## Simple connectedness

EDIT: Here’s some random problem I thought up, about simple connectedness.

Here are some topological remarks relating to simple connectedness of subsets of $\mathbb{C}$. A set is called simply connected if it is path-connected and if every curve within the set can be continuously “shrunk” to a point. That is, every closed curve is homotopic to a point, or equivalently, the fundamental group is trivial at all of its points. Many sets are simply connected, but perhaps our intuition usually causes us to think of some deformation of a convex set, like a star-shaped domain (left), as opposed to something like an annulus (right). Lines and single points are also easily seen to be simply connected.

The annulus is probably the simplest example of a (path-connected) region that fails to be simply connected. We can define something similar to an annulus in any metric space. Given two radii $R > r$ and a point $a \in X$, an annulus is a set of the form $\{ x \in X : r < |x-a| < R \}$. If $X=\mathbb{R}$, the annulus fails to be path-connected; it is a union of two disjoint open intervals. In the case of $\mathbb{R}^n$ ($n > 1$), the annulus is path-connected, but fails to be simply connected only in the case $n=2$ (if I am correct). Recall $\mathbb{C} \cong \mathbb{R}^2$, so for the purposes of topology, we can think of the complex plane just like the real plane.

To see that an annulus in the complex plane is not simply connected, first picture a closed curve which does not “wrap around” the annulus’ “hole”. We can continuously shrink such a curve to a point. However, given a closed curve which does wrap around the hole (for example, a circle whose centre is the same as the annulus’, with radius $\tilde{r}$ for $r < \tilde{r} < R$, where $r$ and $R$ denote the small and large radii of the annulus, respectively) cannot be continuously shrunk to a point, because as we try to contract it, it “catches” around the hole. This means there are two distinct “classes” of closed curves within an annulus: those that wrap around the hole, and those that do not. Hence, its fundamental group has two elements.

A depiction of homotopies in three dimensions and the obvious dependence on the underlying topological space. In this picture we see how adding another "piece" to a topological space can dramatically affect its fundamental group -- by adding a "cap" to this topless-and-bottomless cylinder, all curves now become homotopic to a point.

Simple connectedness does not “play well” (as one might hope) with binary set operations. In general, it is false that the union of two simply connected regions is simply connected. Finding a counterexample is easy; just take the upper and lower halves of the annulus (these are both simply connected but their union is, as previously discussed, not).

The union of two simply connected spaces is in general not simply connected.

This counterexample also serves to illustrate that the intersection of two simply connected regions can also fail to be simply connected (this is more due to the fact, however, that the intersection of two path-connected regions is in general not path-connected). However, I believe the following weaker conjecture to be true, at least when we are working within the complex plane. I give a false “proof” below.

Conjecture. If the intersection of two simply connected regions is path-connected, then it must also be simply connected.

Invalid proof. If $A \cap B$ has a nontrivial fundamental group, then it contains (the image of) a curve which is not homotopic to a point. But $A \cap B \subseteq A$, obviously, hence $A$ also contains such a curve, which contradicts the fact that $A$ is simply connected.

The reason why the proof is false is because (generally speaking) a curve which is not homotopic to a point in one topological space may well be homotopic to a point in another space.

This conjecture does fail, for example, in $\mathbb{R}^3$: consider two spheres who intersect in a circle. The circle is path connected, but not simply connected. However, I can’t think of an example where it fails in the complex case.

It’s interesting that in one dimension, the intersection of any two path-connected subsets is automatically path connected, but this fails in the 2D case. By my conjecture, the path-connected intersection of any two simply connected subsets is also simply connected, but this fails in 3D case. This seems to be because all the concepts of connectedness are the same in one dimension. Once we get to two dimensions, we need to start talking about new stuff like simple connectedness. What about three dimensions? How do we measure the number of higher-dimensional holes in a space? For example, consider a 3-dimensional “annulus” $H$ (obtained by the description I previously gave; this can be viewed as the complement of a small sphere in a large sphere, just as a 2D annulus can be viewed as the complement of a small disk in a large disk). This set $H$ is actually simply connected, but it has a hole in it… This hole goes unnoticed by curve homotopy because curves are effectively one-dimensional (they are, after all, maps from an interval $[0,1]$ into some space).

Anyways, I’m diverging quickly from the topology of $\mathbb{C}$ now, so I’m going to end this post.

just another guy trying to make the diagrams commute.
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### 2 Responses to Simple connectedness

1. yureka says:

just a heads up, the last name portion of your email in the problem statement seems to be incorrect (baker -> rekab, not rakeb).

http://csclub.uwaterloo.ca/~mlbaker/mathematics/problems/complex_problem1.pdf

2. dx7hymcxnpq says:

LOL. Many thanks.