specifying the restriction of a homeomorphism I have topological spaces $A$ and $B$ and subspaces $C \subset A$ and $D \subset B$.
Say I have homeomorphisms $h: C \to D$ and $f: A \to B$. I also know that $f|C : C \to D$ is a homeomorphism. 
Using these facts, can I make a homeomorphism $g: A \to B$ such that $g|C = h$?
 A: Take $A=B=[0,3)$ and $C=D=[1,2]$ (meaning intervals in $\Bbb R$).
The identity map $f\colon A\to B$ is a homeomorphism, and of course so is its restriction $C\to D$. However, consider the map $h\colon C\to D$ given by $h(x)=3-x$.
Suppose we can extend $h$ to a function $g\colon A\to B$. Since $h$ is decreasing on $[1,2]$, $g$ must be decreasing on $[0,3)$. However, $g(1)=h(1)=2$, and by the extreme value theorem, $g(0)$ is some number between $2$ and $3$, contradicting the fact that $g$ is decreasing and surjective.
A: No: Let $A=B=[0,1)$ and $C=D=(0,1)$, let $f$ be the identity map, and let $h(t) = 1-t$.
A: Take $A=B=\mathbb{R}$ and $f(x)=x$. It is an homeomorphism.
Take $C=D=\mathbb{R}^*_+$ and $h(x)=1/x$. Again it is an homeomorphism. $f|C$ is also an homeomorphism. 
You cannot extend $h$ to a continuous function on $\mathbb{R}$, especially in $0$. In particular you cannot find the desired $g$.
A: Exploiting non-homogeneity seems easiest: $A=B=[0,1)$, $f(x)=x$, in the usual topology. $C=\{0\}$, $D=\{\frac12\}$ and $h$ the only map between them. No homeomorphism of $[0,1)$ exists that maps $0$ to $\frac12$ as $0$ is a non-cutpoint of $A$ and $\frac12$ is a cutpoint of $B$, etc. 
