This is a problem from my past Qual
"Let $m,n$ be positive integers and view $S^1$ as the unit circle in $\mathbb{C}$. Let $X$ be the space obtained from the cylinder $S^1\times [0,1]$ by identifying $(z,0)$ and $(w,0)$ if $z^n=w^n$ and identifying $(z,1)$ and $(w,1)$ if $z^m=w^m$.
(a) Use Mayer-Vietoris to compute $H_i(X)$.
(b) Use Seifert-van Kampen to find $\pi_1(X)$."
My solution: (a) Naturally, let $A$ be the upper-half of the cylinder and $B$ be the lower-half (Of course we enlarge it more than a half for them to cover $X$). THen $A\cap B$ deformation retracts to $S^1$. $A$ and $B$ deformation retract to their top and bottom.
I GUESS that the top of $A$ is formed by one $0$-cell, one $1$-cell and one $2$-cell that attaches to the $1$-skeleton by the map $z\mapsto z^m$. Again, this is a guess, it's hard to imagine. With this the chain complex of $A$ is
$$0\to \mathbb{Z} \xrightarrow{m} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$$
and hence $H_2(A)=0,H_1(A)=\mathbb{Z}/m\mathbb{Z}$ and $H_0(A)=\mathbb{Z}$.
(b) About fundamental group, if I'm right, then $\pi_1(A)=\langle a|a^m \rangle = \mathbb{Z}/m\mathbb{Z}$. I'm not good at visualizing, but I GUESS that the induced map $\pi_1(A\cap B) \to \pi_1(A) $ is given by projection $1\mapsto 1$. Hence $\pi_1(X)=\langle a,b|a^m,b^n, ab^{-1} \rangle = \mathbb{Z}/d\mathbb{Z}$ where $d=gcd(m,n)$.
My questions: 1) Is my part (a) correct so far? I don't want to continue before knowing if it's true or not.
- Is my part (b) correct? Is there a way to visualize the induced map in this case?
Best Answer
Expanding on my comment, notice that $z \to z^m$ and $z \to z^n$ are covering maps so really you still end up with a circle on the top and bottom. Thus, $\pi_1(A) = \langle a \rangle \cong \mathbb{Z} \cong \langle b \rangle = \pi_1(B)$. Now, as you pointed out, $A \cap B \cong S^1$ so $\pi_1(A \cap B) \cong \mathbb{Z}$. Let's see where this generator goes. If you push it "up" into the $A$ component, then it winds around $m$ times due to the way the quotienting worked, and similarly pushing it "down" into the $B$ component, it winds around $n$ times. So we have that $a^m = b^n$ or that $a^mb^{-n}$ is a relation. So $\pi_1(X) \cong \langle a,b | a^mb^{-n} \rangle$.
Before continuing with Mayer-Vietoris, it might be helpful to think about what we should expect. $H_1(X)$ is the abelianization of $\pi_1(X)$, so we should expect something like $\mathbb{Z} \oplus \mathbb{Z}$ where $ma = nb$ i.e. $\mathbb{Z}/(\gcd (m,n))$
We have the sequence $$0 \to H_2(A \cap B) \to H_2(A) \oplus H_2(B) \to H_2(X) \to H_1(A \cap B) \to H_1(A) \oplus H_1(B) \to H_1(X) \to 0$$ Note that we don't have to include the $H_0$ because all the pieces are path connected (this is equivalent to doing the calculation with reduced homology). Filling some pieces in, we have $$0 \to 0 \to 0 \to H_2(X) \to H_1(A \cap B) \to H_1(A) \oplus H_1(B) \to H_1(X) \to 0$$ Note that the map $H_1(A \cap B) \to H_1(A) \oplus H_1(B)$ is injective (it is the map $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$ given by $1 \mapsto (m,n)$), so in fact we just deal with the short exact sequence $$0 \to H_1(A \cap B) \to H_1(A) \oplus H_1(B) \to H_1(X) \to 0$$ Now, we have $$0 \to \mathbb{Z} \xrightarrow{(\times m, \times n)} \mathbb{Z} \oplus \mathbb{Z} \to H_1(X) \to 0$$ So $H_1(X) \cong (\mathbb{Z} \oplus \mathbb{Z})$ modulo the relation $ma = nb$ so we have $H_1(X) = \mathbb{Z}/(\gcd(m,n))$, which is precisely what we expected.
Edit: Understanding the Quotient
Alright so first let us understand what it means when saying $z^n = w^n$. In $S^1 \subset \mathbb{C}$, points are of the form $e^{i\theta}$. Thus we identify points if $e^{in\theta_1} = e^{in\theta_2}$. When would this occur? When $n\theta_1-n\theta_2 = 2k\pi$ or when $\theta_1 - \theta_2 = \frac{2k\pi}{n}$. Thus, what is being done is that $S^1$ is being cut into $n$ equal parts and those parts are being identified, so you are essentially gluing by the map $z \mapsto z^n$ as you pointed out. So the quotient is just one of these parts (arcs) with endpoints identified so is just $S^1$ again. If you're comfortable with covering spaces, you'll recall that $S^1$ is a finite covering of itself. This is a manifestation of that fact. It's the same idea as $\mathbb{R}/\mathbb{Z} = S^1$ (indeed $\mathbb{R}$ is the universal cover).
So, the correct way to visualize the cylinder is as a regular cylinder with the top twisted up $m$ times and the bottom twisted up $n$ times.