Re: Extending a function

From: Daniel Grubb (grubb_at_lola.math.niu.edu)
Date: 08/03/04 

Date: 3 Aug 2004 13:52:45 GMT

>>>>>> Let X and Y be topological spaces, let A be a closed subset of X, and
>>>>>> suppose f:A-->Y is continuous. Can f be extended to a continuous
>>>>>> function on X?
>>>>
>>>>>OK with good conditions. Maybe X is normal, Y is completely regular.
>>>>
>>>>
>>>>No. Not even then. Let X be the closed disk in R^2 and A=Y=unit circle.
>>>>Let f be the identity. Then f has no extension to X. All of these spaces
>>>>are about as nice as you want in terms of compactness, connectedness
>>>>and separation properties.
>>
>>>huh. we all know the result is true under various conditions on X if
>>>Y = R. this raises the question of what's so special about R; there's
>>>-some- topological condition on Y that suffices...
>>
>>
>>Define Y to be an 'absolute retract' if whenever A is a closed subset
>>of a normal space X and f:A->Y is continuous, then f can be extended
>>to X. The Tietze extension theorem says that [0,1] is an absolute retract.
>>As you pointed out, so is the real line. Products of absolute retracts
>>are again absolute retracts. Also, an absolute retract *is* a retract
>>of every normal space it is embedded in as a closed subspace, i.e. if
>>Y is homeomorphic to a closed subspace A of the normal space X, then
>>there is a map p:X->A with p(a)=a for all a in A.

>and hmm, a retract of an absolute retract is an absolute retract...

True. So this means any space that is a retract of some cube is
an absolute retract. This gives quite a few spaces.

>[but as far as i can see what this says about my question is that
>absolute retracts are absolute retracts.]

Yes, indeed :). On the other hand if Y is an absolute retract
such that Yx[0,1] is normal, then Y is contractible. Just consider
Y as a subset of it's cone and extend the identity function. My
copy of Spanier has an exercise that says the converse holds for
absolute neighborhood retracts, but I haven't figured out the
proof. So for ANRs contractibility is equivalent to being an AR, it
seems.

We can also approach the topic from the other side: Which pairs
(X,A) have the property that every f:A->Y extends to X? Here, it
is easy to see that if A is a retract of X, we're good. The converse
also holds by letting Y=A. This extends the previous examples
where A is clopen in X.

>>There are many variants of the idea of 'absolute retract' such as
>>'absolute neighborhood retract' and 'absolute Euclidean neighborhood
>>retract'. All of these are defined similarly.
>>
>>--Dan Grubb

>David C. Ullrich

--Dan Grubb

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