Computing the fundamental group of a quotient space

Question

I have been reading some notes of former students of my Algebraic topology lessons and I have seen this excercise:

Let denote $D^2:= \{(x,y) \in \Bbb{R}^2 : x^2 + y^2 \leq 1 \}$ and $S^1=\partial B^2$. Given $n \in \Bbb{N} - \{0\}$, take a subset $A_n \subset S^1$ such that $| A_n | = n$, consider the equivalence relation that identifies all points of $A_n$ to one and $X=B^2 / A_n$ the correspondient quotient space. Compute $\Pi_1(X)$ and study if it is free or not.

I have an idea to calculate this but I am not sure about it. Identify $n$ points to one is similar to remove $n-1$ points placing all of them on another unique point. Therefore, $X$ would be homeomorphic to $B^2$ removing $n-1$ points, and its fundamental group is easy to calculate using that $S^1$ is a deformation retract of $B^2$ removing a single point and the Seifert-Van Kampen.

Is this fine? If it is not, does anyone know how to proceed? Thanks in advance :)

Answer 1

$X$ is not homeomorphic to $B^2-F$ for any finite set. That's because $B^2-F$ is not compact. And if $F\subseteq S^1$ (finite or not) then $B^2-F$ won't even be homotopy equivalent to $X$, because $B^2-F$ is still contractible while $X$ (as we will see soon) is not.

It is true however that $X=B^2/A_n$ is homotopy equivalent to $B^2-F_{n-1}$ for some finite set $F_{n-1}\subseteq int(B^2)$ of size $n-1$. But that is not that easy to show. Plus you would still need to calculate $\pi_1$ of that space, and so you don't gain much.

So what I suggest is to use Seifert-van Kampen directly on $X$. Have a look at this drawing:

This is for $n=7$. We are of course considering all of that for the quotient space $X$, i.e. red points are all collapsed. Let $U_1$ be the blue set on the left, and $U_2$ the green set on the right. So blue set is composed of small neighbourhood around each point and connects only two of them through a bigger neighbourhood. While green set connects all points through a neighbourhood, except one which is isolated (in $B^2$, not in the quotient of course). These are supposed to be open by the way (even though the drawing doesn't suggest it, those sets should not contain their own boundaries). It is also important that the isolated green point is one of those two connected blue points.

It is not hard to see that in the quotient space $U_1\cap U_2$ is contractible, while $U_1$ is homotopy equivalent to $S^1$. The core idea here is that the green set $U_2$ is actually homotopy equivalent to $B^2 / \{x_2,\ldots,x_n\}$, i.e. the quotient by one point less. That is because the neighbourhood of the isolated red point can be contracted to the red point, and ultimately removed. So if we apply Seifert-van Kampen, then we get

$$\pi_1(X)\simeq \pi_1(S^1)*\pi_1(B^2/\{x_2,\ldots, x_n\})$$

Note that for $n=1$ our $B^2/\{x_n\}$ is homeomorphic to $B^2$ and thus contractible. And so it is a matter of applying induction now to conclude that

$$\pi_1(X)\simeq \mathbb{Z}*\cdots *\mathbb{Z}\ \ \ n-1\text{ times}$$

or alternatively $\pi_1(X)$ is a free group on $n-1$ generators.

Of course this is a bit of hand waving, and the drawing is not a formal proof. But it should give you enough intuition to fill the details. Even though this will be time consuming.

Answer 2

It is not clear to me why identifying boundary points is the same as removing interior points. Note boundary points of the disc are topologically distinguishable from interior points; there can be no homeomorphism of the disc mapping a boundary point to an interior point. You can prove this with local homology.

$$X\simeq\bigvee_{k\in[n-1]}S^1,\,\pi_1(X)\cong\star_{k=1}^{n-1}\,\,\Bbb Z$$

The first image: the purple nodes are the points of $A_4$, all dotted lines are fictional. Each sector between two nodes deformation retracts onto the boundary (black) lines; the green arrows show you how this works, where we draw a green dotted median line and push either side of it (and the median line is pushed straight down into the centre). One way to formally convince yourself such a deformation exists is to place a red sun outside of the circle and draw rays from the sun; for a point in the sector, take the line between that point and the sun and extend the line until it hits the black dotted boundaries. Now slowly move that point over time along the line until it hits the boundary.

Very importantly, this deformation retract fixes the nodes and induces a deformation in the quotient space.

The second sketch shows the result; we get a mere skeleton connecting the nodes to the centre. If you actually apply the quotient relation you see you get coils of wire as in the third image. Now it's a theorem: if you quotient a CW complex by its contractible subcomplex, you get a homotopy equivalence. So we kill one of those coils and obtain a bouquet of $(n-1)$ circles.

You can follow all the same basic diagrams and ideas but using SvK, instead of a homotopy equivalence, to obtain the same result.