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Tietze's extension theorem says:

''If $A$ is a closed subset of $X$ a normal space, and $f:A\to \mathbb{R}$ continuous, then we can extend $f$ to a continuous function $g:X\to \mathbb{R}$."

I know that it can be generalized by changing $\mathbb{R}$ to closed subsets of $\mathbb{R}^n$. But, can we find further generalizations by changing $\mathbb{R}$ to a more general space $Y$. To be precise, I'm looking for generalizations for an arbitrary polish space $Y$, and if it's worth it, $X$ is a zero dimensional polish space.

The thing is, I want continuous functions, but I really only care how they behave in a closed set. For any particular case, I can use a nice Cantor scheme and get the function, but I don't want to keep repeating me.

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    There are some results about extending continuous maps or homeomorphisms between completely metrizable spaces, such as [Lavrentieff's Theorem](http://math.stackexchange.com/questions/152680/extending-a-homeomorphism-of-a-subset-of-a-space-to-a-g-delta-set) or Kuratowski theorem. See e.g. [section 4.3](http://thales.doa.fmph.uniba.sk/sleziak/texty/rozne/engel/engel.pdf) in Engelking's book or [Section 3.B](http://books.google.com/books?id=pPv9KCEkklsC&pg=PA15) in Kechris. Are perhaps some of these results sufficient for your needs?2012-06-07
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    If $Y=\mathbb S^1$, there is no way to extend the identity map $\mathbb S^1\to\mathbb S^1$ to a map $\mathbb R^2\to \mathbb S^1$. So you can't have a *generalization of Tietze*; the particular form of $X$ (as you mentioned) will be necessary.2012-06-07
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    @LeonidKovalev: You gave a specific example, which fundamentally depends on the fact that the $\mathbb{R}^2$ has trivial fundamental group and $\mathbb{S}^1$ has fundamental group $\mathbb{Z}$. I'm not sure how this shows that you can't have *any* generalization of Tietze because of this. In particular, my thought is to have some condition on the induced homomorphism $f_*:\pi_1(A)\to\pi_1(Y)$ (perhaps that it is trivial?). However, I know this would introduce other (possibly undesirable) constraints on the spaces, such as path-connectedness.2012-06-07
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    @J.Loreaux OK, let's leave the choice of words besides. In order to have the extension from any closed subset $A$ of any normal space $X$, one needs $Y$ to be an *absolute retract*, this is both necessary and sufficient. Of course, if one imposes additional assumptions on $X$, then there may be more freedom to choose $Y$.2012-06-07
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    @MartinSleziak: My sets are already $G_\delta$, so Lavrentieff's wouldn't really help. In fact, I only really need $X$ to be any zero-dimensional Polish space, so I guess I could only take $X$ to be my set $A$ and call it a day, but I was striving for a more elegant approach and have my function from the whole Cantor space.2012-06-10

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