The first observation one can make, is that S^2 is non-contractible and simply connected (trivial fundamental group), yielding that for n>=3 such subsets exist.

Secondly, for n=0,1 all connected subsets are immediately simply connected.

This leaves the case n=2. The example I came up with is the topoligists sine, with an extra piece adjoined:

**Spoiler:**

That is, an extra arc is added from some place on graph, to the origin.

In polar coordinates: r(\theta)=2+sin(4\pi^2/ \theta), \theta\in(0,2\pi]; Let X=r((0,2\pi])

It is pretty easy to show that there are no paths going through the origin (or (2,0) for the radial version ) , yielding that these have a trivial fundamental group. But whereas intuition tells me that this should be non-contractible, an actual prove has elluded me.

Then yesterday my friend suggested the following map (in polar coordinates):

Code: Select all

`F: X\times [0,1] \to X, `

(\theta, r(\theta), t)\mapsto { (\theta, r(\theta)) if \theta/2\pi <= 1-t

(\theta + (t-(1-\theta/2\pi)) * (2\pi-\theta)/(\theta/2\pi), r(...) else

Basically, keep a coordinate fixed "until its time has come" and than move it at a constant speed along the curve such that it ends up at \theta=2\pi exactly at t=1. This seemed surprisingly a contraction for a minute until we realized that the same trick than should work for circles as well.

So the question is:

Why doesn't "keep points fixed until they have to move; then move at constant speed" contract the circle? I know it doesn't because the circle has a non trivial fundamental group Z, but that doensn't help in the case I'm trying to study....