algebraic-topologyfundamental-groups
I had a question about computing fundamental groups of certain covering spaces. In Hatcher's book, he claims that all of these groups can be computed from Van Kampen. I could figure them out for a few, but am not sure how to show this for covering space 11. I am not sure how to pick the open sets $A_{\alpha}$ for Van Kampen. Thanks for the help!
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).
You have exactly the right idea! We know the fundamental group of your figure is going to be
$$\langle \text{generators of $\pi_1(V)$} \mid \text{relations of $\pi_1(V)$}, cdb^{-1}aab^{-1} \rangle$$
So what we want to do now is compute $\pi_1(V)$. For this (as with all topology problems), a picture is the most important tool we have. Let's label our hexagon and its vertices:
Now. We know that we're identifying the $b$ edges. So their endpoints must be the same too! This tells us that, actually, $v_3 = v_0$ (since these are both the "back" of $b$) and $v_2 = v_5$ (since these are both the "front"). Analogously, we can identify the fronts and backs of the $a$ edges. This tells us that $v_3 = v_4$ and that $v_4 = v_5$. Putting these equalities together, we find that actually $v_0 = v_2 = v_3 = v_4 = v_5$!
So we can find out what our graph actually looks like by drawing each of these repeated vertices exactly once. This gives:
But now it's easy to read off the fundamental group from this picture. If we take $v_{0,2,3,4,5}$ as our basepoint, we find
$$\pi_1(V) = \langle a,b, cd \rangle$$
is the free group $F_3$ (which is not surprising, since the fundamental group of every graph is free. Do you see why?).
So now we find our final answer (writing $x$ for $cd$):
$$\pi_1(Y) = \langle a,b,x \mid xb^{-1}aab^{-1} \rangle$$
I hope this helps ^_^
Best Answer
Apply Van Kampen to the intersection of the graph with a blob around each $a$ loop which covers slightly more than half of the neighbouring $b$ intervals, I call these subsets "pieces".
Each piece has fundamental group $\mathbb{Z}$ and the intersections of neighbouring pieces are $1$-connected.
So starting with one piece we have $\pi_{1} = \mathbb{Z}$, and every we time we add a neighbouring piece we make a free product with $\mathbb{Z}$ (since the intersection is $1$-connected) giving after $k$ steps the free group of $k$ generators; $F_k$.
In the end we get a free group on a countably infinite generating set. It is a nice geometric was to see that $F_2$ has a non finitely generated subgroup (generated by elements of the form $b^n a b^{-n}$)