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Let $X, Y$ be topological spaces and let $S \subseteq X$ be equipped with the subspace topology. Suppose $f: S \rightarrow Y$ is continuous. Under what conditions can we guarantee there exists a continuous function $\tilde{f}: X \rightarrow Y$ extending $f$ (i.e. $\tilde{f}|_S = f$)?

It appears to me that this isn't always possible. For example, let $\Omega$ be the least uncountable ordinal, and let $X = \Omega \cup \{ \Omega \}$ the be set of all ordinals up to and including $\Omega$ equipped with the order topology. Then it seems that the identity map $1: \Omega \rightarrow \Omega$ can't be extended to $\tilde{1}: \Omega \cup \{\Omega\} \rightarrow \Omega$.

I'm wondering what the obstruction is here, exactly, because it seems like the same sort of argument could work for a one-point compactification (of, say, $\mathbb{R}$), which is a nicer space than these ordinals.

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There is indeed no map extending the identity $1 : \omega_1 \rightarrow \omega_1$ to $\tilde{1} : \omega_1 \rightarrow (\omega_1 + 1)$. This is because if $\tilde{1}$ existed, it's image would be some $\alpha < \omega_1$ and then the inverse image $\tilde{1}^{-1}[(\leftarrow, \alpha+1)]$ is countable (as $1$ is 1-1 and so $\tilde{1}$ is at most 2-to-1, and the neighbourhood is countable) but should be an open subset containing $\omega_1$, all of which are uncountable.

A function from $X$ to some space $Y$ can be extended to the one-point compactification of $X$, $\alpha X$, if it exists, only under some conditions on the function. It should have a "limit at infinity" in $Y$. For the reals e.g. this means that $\lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow -\infty} f(x)$ should exist and be finite.

There is one special compactification of $X$, called $\beta X$, which exists for all completely regular $T_1$ spaces (so for all metric ones in particular), so that we can extend any continuous map from $X$ to any compact Hausdorff space $Y$ to $\beta X$ (and $\beta X$ is compact Hausdorff and contains $X$ as a dense subspace). This extension property characterises $\beta X$, in fact.

If $Y$ is nice (like the reals or all Euclidean spaces, or many topological vector spaces, or compact balls...) then we extend a continuous $f$ from a closed subset $A$ of $X$ to $Y$ to all of $X$ to $Y$. This generalises the Tietze-Urysohn theorem.

For nice spaces (at least metric) we can extend a continuous $f$ from a closed subset $A$ of $X$ to $\mathbb{S}^n$ to all of $X$ iff $\dim(X) \le n$ (alexandroff theorem), so there an extension property to a special space characterises the topological dimension.

For metric spaces we have a classical theorem that we can extend a map from $(X,d)$ to a complete metric space $Y$ from $X$ to a $G_\delta$ around $X$, in the completion $(\tilde{X},\tilde{d})$ of $(X,d)$. This is a metric analogue of the compactification theory, in a way.

So also in general topology such problems have been studied a lot.

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  • $\begingroup$ Thanks for such a comprehensive response. I'm noticing that the "extension problem" is particularly relevant whenever you are considering a compactification or completion of a space. $\endgroup$ – Axesilo May 10 '16 at 3:10
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Have a look at Davis and Kirk's Lecture Notes in Algebraic Topology (you can freely download it). Chapter $7$, page $165$.

Roughly speaking, at least when we are dealing with CW-complex (i.e in your case we must assume $(X,S)$ a relative CW-complex). The obstruction for extending a map from the $n$ skeleton to the $n+1$-skeleton is a certain cocycle in $C^{n+1}(X,A;\pi_nY)$ which is explained in theorem $7.1$, page $169$.

Clearly the condition to the lack of this obstruction is not easily satisfied: Just think about this example to see a (very simple) map which cannot be extended.

Take the identity $$Id \colon S^1 \to S^1$$ and consider $S^1\subset D^1$ the usual disk. If the identity extends to a map $D^1\to S^1$ then it would be a null-homotopic map, which is not true (it's the generator of the fundamental group of the circle which is $\mathbb{Z}$). On the other hand, any map $$f\colon S^1 \to D^1$$ can be extended to a map $D^1 \to D^1$ using the fact that $D^1$ has vanishing fundamental group. This family of examples can be generalised to map form spheres and in general are fairly well-understood, motivating why working with CW-complexes (which are glued spheres) is "extremely" useful in these contexts.

Side note: If you relax your hypothesis on the extension problem (so extensions only in specific cases and with specific families of maps) then something more can be said in the case $(X,A)$ CW-pair, just have a look at the notion of cofibration.

I don't know much in the realm of non-CW spaces (i.e. Pathological spaces) though, so i can't provide you with references for this case.

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  • $\begingroup$ I see. So you might say one of the points of working with cocycles is to indicate this obstruction? Are they easy to compute in the case of cell complexes? $\endgroup$ – Axesilo May 7 '16 at 15:12
  • $\begingroup$ I don't think they are easily computable, I mean theoretically you can compute them inductively using the attaching maps of the cells but I don't think is doable (I'm interested in knowing if there are ways to actually compute them). The main problem is that the cocycle complex is huge, and wild. That's why we have cohomology and that's why the second statement (weaker!) of theorem $7.1$ is interesting. Dropping a condition on the extension lets you work with the class of this cocycle. Usually we understand the cohomology well enough to say something in this case $\endgroup$ – Riccardo May 7 '16 at 15:22
  • $\begingroup$ I mean, the problem of extending maps in indeed a huge problem, and lots of research is done in this field. If you are more specific with the map you want to extend there are more specific result, for example the theory of Stiefel-Whitney classes related as the extensions of frames in a vector bundle $\endgroup$ – Riccardo May 7 '16 at 15:23
  • $\begingroup$ @Riccardo, your answer is very interesting, but unfortunately I am not familiar (yet) with the subject you have written about. I am also curious to know when a continuous map can be extended from a sub-space to the space, though I only care about existence (and not about uniqueness). Please, can you tell me if the theorem that you have mentioned will say more if we will deal with a topological ring, in particular, is the extended map is also a ring homomorphism? Also, I am curious to know which cases you thought about in your 'side note'. $\endgroup$ – user237522 Apr 26 '17 at 2:08
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    $\begingroup$ @Riccardo, thanks again! Perhaps in a few days I will post a question about my above comments, but I am not expecting to get a positive answer (only in very special cases if at all). $\endgroup$ – user237522 Apr 28 '17 at 9:24

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