0
$\begingroup$

I'm trying to prove that the dunce cap is simply connected via Seifert- Van Kampen Theorem. I choose to be my open sets $U$ and $V$ the open disk and the punctured surface below, then $U\cap V$ is the annulus.enter image description here

I'm having problems to find the fundamental group of $V$

I need help.

Thanks

  • 0
    Dunce cap is the triangle with three edges identified in your figure?2012-11-26
  • 0
    yes, the triangle with the three edges identified with its interior.2012-11-26
  • 0
    This space is a circle $\mathbb{S}^1$ with a disk glued in via the degree $3$ map $\partial \mathbb{D}^2\ni z\mapsto z^3\in \mathbb{S}^1$. First cellular homology is $\mathbb{Z}_3$ so the space can't be $1$-connected.2012-11-26
  • 0
    The dunce cap is indeed simply connected. The space you have drawn, whch is *not* the dunce cap, has fundamental group $\Bbb{Z}/3\Bbb{Z}$.2012-11-26
  • 0
    @ChrisEagle Concerning the dunce cap, do you have any hint to prove this is simply connected? Thank you, and sorry about my mistake2012-11-26

2 Answers 2

8

Duncehat

I think you have the figure for the Dunce Hat wrong, see above, where all the arrows have the label $a $, say. So you have one $1$-cell, giving $S^1$, and one $2$-cell attached by a map described by $a+a-a$, which gives a group with one generator $a$ and one relation $a+a-a=a$.

Your figure would give the group with generator $a$ and relation $a^3$, as said by others.

[The figure is taken from Topology and Groupoids. ]

  • 0
    Thank you for your answer, however this question is a question of a book of amstrong Basic topology: Amstrong ask how to prove using Van-Kampen theorem that this space is simply connected. page 140 question 222012-11-26
  • 0
    @RafaelChavez Either you have incorrectly defined the "Dunce Cap", or the exercise is incorrect. The space you defined in your question is not simply connected.2012-11-26
  • 0
    @Neal yes, I've defined the wrong space.2012-11-26
  • 0
    Attaching a $2$-cell to a wedge $X^1$ of spheres by a map defining an element $r$ in $F=\pi_1 X^1$ gives a relation $r$ in the free group $F$. This is a consequence of the Seifert-van Kampen Theorem.2012-11-26
3

I am not sure about what is the U∪V, and I guess it's the triangle with three edges identified as in your picture. Then you've give the U and V. The fundamental group of U is trivial. And V can be deformation retract to the edges identified,i.e, a circle, thus the fundamental group of V is Z. And U∩V is annulus as you pointed out. So the fundamental group of it is also Z. Now you may use S.V.K theorem to conclude that the fundamental group of U∪V is Z/3Z. Because the generator of π(U∪V) is three times of the generator of π(V).

  • 0
    sorry, but this space is simply connected. However, thank you for your try :)2012-11-26
  • 0
    @Rafael Chavez Really? But I don't know what's wrong with my reasoning... For example, an edge is a loop as the vertices are identified to be one. And the edge is untrivial(as far as I concerned). And you can give the CW-complex structure, then by direct compute you can still show the fundamental group is Z/3Z.2012-11-26
  • 1
    Your reasoning is fine, Rafael has drawn the wrong space.2012-11-26
  • 0
    @ChrisEagle yes, I've drawn the wrong space, thanks2012-11-26