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Let $X$ be the quotient space of the disk, $\{(x,y)\in \mathbb R^{2} \ | \ x^{2}+y^{2}\leq 1 \}$, obtained by identifying points on the boundary that are $120$ degrees apart. How can we find the fundamental group of $X$ ?

Raeder
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hebele
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  • No, it is a past prelim question, and I think I need to use the Van Kampen theorem but I don't know how – hebele Jun 03 '11 at 20:22
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    It's cyclic of order 3. You can use Van Kampen's theorem directly or you can just put a cell structure on the quotient. – Jim Belk Jun 03 '11 at 20:28
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    I suppose you learned how to compute $\pi_1$ of a cell complex. In this case th 1-skeleton $X^1=S^1$ and $X$ is obtained by gluing a disk with the boundary circle going 3x around $X^1$. Is it helpful enough? – user8268 Jun 03 '11 at 20:33
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    @hebele - if you have done the projective plane before (identify antipodal points - i.e. 180 degrees), you should able to do this! (and you should be able to guess the answer straight away) – Juan S Jun 03 '11 at 22:56
  • @JuanS well this is a little bit different from the case of projective plane... – JacobsonRadical Dec 12 '19 at 05:26

1 Answers1

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Let $A\subset X$ be the closed disk of radius $1/2$ centered at $0$, and let $B$ be the complement of its interior in $X$. Then $A\cap B=\mathbb{S}^{1}$, $A$ is contractible, $B$ is homotopic equivalent to a circle, and $$X=A\cup _{A\cap B}B=A\cup_{\mathbb{S}^{1}}B.$$

Let $\alpha:A\cap B\rightarrow A$ and $\beta:A\cap B\rightarrow B$ be the inclusion map.

Then by Van Kampen Theorem, $$\pi_{1}(X)=\pi_{1}(A)*_{\pi_{1}(\mathbb{S}^{1})}\pi_{1}(B)=\{e\}*_{\mathbb{Z}}\mathbb{Z},$$ where the free product with amalgamation is via the induced homomorphism $\alpha_{*}$ and $\beta_{*}$.

Then $\pi_{1}(X)=\{a|\alpha_{*}([\gamma])=\beta_{*}([\gamma])\}$, but $\alpha_{*}([\gamma])=e$, and $\beta$ is a degree $3$ map, so we have $$\pi_{1}(X)=\{a|a^{3}=1\}=\mathbb{Z}/3\mathbb{Z}.$$

JacobsonRadical
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  • Could you please explain a little more about: how do we know $B$ is homotopic to a circle? Is the quotient map a homotopy equivalence? Thank you! –  Nov 24 '20 at 19:05