# Are all continous functions topologically conjugate to some piecewise linear function?

We've had a number of interesting questions on topological conjugacy on this site. This answer shows how to prove the tent map is topologically conjugate to the quadratic map, and this answer shows that another piecewise linear function can be shown to be topologically conjugate to $$F(x) = 4x^3 - 3x$$. This got me wondering, given any continuous function, can I find some piecewise linear function that is topologically conjugate to it?

• Hint: Think of a self-map $[0,1]\to [0,1]$ whose fixed-point set is the Cantor set. Apr 17, 2022 at 10:06
• This is an intriguing example, but I worry that I'm not getting the hint. It seems to me that a piecewise linear function could be designed to do this. Though in this case, it would need infinitely many 'pieces' to get the job done? Apr 17, 2022 at 11:06
• Then what is your definition of a piecewise-linear function? Apr 17, 2022 at 13:46
• Thinking about this a little more I see your point. If I let my piecewise-linear function have infinitely many segments the question is trivial since I can approximate any function arbitrarily closely. Otherwise, I can't account for maps with infinitely many fixed points. Do counterexamples with finitely many fixed points also exist? Apr 17, 2022 at 17:21
• Of course: Take a function $f: [0,1]\to [0,1]$ such that $f^{-1}(0)$ is the Cantor set. Such a function can be chosen to have only one fixed point. Apr 18, 2022 at 12:16