Sets homeomorphic to convex sets

Question

I have recently encountered the "Brouwer fixed-point theorem". The theorem states (per Wikipedia) that any convex compact set has the fixed point property. This statement struck me as odd as the theorem requires convexity, which isn't a topological property (i.e preserved by homeomorphisms) while the fixed-point property is.

My question then: Is there a simple or well known topological property which is equivalent to "homeomorphic to a convex set"

In the case of $A \subseteq \mathbb{R}$ it's not hard to see that $A$ is homeomorphic to (and is) a convex subset of $\mathbb{R}$ if and only if it is connected. My intuition tells me that in the case of $\mathbb{R}^2$ this extends to simply connected sets, and in higher dimensions to sets where all homotopy groups are trivial, but I don't know if my intuition is right on this.

EDIT: thinking on the comments and answers so far it seems another necessary condition is that the set also be locally connected. However that is still not sufficient as any 'Y shaped' subset of $\mathbb{R}^2$ still isn't homeomorphic to a convex set. Would being a simply connected manifold, possibly with boundary, be sufficient in $\mathbb{R}^2$?

Please note when answering that I'm mostly familiar with point-set topology and only know very little about algebraic topology.

Answer 1

If $A\subset\mathbb{R}^2$ open and simply connected, then by Riemann mapping theorem it is homeomorphic to the unit disc.

Let $A\subset \mathbb{R}^n$ open and homeomorphic to a convex set $K\subset \mathbb{R}^n$. Then $K$ has to be open as well (invariance of domain) and by this argument it has to be homeomorphic to $\mathbb{R}^n$. Hence $A$ is a contractible manifold. However, for $n\ge 3$ there are contractible manifolds (such as the Whitehead manifold - see last link) which are not homeomorphic to $\mathbb{R}^n$.

Answer 2

If I haven't made a mistake, the following characterization works:

A subset $X$ of $\mathbb{R}^n$ is homeomorphic to a convex set if and only if it is homeomorphic to the union of an open disc with a subset of its boundary.

The "if" direction is pretty clear, and here's a sketch of a proof of the "only if":

Without loss of generality, assume $X$ is bounded (we can map all of $\mathbb{R}^n$ to the unit sphere homeomorphically).

First, if $X$ is contained in a hyperplane of dimension $n-1$, proceed by induction on $n$. Otherwise, the set must contain at least $n+1$ points that are not contained in a hyperplane of dimension $n-1$, so by convexity we can see there's an open ball in $X$ (contained in the convex hull of these points).

Choose a point $p$ in this ball. We can think of $X$ as a union of intervals emanating from $p$, such that their only point of intersection is $p$. We can identify this set of intervals with $S^{n-1}$, call this mapping $s$. Essentially here we are making $p$ the "center" point, and drawing all the line segments from it to the other points in the set $X$, and a point in $S^{n-1}$ corresponds to a "direction" of a segment coming out of $p$.

Now consider the map $d:S^{n-1}\rightarrow \mathbb{R}$, which takes each "direction" to the length of the segment from $p$ emanating in that direction. This map is convex and therefore continuous. You can thus apply a mapping to $X$ by shrinking each point's distance to $p$ proportionally to the full length of the interval it occupies, and this map is a homeomorphism.

Under this transformation, each interval is transformed into either a closed or half-open interval emanating from $p$ of length 1. Since $p$ was an interior point of $X$ such intervals emanate in all directions, so the set is exactly an open ball together with a subset of its boundary.