Ok, here’s an idea for constructing such a map, with a few key details left unproven; let me know if people see any immediate flaws in the approach, before I spend time filling in the holes.
Let X be a countable collection of open intervals (eg X={x∈R,x∉N}), given the usual topology. Let I=[0,1] be the closed unit interval, and C(X,I) the set of continuous functions from X to I. Give C(X,I) the compact-open topology.
By the properties of the compact-open topology, since I is T3.5 (Tychonoff), then so is C(X,I). I’m hoping that the proof can be extended, at least in this case, to show that C(X,I) is T4 (normal Haussdorff).
It seems clear that C(X,I) is second-countable: let V(K,U) consist of all functions that map K into U, where K⊂X is the intersection of X with a closed interval with rational endpoints, and U⊂I is an open interval with rational endpoints. The set of all such V(K,U) is countable, and forms a subbasis of C(X,I). A countable subbasis means a countable basis, as the set of finite subsets of countable set, is itself countable.
If C(X,I) is T4 and second countable, then it is homeomorphic to a subset of the Hilbert Cube. To simplify notation, we will identify C(X,I) with its image in the Hilbert Cube.
Take the closure ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C(X,I) of C(X,I) within the Hilbert Cube. This closure is compact and second countable (since the Hilbert Cube itself is both). It seems clear that C(X,I) is connected and locally connected; connected will extend to the closure, we’ll need to prove that locally connected does as well.
A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.
So there is a continuous surjection ϕ:I→¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C(X,I). Pull back C(X,I), defining ϕ−1(C(X,I))⊂I. If C(X,I) is open in ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C(X,I) (with the subspace topology), then ϕ−1(C(X,I)) is open in I. Even if ϕ−1(C(X,I)) is not open, we can hope that, at worst, it consists of a countable collection of points, open intervals, half-closed intervals, and closed intervals (this is not a general property of subsets of the interval, cf the Cantor set, but it feels very likely that it will apply here).
In that case, these is a continuous surjection s from X to ϕ−1(C(X,I)), mapping each open interval to one of the points or intervals (“folding” over the ends when mapping to those with closed end-points).
Then ϕ∘s:X→C(X,I) is the continuous surjection we are looking for.
Note: I’m thinking now that C(X,I) might not be connected, but this would not be a problem as long as it has a countable number of connected components.
EDIT: This idea is wrong: https://www.lesswrong.com/posts/eqi83c2nNSX7TFSfW/no-surjection-onto-function-space-for-manifold-x
Ok, here’s an idea for constructing such a map, with a few key details left unproven; let me know if people see any immediate flaws in the approach, before I spend time filling in the holes.
Let X be a countable collection of open intervals (eg X={x∈R,x∉N}), given the usual topology. Let I=[0,1] be the closed unit interval, and C(X,I) the set of continuous functions from X to I. Give C(X,I) the compact-open topology.
By the properties of the compact-open topology, since I is T3.5 (Tychonoff), then so is C(X,I). I’m hoping that the proof can be extended, at least in this case, to show that C(X,I) is T4 (normal Haussdorff).
It seems clear that C(X,I) is second-countable: let V(K,U) consist of all functions that map K into U, where K⊂X is the intersection of X with a closed interval with rational endpoints, and U⊂I is an open interval with rational endpoints. The set of all such V(K,U) is countable, and forms a subbasis of C(X,I). A countable subbasis means a countable basis, as the set of finite subsets of countable set, is itself countable.
If C(X,I) is T4 and second countable, then it is homeomorphic to a subset of the Hilbert Cube. To simplify notation, we will identify C(X,I) with its image in the Hilbert Cube.
Take the closure ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C(X,I) of C(X,I) within the Hilbert Cube. This closure is compact and second countable (since the Hilbert Cube itself is both). It seems clear that C(X,I) is connected and locally connected; connected will extend to the closure, we’ll need to prove that locally connected does as well.
Then we can apply the Hahn-Mazurkiewicz theorem:
A non-empty Hausdorff topological space is a continuous image of the unit interval if and only if it is a compact, connected, locally connected second-countable space.
So there is a continuous surjection ϕ:I→¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C(X,I). Pull back C(X,I), defining ϕ−1(C(X,I))⊂I. If C(X,I) is open in ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯C(X,I) (with the subspace topology), then ϕ−1(C(X,I)) is open in I. Even if ϕ−1(C(X,I)) is not open, we can hope that, at worst, it consists of a countable collection of points, open intervals, half-closed intervals, and closed intervals (this is not a general property of subsets of the interval, cf the Cantor set, but it feels very likely that it will apply here).
In that case, these is a continuous surjection s from X to ϕ−1(C(X,I)), mapping each open interval to one of the points or intervals (“folding” over the ends when mapping to those with closed end-points).
Then ϕ∘s:X→C(X,I) is the continuous surjection we are looking for.
Note: I’m thinking now that C(X,I) might not be connected, but this would not be a problem as long as it has a countable number of connected components.