Suppose $X, Y, Z$ are metric spaces and $Y$ is compact. Let $f$ maps $X$ into $Y$. Let $g$ be continuous one-to-one map $Y$ into $Z$ and put $h(x)=g(f(x))$ for $x$ in $X$. If $h(x)$ is uniformly continuous then $f$ is
a) Continuous
b) Uniformly continuous
c) May be discontinuous
d) strictly increasing
My try:
As $g$ is continuous and $Y$ is compact. As we know continuous Image of compact set is compact $g(Y)$ is compact and hence uniformly continuous.$\qquad(1)$
$h(x)$ is uniformly continuous means for every $\varepsilon>0$ there exists $δ>0$,which only depends on $\varepsilon $ such that $ |h(x)-h(y)|<\varepsilon $ whenever $ |x-y|<δ$.
And from $(1)$, if $u, v$ in $Y$ such that $ |g(u)-g(v)|<\varepsilon$ whenever $ |u-v|<δ $.
Put $u=f(x)$ and $v=f(y)$ we get $ |f(x)-f(y)|<\varepsilon $ whenever $ |x-y|<δ $ . Hence $f(x)$ is uniformly continuous.. I don't know whether I am going in right way or not.
It is true that $f$ is uniformly continuous but the argument requires some clarity. Define $g_1:g(Y)\to Y$ to be the inverse of $g$. [When you consider $g$ as a map between the compact spaces $Y$ and $g(Y)$ it becomes a homeomorphism since the domain and range are both compact Hausdorff spaces]. Now $g_1$, being a continuous map on a compact metric space, is uniformly continuous. Composing $h$ and $g_1$ we see that $f$ is uniformly continuous.