Let $X$ be a (non-empty) compact subset of $D(0,M):=\left\{x\in \mathbb{R}^n:\, \|x\|\leq M\right\}$, and let $f:X\rightarrow Y$ be uniformly continuous; for some metric space $Y$. Are there any known result guaranteeing that $f$ can be extended to a uniformly continuous function $F:D(0,M)\rightarrow Y$ such that the modulus of continuity of $f$ is (point-wise) greater-than or equal to the modulus of continuity of $F$?
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asked Mar 14, 2021 at 18:32
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$\begingroup$ If $X$ is the sphere, then possibility of extension means that $n-1$-th homotopy group of $Y$ vanish. Also, it doesn't say in the body of the question that modulus of continuity has to be preserved (as you mentioned in the title). Could you please clarify? $\endgroup$– erzCommented Mar 15, 2021 at 6:12
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$\begingroup$ I added the requested clarification. However, I don't see the homotopic argument. $\endgroup$– Catologist_who_flies_on_MondayCommented Mar 15, 2021 at 10:23
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$\begingroup$ one of the definition of having vanishing $n-1$-th homotopy group is precisely existence of extension of any continuous map $f:S^{n-1}\to Y$ to $F:D_{n}\to Y$ $\endgroup$– erzCommented Mar 15, 2021 at 16:47
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1$\begingroup$ See for example Hatcher - Algebraic topology, chapter on homotopy groups $\endgroup$– erzCommented Mar 16, 2021 at 17:42
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1$\begingroup$ I did not claim it does. It's just a necessary condition $\endgroup$– erzCommented Mar 19, 2021 at 18:11
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