Given a continuum $X$ (compact metrizable connected $X$) let $K(X)$ denote the hyperspace of nonempty compact subspaces of $X$ with the Vietoris topology and let $C(X)$ denote the (closed) subspace of $K(X)$ consisting of connected sets.


It is well known (and has been discussed on [MO before](https://mathoverflow.net/questions/354092/which-compact-metrizable-spaces-have-continuous-choice-functions-for-non-empty-c)) that if there is a continuous choice function $K(X)\to X$ then $X\simeq[0,1]$, indeed any other continua doesn’t even have a continuous choice function for the subspace of $K(X)$ of unordered pairs.


The situation is different when looking at $C(X)$: for example the “star” $X$ obtained by joining $n$ arcs at their endpoint has a continuous selection for $C(X)$ by looking at the tree partial order induced by choosing the common point as root and then mapping $C\in C(X)$ to its $\min$ with respect to this partial order. In a similar way we get a continuous selection for $C(X)$ whenever $X$ is a dendrite.


It is also easy to see that $S^1$ doesn’t have a continuous selection for $C(S^1)$ by using a combination of Borsuk-Ulam, $C(S^1)\simeq D^2$ and $\partial C(S^1)\cong F\_1(S^1)$, where $D^2$ is the unit disk in $\Bbb R^2$ and $F\_1(X)$ is the space of singletons of $X$.


This suggests the following: suppose $X$ is a continuum with a continuous selection $C(X)\to X$. Must $X$ be a subcontinuum of a dendrite? If not is there a characterization of the continua $X$ with a continuous selection $C(X)\to X$?



When does $C(X)$, $X$ a continuum, admit a continuous choice function?
I will post a CW answer to my own question to remove it from the unanswered list since Benjamin doesn't seem interested in turning his comment into an answer, but this is all informations taken from the section of *Hyperspaces* that he suggested.


To begin with let's fix the terminology *selectible* to refer to a continuum $X$ for which there exist a continuous selection $C(X)\to X$.


The simple answer to the question at the end, namely "is there a characterization of the selectible continua?" is no.


What is known (according to the book by Illanes and Nadler, I'm not aware of any recent developement however) is more or less the following:


* If $X$ is a selectible continuum, then $X$ is a dendroid.
* Say that a selection $s\colon C(X)\to X$ is rigid if whenever $A,B\in C(X)$ are such that $s(B)\in A\subseteq B$, then $s(A)=s(B)$ (note that the definition of rigid selection in *Hyperspaces* has a typo, this is the correct definition as given in the original reference *Rigid Selections and Smooth Dendroids* by L.E.Ward). Then there is a characterization of the rigidly selectable continua: those are exactly the smooth dendroids by the paper of Ward mentioned above. The idea here is that if $X$ is a smooth dendroid then the "tree" partial order (with root a point witnessing smoothness) is closed in $X\times X$, so the map sending each subcontinuum to its minimum is a rigid selection. Conversely if $s\colon C(X)\to X$ is a rigid selection, then $\{\{s(A)\}\times A\mid A\in C(X)\}$ is a closed partial order on $X$ which agrees with the "tree" order with root $s(X)$, so that $s(X)$ is a point witnessing the smoothness of $X$.
* Not all dendroids are selectible. Given two points $x,y$ in a dendroid let $[x,y]$ denote the unique arc joining them and let $(x,y)$ denote $[x,y]\setminus\{x,y\}$. Following definition 75.10 in *Hyperspaces* say that a dendroid $X$ is of type $N$ (a drawing justifies this name and clarifies the definition) if there are two points $p,q\in X$ and two sequences of arcs $[p\_np\_n'],[q\_n,q\_n']$ with points $p\_n''\in(q\_nq\_n')$ and $q\_n''\in(p\_n,p\_n')$ such that:
$$\begin{align} [p,q] &=\lim [p\_n,p\_n']=\lim [q\_nq\_n'] \\
 p &= \lim p\_n = \lim p\_n' = \lim p\_n'' \\
 q &= \lim q\_n = \lim q\_n' = \lim q\_n''\end{align}.$$
Then any dendroid of type $N$ is nonselectible.  

A simple example of a dendroid of type $N$ is the doubly infinite broom: start with the segment between $(0,0)$ and $(1,0)$ in $\Bbb R^2$ and add segments between $(0,0)$ and $(1,1/n)$ and between $(1,0)$ and $(0,-1/n)$ for all $n\in\Bbb N^+$.
* More surprisingly there are even contractible dendroids that are not selectible, this is example 75.9 in *Hyperspaces* and the construction is originally due to Maćkowiak.
* selectibility is not preserved by neither monotone nor open maps. Nonselectibility isn't either, this is exercise 75.28 in *Hyperspaces*.
* There are various open questions about selectibility apart from a classification of selectible continua, some of them are in chapter 76 of *Hyperspaces*.


