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→XX, and A is a retract of X, then there’s also a continuous surjective map g::X→AX.
Proof: If A is a retract of X then we have a retraction r::X→A and a section s::A→X with r∘s=1A. Construct g(x):=r∘f(x). To show that g is a surjection consider an arbitrary q∈AX. Thus, s∘q::X→X. Since f is a surjection there must be some x with f(x)=s∘q. It follows that g(x)=r∘f(x)=r∘s∘q=q. Since q was arbitrary, g is also a surjection.
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→XX, and A is a retract of X, then there’s also a continuous surjective map g::X→AX.
Proof: If A is a retract of X then we have a retraction r::X→A and a section s::A→X with r∘s=1A. Construct g(x):=r∘f(x). To show that g is a surjection consider an arbitrary q∈AX. Thus, s∘q::X→X. Since f is a surjection there must be some x with f(x)=s∘q. It follows that g(x)=r∘f(x)=r∘s∘q=q. Since q was arbitrary, g is also a surjection.