When does an Embedding extend into a Homeomorphism? This is from a post in sci.math that did not get a full answer; I may repost it for the OP there:
I am  interested on the issue I read in another site of when an embedding from a closed set extends into a homeomorphism, i.e., if $C$ is closed in $X$, and $f:C \rightarrow Y$ is an embedding, when can we extend $f$ to $F$ so that $F:X \rightarrow Y$ is a homeomorphism, and $F|_C=f$ (i.e., $f=F$ in $C$ )?  Of course there are trivial cases
like when $f$ is the identity on $C$:
I know of, e.g., Tietze Extension, and I think there are results about extending
maps from a space into its compactification; I think the map must be regular (inverse
image of a compact set is compact). But I don't know of any general result.
I will learn Latex as soon as I can; my apologies for using ASCII
 A: gary, I think your question is too general. For instance, there is something called 'cofibration' in topology, which deals with this type of problem under a strong condition: Namely, $f:A\rightarrow Y$ is continuous and has an extension iff every $g\colon A\rightarrow Y$ satisfies this property where $g$ is homotopic to f. 
If a space $(X,A)$ has homotopy extension property with respect to $Y$ then it is easier to check if a map $f\colon A\rightarrow X$ has an extension, because , now, there are billions of maps that should simultaneously have extensions. However, even under this condition, there is no general theorem that I know.  
A: This will never happen unless $C=X$. Consider the identity embedding $C \to C$; by your hypothesis, this extends to a homeomorphism $X \to C$ extending $C \to C$. This is an inverse to the inclusion $C \to X$. Thus $C=X$.

In general, this is unlikely. Consider any $Y$ between $X$ and $C$. Then there is an embedding $C \to Y$ which by your condition extends to a homeomorphism $X \simeq Y$. So $X$ is homeomorphic to all its closed subspaces that contain $C$ (with a homeomorphism that extends $1: C \to C$). This generally will not happen in natural examples (though it will if $X$ is a [Toronto space][1] and $C$ has the same cardinality as $X$): for instance, it means that $X/C$ is homeomorphic to any of its closed subspaces.

A: Thanks both for your comments.
I am sorry if I did not make this clear; I don't know if you were assuming this,
   but the embedding is into Y, and not onto it, i.e., we have an embedding
   e:C-->e(C) , and not e:C-->Y, which,  as you said, cannot be 1-1. I can see
   how my using the identity on C was misleading, sorry; what I meant was if, e.g.,
   we had a map from (0,1) as a subspace of R embedded by identity into another
   copy of (0,1) in R. Then we would define f^ (R-(0,1))->R-(0,1) by f(x)=x (though
   there are many other options, using, e.g., bump functions and partitions of unity.)
   But R, R^n are too nice in too many ways, and it would be great to have more general
   results.
And, re your second question, Akhil: I would like , if workable, to fix Y, and
   not look for a result for all possible spaces Y, which may be too ambitious for
   me at this point. so maybe we could look for properties that Y may have to be
   able to have this extension.
