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.


1 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 $R$.

  • $\begingroup$ Thank you so much! my qual's are coming up and I've been having a hard time figuring certain things out. Would you care to spend a moment explaining to me gluing a Mobius strip to a circle (boundary of a disc usually) by an n:1 covering, and the effect this has on the map induced by inclusion from the circle back into the Mobius band? If you have a loop that goes around the whole boundary of the mobius band, then does that represent the generator of the infinite cyclic group $<a>=H_1(MB)$? Or does it represent the generator squared? $\endgroup$
    – user637978
    Jul 31, 2019 at 18:23
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    $\begingroup$ In my previous answer I assumed that the boundaries of $MB$ and $X$ get identified by a $1:1$ map, I misread your first paragraph, sorry for that. I've made an edit. $\endgroup$
    – PR_
    Aug 1, 2019 at 12:25
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    $\begingroup$ The effect of an $n:1$ covering is that you move $n$ times around the boundary of $X$ while moving once around the boundary of $MB$ (not the other way round). This maps a generator of $\pi_1(S^1)$ to $(aba^{-1}b^{-1})^n\in\pi_1(X)=\langle a,b\rangle$. On the level of homology all these expressions are trivial, so it doesn't make a difference (but for the fundamental group it does). $\endgroup$
    – PR_
    Aug 1, 2019 at 12:30
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    $\begingroup$ We have two different coverings of $S^1$ involved here: The inclusion $\partial MB\hookrightarrow MB$ corresponds to a $2:1$ covering of $S^1$, and the map $f\colon \partial MB\to \partial X$ which is a $2:1$ covering of $S^1$, too (or $n:1$ as you assumed in your comment). From the first covering we obtain that $b(1)$ is $2$ on the first factor (always and independently of $n$), from the second one we obtain that $b(1)$ is $(aba^{-1}b^{-1})^n$ on the second factor, which is always trivial in homology. $\endgroup$
    – PR_
    Aug 2, 2019 at 13:18
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    $\begingroup$ Yes, that's correct. $\endgroup$
    – PR_
    Aug 3, 2019 at 10:39

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