Okay the problem goes like this:
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. Hint : $g^{-1}$ has compact domain $g(Y)$, and $f(x)=g^{-1}(h(x))$. Prove also that $f$ is continuous if h is continuous.
Show (by modifying Example 4.21, or by finding a different example) that the compactness of Y cannot be ommited from the hypotheses, even when X and Z are compact.
The problem is in general I cannot seem to understand why Y needs to be compact. I know does because I found a counter example namely:
Let $X=[0,1]$, $Y=[0,2\pi)$, and $Z=${$(x_r,y_r)\in \mathbb{R^2}:x^2+y^2=1 $}. If $f:X\to Y$ where, $f(x)=2\pi x$ for all $x\in X$ except for $x=1$ ,where $f(1)=0$, then $f$ discontinuous at $x=1$. Also if $g:Y\to Z$, where $g(y)=(\cos{y},\sin{y})$ then $g$ is continuous and injective, and the mapping $h=g\circ f$ is continuous.
But as far as why $Y$ must be compact in general for $f$ to be continuous (when $h$ and $g$ are continuous of course), I am sort of stuck.
Specific questions
- Might I need to use the compactness of $Y$ to show that for every open set $U\subset Y$ the set $h^{-1}\circ g(U)$ is open? If so please point me in some direction as to how to use compactness for that.
