Taking the maximum/supremum under the inverse image of a proper map

Question

Let $f: [0,\infty) \rightarrow [0,\infty)$ be a proper (inverse image of compact is compact) and continuous map, with $f(0)=0$. Then there is a well-defined map in the reverse direction, settting $g(y) := \max(f^{-1}[0,y])$. Is $g$ then continuous?

Answer 1

Let $f(x)$ be $x$ on $[0,2]$, then $4-x$ on $[2,3]$ and then $x-2$ on $[3,+\infty)$. For $y\in[0,1)$ we have $g(y)=y$, but $g(1)=3$.