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A topological space $Y$ has the universal extension property if for every normal space $X$, every closed subset $A$ of $X$, and every continuous function $f:A\rightarrow Y$, we can extend $f$ to a continuous function $g:X\rightarrow Y$. Now for any $J$, the product space $\mathbb{R}^J$ has the universal extension property, and so does every retract of $\mathbb{R}^J$.

But my question is, what is an example of a topological space which has the universal extension property but is not homeomorphic to $\mathbb{R}^J$ or any of its retracts? Or does no such example exist?

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  • $\begingroup$ Absolute Extensors are equivalent to Absolute Retracts so if $Y$ can be embedded into $\mathbb{R}^J$ as a closed subset then it has to be a retract of $\mathbb{R}^J$. But by Dugundji every locally convex metrizable topological vector space is an AR. This leads to a simple example: an infinite dimensional normed space. If you extended your question to "normed spaces" instead of $\mathbb{R}^J$ then the answer is "no such example exists", at least not under assumption that $Y$ is metric. $\endgroup$ – freakish Nov 9 '18 at 12:04
  • $\begingroup$ That's because every metric space is homeomorphic to a bounded metric space. And every bounded metric space is isometric to a closed subset of a Banach space (Kuratowski–Wojdysławski theorem). And in this situation my previous comment applies. $\endgroup$ – freakish Nov 9 '18 at 12:08
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Your question is whether any $Y$ which has the universal extension property admits a closed embedding into some $\mathbb{R}^J$.

The answer is "no". Let $Y$ have more than one point and the trivial topology. Then it has the universal extension property, but cannot embedded into any $\mathbb{R}^J$.

To obtain a closed embedding, you therefore need additional assumptions on $Y$. For example, if $Y$ is compact, then it embeds into some Tychonoff cube $[0,1]^J$ which is contained in $\mathbb{R}^J$.

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