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I am fairly new to algebraic topology so please bare with me if this seems simple

I am trying to find the fundamental group of the unit disk with the identification on the boundary z = (cos(θ), sin(θ)) being mapped to (cos(θ+2π/n), sin(θ+2π/n)).

For n=1 it is just the disc so the fundamental group is trivial (since the disk is convex).

Therefore, I was trying to solve it for n=2 to begin. With n=2 each point on the boundary is being identified to its antipodal point.

I was trying figure out a way to use van kampen's theorem.

As I know it, van kampen's theorem states that if X = A∪B (A and B open) and A∩B path-connected, then π1(X) = [π1(A) * π1(B)] / π1(A∩B)

(where * is the free product and you quotient out by the fundamental group of the intersection)

I think that there probably is a way by letting A equal the interior of the disk, but I am not sure what to make B since I am even having trouble seeing what the open sets in the quotient space are.

Thanks for the help

slim
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  • Formatting tips here. – Em. Dec 11 '16 at 01:09
  • Say $n=2$, and $D_r(a) = {|z-a| < r}$. You have the open unit disk $A$ and you add some new open sets of the form $U_r(e^{i \theta}) = (D_r(e^{i \theta}) \cap A)\cup (D_r(e^{-i \theta}) \cap A)$. If you look at $B =\bigcup_{\theta \in [0,\pi) , r < 1/4} U_r(e^{i \theta}) $ then it is homeomorphic to an annulus $3/4 < |z|<5/4$ – reuns Dec 11 '16 at 01:31

1 Answers1

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You can take $B$ as the image of the annulus $\{ z \ | r< |z| \le 1\}$. This is an open subset in the quotient space, as the image of a saturated open subset ( saturated meaning: any point equivalent to a point in that subset is still in the subset).

(As a side notice, for a surjective quotient map of topological spaces the open subsets in the target space are the images of saturated open subsets in the source).

To apply the Seifert-van Kampen theorem, it's best to think in terms of diagrams and universal objects. $\pi_1(A\cup B)$ is a direct sum with identifications, the way $A\cup B$ is a direct sum of $A$ and $B$ with identifications along $A\cup B$. So one has a category of nice commutative diagrams and $\pi_1(A\cup B)$ is (more or less) an initial object. But basically it's $\mathbb{Z}/n$.

Orest Bucicovschi
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  • Thank you for the answer. I completely understand why B is open in the quotient topology (great explanation). But is it possible to elaborate on how you get to ℤ/n as the fundamental group? The fundamental group of A is obviously trivial since it is convex. But why is the fundamental group of B ℤ (I assume that it is). And why are you modding out by n? Unfortunately, my algebra is (very) shaky so I still don't have a great grasp of the theorem generally. – slim Dec 11 '16 at 03:11