# Extending continuous maps from spheres to Euclidean spaces

Fix $$d\in\mathbb{N}$$. Consider the following sets as topological spaces with the subspace topology from $$\mathbb{R}^{d+1}$$. $$S^d = \{ (x_0,\ldots,x_d)\in\mathbb{R}^{d+1}\mid \sum x_i^2 = 1\}$$ $$D^{d+1} = \{ (x_0,\ldots,x_d)\in\mathbb{R}^{d+1}\mid \sum x_i^2 \leq 1\}.$$

Is it true that for every continuous map $$f:S^d\to\mathbb{R}^d$$, there is a continuous map $$g:D^{d+1}\to\mathbb{R}^d$$ such that $$g|_{S^d}\equiv f$$, and $$g(D^{d+1})\subseteq f(S^d)$$?

• The Hopf map $S^3\to S^2\subset \mathbb{R}^3$ doesn't have such a g, otherwise it would be homotopically trivial, which it isnt.
– Otis
Commented Jan 5, 2023 at 23:16
• @OtisChodosh Every map into $\mathbb{R}^3$ is homotopically trivial regardless of its domain. Commented Jan 5, 2023 at 23:52
• In fact every map $f:S^d\rightarrow \mathbb{R^d}$ extends over $D^{d+1}$. The condition $g(D^{d+1})\subseteq f(S^d)$ will not be possible in general. Commented Jan 5, 2023 at 23:52
• @Tyrone, Otis was referring to the nontriviality of the Hopf map to the sphere. This obstructs a filling of the form desired by the OP Commented Jan 6, 2023 at 1:10
• Thanks for the replies. So as I understand it, the Hopf map can be extended, but the image of $g$ can never be contained in the image of $f$. This is a counterexample when d=3, the OP obviously holds for d=0,1, what about d=2? Commented Jan 6, 2023 at 9:49

As requested, here is my comment in answer form:

This fails for $$d=3$$. The Hopf map $$f : S^3\to S^2\subset \mathbb{R}^3$$ cannot be homotoped (through maps to $$S^2$$) to a constant map. The existence of $$g : D^4 \to \mathbb{R}^3$$ with $$g(D^4)\subset f(S^3) = S^2$$ would precisely be such a homotopy.

For $$d=2$$ the question asks if a map $$f:S^2 \to \mathbb{R}^2$$ can always be homotoped to a trivial map inside of its image. I think the answer to this is "yes" it can: Ian Agol's answer here points to a paper of Cannon-Conner-Zastrow proving that $$\pi_2(f(S^2)) = 0$$. This yields the desired homotopy.

(Of course, as pointed out in the comments, for $$d=1$$ a continuous map $$f:S^1 \to \mathbb{R}$$ has an interval as an image, and thus is homotopically trivial in its image.)