The question is as such:
If two continuous mappings $f$ and $g$ of a closed interval into itself commute, that is, $f\circ g=g\circ f$, then they do not always have a common fixed point.
-- Zorich Mathematical Analysis I
How prove this they have a common fixed point
This question has been asked once but has been phrased incorrectly since the author corrected the exercise in the second edition.
Also, I checked the chinese translation, it explicitly states that it should be closed interval, so I changed the question accordingly, which you can check the link. And in the question prior to that, it asked to show a mapping $[0,1]\to [0,1]$ always have a fixed point.And my proof is the following:
Let $f$ be a continuous function ,$g(x)=f(x)-x,$ we have that $g$ is continuous also.
If $f(0)=0\vee f(1)=1$, then the condition works.
If $f(0)\ne 0\wedge f(1)\ne 1$，and since that $f([0,1])\supset [0,1]$， We must have $f(0)>0$And $f(1)<1$
We thus get $g(0)>0$and $g(1)<0$Therefore there exists $c\in (0,1)$such that $g(c)=0\iff f(c)=c$