I want to calculate the fundamental group of the space $(S^1 \times [0,1])/$~ where $(z,0)$ ~ $(e^{2\pi i/n}z, 0)$ and $(z,1)$ ~ $(e^{2\pi i/m}z, 1)$. My idea is to find a pushout and then use the van – Kampen – theorem.
We get a pushout from the inclusions $S^1 \times \{1/2\} \hookrightarrow (S^1 \times [1/2, 1])/$~ and $S^1 \times \{1/2\} \hookrightarrow (S^1 \times [0, 1/2])/$~ and since $(S^1 \times [1/2, 1])/$~ and $(S^1 \times [0, 1/2])/$~ are homotopy-equivalent to $S^1$, van – Kampen suggests that the sought-after fundamental group is isomorphic to $\mathbb{Z}$, specifically $\langle a, b| a=b \rangle$, which is obviously wrong.
So where are my mistakes? I think the induced inclusions are wrong, but how do I see the correct ones?
Best Answer
Your approach is fine, but you need to be careful in calculating the maps on fundamental groups that are induced by $\iota_1: {\mathbb S}^1\times \{1/2\}\to ({\mathbb S}^1\times [1/2,1])/_\sim\simeq{\mathbb S}^1$ and $\iota_2: {\mathbb S}^1\times \{1/2\}\to ({\mathbb S}^1\times [0,1/2])/_\sim\simeq{\mathbb S}^1$ (for the concrete description of the homotopy equivalences involved here, see below):
If you push the loop around ${\mathbb S}^1\times \{1/2\}$ to ${\mathbb S}^1\times \{1\}$ and follow it along ${\mathbb S}^1\times\{1\}\to ({\mathbb S}^1\times\{1\})/_\sim\cong{\mathbb S}^1$, you get the $m$-fold of the loop around ${\mathbb S}^1$. Hence, $$\pi_1(\iota_1): {\mathbb Z}\cong\pi_1({\mathbb S}^1\times\{1/2\})\to\pi_1(({\mathbb S}^1\times [1/2,1])/_\sim)\cong{\mathbb Z}$$ is the multiplication by $m$. Similarly, $\pi_1(\iota_2)$ is the multiplication by $n$.
Can you finish the computation on your own now?
Addendum The above builds on the homotopy equivalence $({\mathbb S}^1\times [1/2,1])/_\sim\xrightarrow{\simeq}{\mathbb S}^1$ given by the deformation retraction $({\mathbb S}^1\times [1/2,1])_\sim\to ({\mathbb S}^1\times\{1\})/_\sim, \overline{(x,t)}\mapsto\overline{(x,1)}$, composed with the homeomorphism $({\mathbb S}^1\times\{1\})/_\sim\cong {\mathbb S}^1, (z,1)\mapsto z^m$. In particular, it sends the single fold loop around ${\mathbb S}^1\times\{1/2\}$ to the $m$-fold loop around ${\mathbb S}^1$. For $({\mathbb S}^1\times[0,1/2])/_\sim\xrightarrow{\simeq}{\mathbb S}^1$ it's analogous.