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Recall from topology that a space $Y$ has the Universal Extension Property if there is a normal space $X$, a closed subset $A \subset X$, and a continuous map $f: A \rightarrow Y$, such that $\exists g:X \rightarrow Y$, an extension of $f$. Then how is it that $Y$ is connected? I'd really like a nice proof of this.

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Do you mean for every normal space $X$ and every closed subset $A$ and every $f:A\rightarrow Y$ there is an extension to $X$? Without those assumptions, I think it's false. For example, take $Y$ to be 2 discrete points, $A = X$ the unique one point set and any $f$ you wish. More generally, every $Y$ would the the Universal Extension Property by using the same example. If you insist that $A$ be a proper subset of $X$, choose $X$ to be 2 discrete points, $A$ a subset consisting of $1$ point, and let $f$ be anything. – Jason DeVito Nov 29 '12 at 0:24
As @Jason says, that’s not the UEP; $Y$ has the UEP iff for each normal space $X$ and closed $A\subseteq X$, each continuous $f:A\to Y$ has a continuous extension to all of $X$. – Brian M. Scott Nov 29 '12 at 0:28
Yes, I meant it for every normal space $X$ and every closed subset $A$, and every continuous map $f: A \rightarrow Y$. With regard to Brian, thanks for the clarification. – Libertron Nov 29 '12 at 0:30
Nice question: I never knew this fact before. – Brian M. Scott Nov 29 '12 at 0:45
Are there any easy nontrivial examples of UEPs? Of course if $Y$ is a singleton, it has the UEP. – Jason DeVito Nov 29 '12 at 1:54
up vote 3 down vote accepted

Suppose that $Y$ has the UEP. Take $X=[0,1]$ and $A=\{0,1\}$. Let $y_0$ and $y_1$ be any two points of $Y$, and let $f:A\to Y$ take $i$ to $y_i$. Clearly $f$ is continuous, so it has a continuous extension $\hat f:X\to Y$. Then $\hat f[X]$ is a connected set containing $y_0$ and $y_1$, so $y_0$ and $y_1$ must lie in the same component of $Y$. But $y_0$ and $y_1$ were arbitrary, so $Y$ has only one component and is therefore connected.

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Just to make it explicit: this proves that $Y$ is even path connected. – nonpop Dec 2 '12 at 8:21

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