We say that a topological space $(X,\tau)$ is *flexible*, if for every closed discrete subset $D\subseteq X$ and every map $f: D\to X$ there is a continous map $f^X:X\to X$ such that $f^X|\_D = f$.


$\mathbb{R}$ with the Euclidean topology is flexible.


Is every [homogeneous](https://en.wikipedia.org/wiki/Homogeneous_space) $T\_2$-space flexible?


**Note.** I called spaces with the above property "flexible" because they are in some sense the opposite of [(strongly) rigid](https://mathoverflow.net/questions/188729/strongly-rigid-hausdorff-spaces) spaces; flexible spaces admit a lot of continuous self-maps, and strongly rigid spaces only very few.



Are homogeneous $T_2$-spaces flexible?
I like this question a lot. 


The answer is no. Let $X$ consist of two disjoint copies of the real line. This is $T\_2$ and homogeneous, in the sense that for any two points, there is a homeomorphism taking the first to the second — all points look alike.


But it is not flexible, since if $D$ consists of two points $a$ and $b$, both in the same copy, and we let $f(a)$ be in one copy and $f(b)$ in the other, we cannot extend this to a continuous map on the whole space, since the image of the copy of $\mathbb{R}$ from which $a$ and $b$ came must be connected, but it cannot be.


(This answer suggests that one should consider a stronger notion of homogeneity, for example, insisting not just that the group of self-homeomorphisms acts transitively, but that every finite permutation extends to a self-homeomorphism. For example: if a space is flexible with respect to finite sets $D$, then is it flexible with respect to all closed discrete sets?) 


