# Compute fundamental group _visually_ by the polygonal representation of the space

Some time ago, before I learnt about covering spaces and Seifert-Van Kampen theorem, I tried to compute visually the fundamental group of some spaces. For example I figure out by myself that the fundamental group of the projective plane $$\mathbb{RP}^2$$ is $$\mathbb{Z}_2$$. I found it by drawing paths on the polygonal representation of this surface. Here is what I wrote

I'm sorry, it's in italian, but the important parts are the drawings. In (1) I show that the paths in interior of rectangle are trivial; in (2) I show that the paths which cross the frontier one time are not trivial; in (3) I show that the paths which cross the frontier two time are again trivial.

I know it is not a proof, but that's not really important: I made it to develop intuition. So, what's my question?

I want to do this for other spaces that can be obtained by gluing a polygon. I know how do this with Seifert-Van Kampen, but I want to visualize this as in the example above. The problem is that when I try with different spaces, I don't know what I can do and what I can't do. For example, I tried with the Dunce Hat (see below). I would like to show that every paths is trivial, but when I draw paths that cross the frontier I am not able to trivialize them, and this is because they "appear" on three sides (the three identified) and I don't under which conditions I can push them back in the interior.

How can I deal, in general, with the polygonal representation of a space? How can I visualize his fundamental group from it? What can I do? How can I treat the frontier? When can I push back the paths in the interior? And, for example, how can I see in this way that every path on the Dunce Hat is trivial?

EDIT

For example, how can I show that this is trivial?

• It is possible to construct a polygonal representation of a space where you cannot decide if its fundamental group is trivial. However, your arguments can be justified using the Seifert-Van Kampen Theorem to give a presentation of the fundamental group. This is a standard procedure that almost any low dimensional topologist could explain. – Charlie Frohman Feb 24 at 14:32
• Could you expand more your first sentence? What example? And what do you mean with "decide"? In which way? For the second sentence, I wrote: "I want to do this for other spaces that can be obtained by gluing a polygon. I know how do this with Seifert-Van Kampen, but I want to visualize this as in the example above". – Marco All-in Nervo Feb 24 at 15:46
• @CharlieFrohman I think Marco may have meant to @ you. – Mark S. Feb 24 at 16:36
• @marco All-in Nervo I could describe a polyhedral space and present it’s fundamental group so that the relations say, you can’t tell if this is the trivial group. – Charlie Frohman Feb 24 at 21:43
• The rules you are using are commonly accepted. You just need to be around some low dimensional topologists. – Charlie Frohman Feb 24 at 21:44

Maybe this is the picture you are looking for. I refer to a path tracing the oriented edge as $$a$$, and its inverse as $$a^{-1}$$. Note that every continuous path on the dunce cap (any CW-complex really) can be continuously deformed to trace only edges (of the given CW-structure). This should be somewhat visually intuitive, by straightening your path along edges, vertex to vertex. Thus if the path $$a$$ is trivial, so is any other path (do you see why?).

A comment as to what deformations are allowed - you are free to push your path anyway you want, and over edges (reemerging on any of the copies of the edge, noting orientations). This is what happens between the first and the second image. The one important rule is that you CANNOT ever move the basepoint (=initial and final point of your path).

Note that I did not move the basepoint passing from the first to the second picture, as the three corners of the triangle are literally the same point of the dunce cap.

Edit: As per your edit, one way to do it directly is this. Usually, you would just say the dunce cap is connected, thus its fundamental group relative to some basepoint is trivial if and only if it is trivial for some other basepoint (which we have shown above).

• Thanks for the reply, very very helpful. Can I ask you the drawing also for the new image I have inserted (I have edited the question)? I would be very thankful – Marco All-in Nervo Feb 28 at 18:04
• In relation to your triviality comment: There are spaces in which there exist base-free loops which are not homotopic take $S^1\vee S^1$ for instance with loop on one side and another on the other. Moreover you can find loops that are base-non-homotopic and non-base homotopic ($ab$ and $ba$) in my example. – Radost Feb 28 at 22:55
• thanks for the correction Radost, what I said was of course very wrong. removed the offending statement – amueller Mar 1 at 14:06
• @amueller I posted an answer with another solution, without changing basepoint and without jumping from one side to another side of the triangle. Maybe you are interested. Thanks a lot, you have been very helpful. – Marco All-in Nervo Mar 1 at 21:37

Many many thanks to amueller, who enlighted me.

I have changed his solution, in order to avoid to (1) change basepoint (2) "jump" from one side to another. I find the solution, inspired by amueller, just moving on the frontier back and forth. I have also tried with another path. Here the drawings ("esempio" means "example")