kampen to determine the fundamental group once you found out what the maps of $S^1$ to $MB$ and $X$ do on fundamental groups. Then the fundamental group of $R$ is the corresponding amalgamated product of $\pi_1(MB)$ and $\pi_1(X)$ (are you familiar with that?).

We identify the boundary of $MB$ with $S$ by a $2:1$ covering. This means that we identify every $x\in S^1\cong\partial MB$ with $f(x)\in S\subseteq X$, or equivalently, we identify $g(x)$ with $f(x)$ for all $x\in S^1$, where $g\colon S^1\to\partial MB$ is the corresponding diffeomorphism. Now for the fundamental group and the homology we have to consider the induced maps of $f$ and $g$.

The inclusion $S^1\cong\partial MB\hookrightarrow MB$ corresponds to a $2:1$ covering of $S^1$, hence the induced map $\mathbb{Z}\cong\pi_1(S^1)\to\pi_1(MB)\cong\mathbb{Z}$ of $g$ is multiplication by $2$. By the theorem of Hurewicz the first homology is the abelianization of the fundamental group (naturally!) and hence the map on the first homology is multiplication by $2$, too.

The fundamental group of $X$ is $\langle a,b\rangle$, the free group in $2$ generators (each generator corresponds to one side of the square with opposite sides identified). The induced map of $f\colon S^1\to X$ maps $1$ to $(aba^{-1}b^{-1})^2$, that is exactly what you obtain when you move twice around the square with opposite sides identified. When considering the first homology we have to abelianize, which makes $(aba^{-1}b^{-1})^2$ trivial, so you are correct that $1\in H_1(S^1)$ gets mapped to $(0,0)\in H_1(X)$.

Hence, $b(1)=(2,(0,0))$. From this you can calculate all homology groups of $R$.