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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

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3 Answers 3

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.

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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.
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what is $A$ supposed to be in your answer? I don't see it in the question. –  Pete L. Clark Feb 4 '11 at 3:18
    
@Pete: Dear Pete, thanks for the correction: I meant $C$. –  Akhil Mathew Feb 4 '11 at 3:19
    
okay, but I still feel like I am missing something. If we start with a homeomorphism $F: X \rightarrow Y$, restrict $F$ to a closed subset $C$ of $X$ and call if $f$, then it seems to me that we have an affirmative answer to the OP's question not covered by your response. No? –  Pete L. Clark Feb 4 '11 at 3:23
    
@Pete: I was under the assumption that the OP wanted the result to hold for all spaces $Y$. I may have been mistaken -- can the OP clarify? –  Akhil Mathew Feb 4 '11 at 3:27
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I was considering explaining my situation more clearly, but after the 2-3 downvotes I don't see the point. Specially 2 downvotes to someone who was at that point clearly a newbie. –  gary Jul 25 '11 at 22:15

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.

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Are you the same user as gary? If so, would you like me to merge your accounts? –  Zev Chonoles Jul 10 '11 at 15:14

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