# 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 $$g|C = h$$?

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.

No: Let $$A=B=[0,1)$$ and $$C=D=(0,1)$$, let $$f$$ be the identity map, and let $$h(t) = 1-t$$.

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 $$g$$.

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.