# Arcwise-connectedness generalized to higher connectivity?

This is a crosspost from stackexchange. I'm not completely sure whether the question below is research-level, but I have not yet found an obvious answer, and what I have found thus far suggests that it might be research-level.

A topological space $X$ is path-connected if for points $a, b \in X$, there is a continuous function $f : [0, 1] \to X$ such that $f(0) = a$ and $f(1) = b$. This condition can be strengthened to arcwise-connectedness, which additionally requires that $f$ is a topological embedding, or homeomorphic onto its image. Clearly if $X$ is arcwise-connected, then it is also path-connected. If $X$ is Hausdorff, then the converse also holds. There is a not-so-trivial proof of the converse in Chapter 31 of Willard's General Topology on Peano spaces (there is also some related discussion on nlab).

A space is called $n$-connected if all its homotopy groups vanish up to $n$, or $\pi_i(X) = 0$ for all $i \le n$. Equivalently, $X$ is $n$-connected for any $i \le n$, and continuous function $f : S^i \to X$, we can extend it to a continuous map $F : D^{n+1} \to X$. This generalizes path-connectedness, it is the same as $0$-connectedness.

Can the "path-connectedness implies arcwise-connectedness" property be generalized to higher connectivity? That is, suppose we say that $X$ is "$n$-arcwise-connected" if for any $i \le n$ and embedding $f : S^i \to X$, there is an embedding $F : D^{i+1} \to X$ extending $f$. Are there sufficient conditions on $X$ (like $X$ is Hausdorff) for which we have $n$-connectedness implying $n$-arcwise-connectedness? Are there known results on this?

My motivation is: given an embedding $S^i \to X$ in an $n$-connected space, I want to extend it to an embedding $D^{i+1} \to X$. $n$-connectedness alone doesn't immediately give me this.

• If I understand the question correctly, then even such a nice space as $\mathbb{R}^3$ fails to be 2-arcvise connected, because of stuff like a horned sphere. – erz Apr 23 '18 at 5:34
• What is $n$-connectedness? If you give a definition for path-connected, I think you should give a definition for this lesser known concept. – Jean Duchon Apr 23 '18 at 9:46
• @JeanDuchon I added a definition of $n$-connected. – Vikram Saraph Apr 23 '18 at 13:32
• @erz could you say a bit more about why exactly the alexander horned sphere shows that $\mathbb{R}^3$ would not be $2$-arcwise connected? Which embedding of $S^2$ is it exactly that fails to extend to $D^3$? I know that the horned sphere is a kind of pathological embedding of the 2-sphere in $\mathbb{R}^3$, but doesn't it still extend to the $3$-ball? – Vikram Saraph Apr 23 '18 at 13:34

Take $X= \mathbb{R}^3$. This space is as nice a space as you could ever hope for. It is also contractible, so in particular it is simply connected (1-connected).
Now take an embedded $S^1$ in $\mathbb{R}^3$, also called a knot. Your question asks whether it is always possible to find an embedded disk which bounds the given knot. The answer is an emphatic no!, unless the knot is the unknot. In fact in some sense the whole field of knot theory is the study of the failure of the property you seek (in the easiest case of simple connectivity).