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I'm looking for help with the following question, specifically the Fundamental Group part. I won't complain if the Homology Groups are calculated as well, but I think I should be okay on this part:

Identification diagram

I think van Kampen is the way to go with $U$ being the space minus a point, and $V$ being a neighborhood about the point, but I'm having difficulty grinding through it. When calculating things like the Torus or Klein bottle, one is easily able to identify what the respective $U$ deformation retracts to. What does our $U$ deformation retract to in this case?

Edit: Actually, we could kill questions like this in general if there is a standard way to recognize what the $U$ deformation retracts to in the general case. I remember reading somewhere, maybe on this site, that you can realize the fundamental group of all spaces built in this fashion using this setup of applying van Kampen.

Edit 2: This isn't HW, I'm studying for a qual.

Edit 3: I clarified that I am mostly interested in knowing the solution for the fundamental group part. This editing is getting out of control, sorry about that.

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  • $\begingroup$ In a previous edit (link) I included the image in your post and did some minor formatting which you undid. Is this due to cross-editing or did you undo my edits on purpose? In the former case I could do it again, but I'd like to ask you for your consent before doing so. $\endgroup$
    – Martin
    Jun 16, 2013 at 8:55
  • $\begingroup$ Sorry about that, I didn't realize you had done anything. My comment about the editing is hilarious in this context. Have at it. It wouldn't let me include the image because of my low rep. $\endgroup$
    – user82668
    Jun 16, 2013 at 9:19
  • $\begingroup$ No problem :-) There you go. $\endgroup$
    – Martin
    Jun 16, 2013 at 9:20

1 Answer 1

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$U$ retracts to the boundary of a hexagon $H$ modulo your equivalence relation $\sim$. Under $\sim$, there are two equivalence classes of vertices of $H$: The ones being pointed at and the ones being pointed away from. These two equivalence classes are connected by 6 edges, but only 2 equivalence classes of edges: The single arrowed ones and the double arrowed ones. Hence $U$ retracts to a space consisting of two points, connected by two paths, i.e. $S^1$.

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  • $\begingroup$ So, you're saying it's $\mathbb Z_3$? By the way what you are saying makes perfect sense and for sure works in other settings. It makes things like the Real Projective Plane with its square identification easier to stomach. $\endgroup$
    – user82668
    Jun 16, 2013 at 10:27
  • $\begingroup$ You are absolutely right. The fundamental group is indeed $\mathbb{Z}_3$. I'm glad you found my answer understandable and useful :). $\endgroup$
    – Abel
    Jun 16, 2013 at 12:29

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