Let $f,g: I \to I := [0,1]$ be continous functions satisfying $f \circ g = g \circ f$. Does 
there exist $x\_0 \in I$ such that $f(x\_0) = g(x\_0)$ ? 


Background: In a homework the problem was posed with $g=\operatorname{id}$ (where 
it can easily be solved with the help of the intermediate value theorem). The 
lecturer said the stronger statement above is true, but he didn't know a proof. 
I googled a little around, but could only find something about the "commuting 
function problem" (existence of a common fixed point of $f$ and $g$) that is 
known to be false. 



Have commuting functions a common value ? 
Yes. The set of fixed points of $g$ is closed, nonempty, and is mapped into itself by $f$. Letting $a\le b$ be, respectively, the minimum and maximum fixed points of $g$, we have $f(a)\ge a=g(a)$ and $f(b)\le b=g(b)$. So, by the intermediate value theorem, there is an $x\in[a,b]$ with $f(x)=g(x)$.


Also, to reiterate the points made in the comments, this is a difficult problem for more general domains. The case of commuting maps on the closed disc has been [asked before](https://mathoverflow.net/questions/3332), and is still open. In fact, even the case of commuting maps on the simple triod (i.e., a capital 'T') appears to be an open problem, according to the contributed problem from Jeff Norden [here](http://notch.mathstat.muohio.edu/balogh_conference/booklet.ps) (*Commuting, coincidence-point-free maps on a triod*).


