Consider $\varphi:M\rightarrow N $ surjection satisfying the following property: exists $c>0$ such that for $x,y\in N, X = \varphi^{-1}(x)$ and $ Y = \varphi^{-1}(y)$, we have $d(X,Y)\leq cd(x,y)$. Then, if $f:N\rightarrow P$ is such that $f\circ \varphi: M\rightarrow P$ is uniformly continuous, prove that $f$ is also uniformly continuous.
I've been thinking of this question for a long time and i couldn't find a way for it. I think that i must take $\epsilon$ realated to the given constant $c>0$ and then use the hyphotesis of $f\circ \varphi$ being uniformly continuous. But i can't see exactly how.
Can someone please help?
Given $\varepsilon>0$, since $f\circ \phi$ is uniformly continuous, there is $\eta>0$ such that for all $z,w\in M$ such that $d(w,z)<\eta$, $d(f\circ\phi(w),f\circ \phi (z)) < \varepsilon$. Pick $0<\delta < \eta/c$. I claim this $\delta$ works for the uniform continuity of $f$.
Indeed, if $x,y\in N$ are such that $d(x,y)<\delta$, then $$ d(\phi^{-1}(x),\phi^{-1}(y)) \leq cd(x,y) < \eta. $$
Since $\eta$ is greater than $\inf \{d(z,w); z\in \phi^{-1}x,w\in \phi^{-1}y\}$, there are $z\in \phi^{-1}(x)$ and $w\in \phi^{-1}(y)$ such that $d(z,w)<\eta$. Therefore, $$ d(f(x),f(y)) = d(f\circ \phi(z),f \circ \phi(w)) < \varepsilon, $$
as desired.