Fundamental group of a quotient space

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?

• Could you clarify the equivalence relation please? This way it appears you're gluing the two boundaries to a point. Do $n$ and $m$ vary? If so it still seems to me the whole boundaries are glued to just one point. Which would make your space the sphere. – Mr.P Aug 15 '15 at 12:59
• Ah I forgot something, pardon, I'll edit. – Cosmare Aug 15 '15 at 13:09
• Am I correct in assuming the equivalence is in fact $(z,0)~(ze^{\frac{2k\pi i}{n}},0)$ for $k\in\mathbb{Z}$? Which would add $n$ loops not equivalent to the trivial loop? – Mr.P Aug 15 '15 at 14:56
• No, you just identify points that are $(360/n)°$ apart. – Cosmare Aug 15 '15 at 15:23
• Your approach is correct, @Cosmare, but the conclusion you draw seems to be wrong. The Van-Kampen theorem says that applying $\pi_1$ to the pushout you describe yields a pushout in Groups. As you noted correctly, the top and bottom half of $X$ deformations retract onto $S^1$. Hence, $\pi_1(X)$ has two generators $a,b$. Note that the inclusions induce multiplication with $m$ and $n$ on $\pi_1$, since the middle circles becomes $m$-times (resp. $n$-times) the top (resp. bottom) circle with regard to the deformation retraction onto $S^1$. Hence, $$\pi_1(X)\cong \langle a, b| a^m=b^n \rangle$$ – iwriteonbananas Aug 17 '15 at 14:33

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.

• My problem is why I need to care about $S^1 \times \{1\}$ at all, because the inclusion is from $S^1 \times \{1/2\}$, how do you end up in $S^1 \times \{1\}$ all of sudden? I guess I don't understand the inclusion maps? – Cosmare Aug 15 '15 at 13:42
• @Cosmare: I added some details - is it clearer now? – Hanno Aug 15 '15 at 16:25
• I think I don't really understand how we know which maps we have to use to apply van Kampen, you end up with an homotopy equivalence, I just see an inclusion into the middle of the quotient space and that's it. Yet I know that it can't be right since it has to be more complex because the space involved is more complex. – Cosmare Aug 15 '15 at 17:16
• I understood the part I mentioned now, thank you very much, but could you please explain why the map $(z,1) \mapsto z^m$ is a homeomorphism? – Cosmare Aug 16 '15 at 11:51
• @Cosmare: You have $z_0^m = z_1^m$ if and only if $z_1 = z_0 \zeta_m^i$ for some $i$, i.e. if and only if $z_0$ and $z_1$ are equivalent with respect to the relation $\sim$. – Hanno Aug 16 '15 at 11:53