Let $f:X\rightarrow Y$, $g:Y\rightarrow Z$ be uniformly continuous functions between metric spaces $X,Y,Z$ with moduli of continuity $\omega\_f$ and $\omega\_g$, respectively. Suppose that we know that $g\circ f$ has modulus of continuity $\omega$ then can we express $\omega\_g$ as a functions of $\omega$ and of $\omega\_f$?


I was thinking: Since $\omega = \omega\_g\circ \omega\_f$ then
$$
\omega\_g = \omega \circ \omega\_f^{-1};
$$
where $\omega\_f^{-1}$ is a right-inverse of $\omega\_f$. However, can we guarantee that such a right-inverse exists?


In the case where $\omega\_f(x)=K|x|$ then can we at-least say that $\omega\_g(x)\leq \max\{1,K^{-1}\}\omega(x)$?



Inferring the modulus of continuity
$\newcommand\om\omega\newcommand\R{\mathbb R}$In general, even the inequality
$$\om\_g\le\om\circ\om\_f^{-1}\tag{0}$$
will not hold, for the right inverse $\om\_f^{-1}$ of $\om\_f$ defined by
$$\om\_f^{-1}(u):=\inf\{t\in\R\colon\om\_f(t)\ge u\}\tag{1}$$
for real $u\ge0$.
Indeed, suppose e.g. that $X=\{0\}\cup(1,3/2]$, $Y=Z=\R$, $f(x)\equiv x$, and $g(y)\equiv y$. Then $\om\_f(1)=1/2$ and $\om\_f(u)\ge1$ for $u>1$, whence $\om\_f^{-1}(1)=1$ and
$$\om\_g(1)=1\not\le1/2=\om\_f(1)=\om\_f(\om\_f^{-1}(1))=\om\_g(\om\_f(\om\_f^{-1}(1)))=\om(\om\_f^{-1}(1))=(\om\circ\om\_f^{-1})(1).$$




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However, if $\om\_f$ is right-continuous, then, by (1), for all real $u\ge0$ we have $\om\_f^{-1}(u)=\min\{t\in\R\colon\om\_f(t)\ge u\}$ and hence
$\om\_f(\om\_f^{-1}(u))\ge u$, which implies
$$\om\_g(u)\le \om\_g(\om\_f(\om\_f^{-1}(u)))=\om(\om\_f^{-1}(u))),$$
so that (0) holds.


In particular, if $\om\_f(t)\equiv Kt$ for real $t\ge0$, then $\om\_f^{-1}(u)=u/K$ for all real $u\ge0$ and hence $\om\_g(u)\le\om(u/K)$ for real $u\ge0$.


