Given $f$ from the cylinder $C$ to the interval constant on one boundary, is there a $r:C\to C$ constant on a boundary with $f\circ r = f$? My question might be trivial, but my lack of knowledge of this particular subject has not enabled me to find the answer. What I want to know is the following. Let $I=[0,1]$ and $C=S^1\times I$ be the cylinder. Given a continuous $f:C\to I$ which is onto and constant on a neighborhood of the top boundary $S^1\times\{1\}$, is there an $r:C\to C$, constant on $S^1\times\{1\}$, such that $f\circ r = f$ ?
Of course, if there is a continuous $s:I\to C$ such that $f\circ s = id_I$, then letting $r = s\circ f$, we get $f\circ r = (f\circ s)\circ f = f$, but such an $s$ does not always exist.
It is easy to see that if the range is $I^2$ instead of $I$, then the answer is negative: take for instance $f$ to be the quotient map collapsing all of $S^1\times[\frac{1}{2},1]$ to a point, this yields a disk homeomorphic to $I^2$. If $f\circ r= f$, then $r$ is the identity on $S^1\times[0,\frac{1}{2})$. But then $r$ cannot send the top boundary to a point for evident homotopic reasons.
This question can be seen in the context of continuous selection properties: define $\Phi$ from $C$ to the closed subsets of $C$ as $\Phi(x) = f^{-1}(\{f(x)\})$. A continuous selection for $\Phi$ is a continuous $r:C\to C$ such that $r(x)\in\Phi(x)$, that is, $f\circ r = f$. But my knowledge of this particular subject is close to nothing, as I have only read parts of classical articles by E. Michael in which it seems that there is an always present assumption that $\Phi$ is lower semi-continuous, which is not the case in my situation.
For information, this question arose while inspecting "stagnation properties" of real valued maps of non-metrizable surfaces.
 A: Posting the question here made me think about it more thoroughly, and I believe that the answer is negative. More precisely, with $C=\mathbb{S}^1\times[0,1]$, there is a $f:C\to I$, constant on $\mathbb{S}^1\times[\frac{1}{2},1]$, such that any $r:C\to C$ satisfying $f\circ r=f$ sends $\mathbb{S}^1\times\{\frac{1}{2}\}$ to a loop which ``turns once around the cylinder'' and is thus non-homotopic to a constant map. (In this answer all functions are assumed to be continuous.) It follows that $r$ cannot send $\mathbb{S}^1\times\{1\}$ (or any non-nullhomotopic circle) to a constant.
The idea for $f$ is the following. The details are a bit tedious, so I'll skip some details in the proofs. (I hope that I did not miss something, by the way.) We see $\mathbb{S}^1$ as $I$ with identified endpoints. For each $n\ge 0$, set $D_n$ to be the subspace $[2^{-n-1},2^{-n}]\times[0,\frac{1}{2}]\subset C$. We define $f$ such that if $f\circ r=f$, then $r(D_n)\subset D_n$ for each $n$, which implies that $r(\mathbb{S}^1\times\{1\})$ turns once around the cylinder.
To this end, we start with families of interval maps. Let $p_u:I\to I$ ($0<u<\frac{1}{2}$) take values $0,\frac{1}{2}+u,\frac{1}{2}-u,1$ at $0,\frac{1}{3},\frac{2}{3},1$, respectively.
It is not difficult to check that if $p_u\circ s = p_v$ for some $s:I\to I$, then $u=v$. We then let $p_{u,n}$ to be as in the graph below, that is $p_{u,n}$ is equal to $p_u$ suitably rescaled by an affine map in each ``box'' $[2^{-k-1},2^{-k}]\times[2^{-k-n-1},2^{-k-n}]$.

Looking at what happens in boxes $[2^{-k-1},2^{-k}]$, we see that if there is $s:I\to I$ with $p_{u,n}\circ s = p_{v,m}$ then $u=v$. We may finally define $\varphi_n:B\to B$, where $B$ is the closed unit ball in $\mathbb{R}^2$, as $p_{\frac{1}{n+3},n}\circ j$, where $j:B\to I$ is $j(x) = 1-|x|$.
Letting $i:I\to B$, $i(t) = \langle 1-t,0\rangle$, then $j\circ i = id_I$.
Hence, if there is $s:B\to B$ with $\varphi_n \circ s = \varphi_m$, then
$$ p_{\frac{1}{n+3},n}\circ j \circ s \circ i = \varphi_n \circ s \circ i =  \varphi_m\circ i = p_{\frac{1}{m+3},m}\circ j\circ i = p_{\frac{1}{m+3},m},$$
hence $n=m$.
To finally define $f$, we take homeomorphisms $\psi_n:D_n\to B$ and define $f$ restricted to $D_n$ to be equal to $\varphi_n\circ\psi_n$ and to $0$ elsewhere. Since $\varphi_n$ is equal to $0$ on the boundary of $B$ and has maximal value $2^{-n}$, $f$ is continuous, and $r(D_n)\subset D_n$ for any $r:C\to C$ with $f\circ r = f$.
