Elementary geometric characterization of spheres?

Question

I've read the following two theorems.

Theorem. A compact connected metric space whose points are cuts points with the exception of at most two is homeomorphic to the unit interval.

Theorem. A compact connected metric space which becomes disconnected upon removing any two points is homeomorphic to the unit circle.

Are there analogous characterizations for $S^n$ using higher-connectedness, i.e triviality of homotopy groups?

Answer 1

In 2-dimensions you might want the Kline sphere characterization theorem.

Answer 2

In higher dimensions this is very difficult and, I think, there is no simple characterization along the lines of the ones in low dimensions. The main difficulty comes from characterization of compact manifolds among, say, compact metrizable topological spaces. The characterizations of spheres in dimensions 1 and 2 are (implicitly) based on the fact that in these dimensions homology manifolds are the same as topological manifolds. This is far from being true in higher dimensions. Until 1995 there was at least a conjecture describing which homology manifolds are topological manifolds (Cannon's conjecture). If this conjecture were true then topological manifolds of dimension $\ge 5$ would be characterized as ANRs which are homology manifolds satisfying the "disjoint disk property". Cannon's conjecture was disproved in the paper "Topology of Homology Manifolds" by J. Bryant, S. Ferry, W. Mio and S. Weinberger (Annals of Math., 1996). Now, as far as I known, there is no even a simple-minded conjecture characterizing topological manifolds and, in particular, spheres, of dimension $\ge 5$. In view of these results, the only (known to me) topological characterization of $S^n$ is the following:

Theorem. Suppose that $X$ is a compact metrizable space. Assume also that $M$ is ANR (an absolute neighborhood retract), is an $n$-dimensional homology manifold, has DDP (disjoint disk property) and $i(X)=1$ where $i$ is Quinn's invariant. Assume also $X-x$ is contractible for some (every) $x\in X$. Then $X$ is homeomorphic to $S^n$, provided $n\ge 5$.

This theorem contains many words (likely) unfamiliar to you, and, depending on your background, you may as well spend 5 years just learning what do they all mean.