by Marcello Herreshoff 250 days ago | Jessica Taylor, Scott Garrabrant and Stuart Armstrong like this | link | parent From discussions I had with Sam, Scott, and Jack: To solve the problem, it would suffice to find a reflexive domain $$X$$ with a retract onto $$[0, 1]$$. This is because if you have a reflexive domain $$X$$, that is, an $$X$$ with a continuous surjective map $$f :: X \rightarrow X^X$$, and $$A$$ is a retract of $$X$$, then there’s also a continuous surjective map $$g :: X \rightarrow A^X$$. Proof: If $$A$$ is a retract of $$X$$ then we have a retraction $$r::X\rightarrow A$$ and a section $$s::A \rightarrow X$$ with $$r\circ s = 1_A$$. Construct $$g(x) := r \circ f(x)$$. To show that $$g$$ is a surjection consider an arbitrary $$q \in A^X$$. Thus, $$s \circ q :: X \rightarrow X$$. Since $$f$$ is a surjection there must be some $$x$$ with $$f(x) = s \circ q$$. It follows that $$g(x) = r \circ f(x) = r \circ s \circ q = q$$. Since $$q$$ was arbitrary, $$g$$ is also a surjection.

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