Exercise 4.26

Question

Exercise 26: Suppose $X,Y,Z$ are metric spaces, and $Y$ is compact. Let $f$ map $X$ into $Y$, let $g$ be a continuous one-to-one mapping of $Y$ into $Z$, and put $h(x)=g(f(x))$
    for $x\in X$. Prove that $f$ is uniformly continuous if $h$ is uniformly continuous. Prove also that $f$ is continuous if $h$ is continuous. Show that the compactness of $Y$
    cannot be omitted from the hypotheses, even when $X$ and $Z$ are compact.

analambanomenos · Accepted Answer

Since $g$ is one-to-one, it defines a function $g^{-1}$ from $g(Y)$ to $Y$, which is continuous by Theorem 4.17. If $h$ is continuous, then $f=g^{-1}\circ h$ is continuous
    by Theorem 4.7, and if h is uniformly continuous, then $f$ is uniformly continuous by Exercise 12.

Let $X=Z=[0,2]$ and $Y=[0,1)\cup[2,3]$. Let $f(x)=x$ for $0\le x<1$ and $f(x)=x+1$ for $1\le x\le 2$. Let $g(x)=x$ for $0\le x<1$ and $g(x)=x-1$ for $2\le x\le 3$. Then $X$ and $Z$
    are compact, $Y$ is not compact, $g$ is continuous and one-to-one, $h(x)=g(f(x))=x$ for $0\le x\le 2$ is uniformly continous, but $f$ is discontinuous at $x=1$.

Exercise 4.26

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Exercise 4.26

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