# A topological space with the Universal Extension Property which is not homeomorphic to a retract of $\mathbb{R}^J$?

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?

• 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. – freakish Nov 9 '18 at 12:04
• 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. – freakish Nov 9 '18 at 12:08

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 $$\mathbb{R}^J$$.