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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 or Kuratowski theorem. See e.g. section 4.3 in Engelking's book or Section 3.B in Kechris. Are perhaps some of these results sufficient for your needs? –  Martin Sleziak Jun 7 '12 at 14:53
    
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. –  user31373 Jun 7 '12 at 15:38
    
@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. –  J. Loreaux Jun 7 '12 at 19:57
    
@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 $Y$. –  user31373 Jun 7 '12 at 20:18
    
@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. –  Rafa Jun 10 '12 at 8:11

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