# Do continuous maps factor through continuous surjections via Borel maps?

Let $$f \colon X \twoheadrightarrow Y$$ be a continuous surjection between compact Hausdorff spaces, and $$g \colon \mathbb{R} \to Y$$ a continuous function. Can you always find a Borel-measurable function $$h \colon \mathbb{R} \to X$$ with $$g=f \circ h$$?

• Yes. In fact there exists a Borel map $s:Y\rightarrow X$ such that $f\circ s=\operatorname{Id}_X$, and you can just take $s\circ h$. The existence of $s$ holds under more general hypotheses, see Bourbaki's General Topology IX, 6, Theorem 5. – abx Apr 25 at 10:09
• Thanks abx, that sounds perfect. However, if I look that theorem up it reads: "Let Y be metrizable and a relatively compact Souslin subspace of X. Then Y is capacitable with respect to every capactiy f on X." This looks nothing like what I'd expect from your comment. Do I have a different version of the book? – Chris Heunen Apr 25 at 16:01
• Oops, sorry, this is Theorem 4 in §6, no. 8. – abx Apr 25 at 16:10
• I don't have a Bourbaki here, but would you confirm that no other assumption are needed? I'd be glad to know it. Thank you! – Pietro Majer Apr 25 at 17:05
• X has to be polish. – Ramiro de la Vega Apr 25 at 17:40

The answer is yes if we also assume $$X$$ to be metrizable (while other hypotheses may be weakened).

The multivalued map $$\psi:\mathbb{R}\to 2^X$$ defined by $$\psi(t):=f^{-1}(g(t))$$ for all $$t \in\mathbb{R}$$, takes values in non-empty closed subsets of $$X$$, because $$f$$is surjective and continuous. It is weakly Borel measurable, that is, for any open set $$U$$ of $$X$$ the set $$\{t\in\mathbb{R}: \psi(t)\cap U\neq\emptyset\}$$ is a Borel subset of $$\mathbb{R}$$. Indeed, for this map $$\psi$$, the latter set is $$g^{-1}(f(U))$$, which is a Borel set provided $$X$$ is a compact metrizable space: then every open $$U$$ set of its is a countable union of compact sets, $$f(U)$$ is an $$F_\sigma$$ and so is $$g^{-1}(f(U))$$). We can therefore apply the Kuratowski and Ryll-Nardzewski selection theorem and state the existence of a Borel measurable selection $$h:\mathbb{R}\to X$$ of $$\psi$$, that is $$h(t)\in f^{-1}(g(t))$$ for all $$t\in\mathbb{R}$$, that is $$f\circ h= g$$.

rmk. As shown in the above argument, what is needed to apply the KRN measurable selection theorem is: $$f$$ is continuous and surjective and for any $$U\subset X$$ open subset, $$g^{-1}(f(U))\in\mathcal{B}(\mathbb{R})$$.

Also, as suggested by user abx in comment, one can directly prove that $$f$$ has a Borel measurable section $$s:Y\to X$$, replacing $$g:\mathbb{R}\to Y$$ in the argument with the identity map $$Y\to Y$$. In order to apply the KRN thm one needs to know that for all $$U\subset X$$ open subset, $$f(U) \in\mathcal{B}(Y)$$ (which is true for metrizable $$X$$).

• Thanks Pietro. Do you know if X being metrizable is necessary? – Chris Heunen Apr 25 at 16:02
• I don't know, maybe not. For instance, the existence of a Borel measurable section $s:Y\to X$ to $f:X\to Y$ does not seem to imply that $f$ maps open subsets of $X$ to Borel subsets of $Y$, because it does not give $s^{-1}(U)=f(U)$, but only the inclusion. – Pietro Majer Apr 25 at 16:49

Theorem. There exists a non-metrizable compact Hausdorff space $$K$$ admitting a continuous surjective function $$f:K\to[0,1]^2$$ to the unit square such that $$f$$ has a Borel section if and only if CH holds. Then taking the Peano square-filling curve $$p:[0,1]\to[0,1]^2$$ we can show that $$p$$ has a Borel $$f$$-lifting $$[0,1]\to K$$ if and only if the Continuum Hypothesis holds.
Skecth of the proof. Consider the split interval $$\ddot{\mathbb I}$$ (which is the lexicographic product $$[0,1]\times\{0,1\}$$ endowed with the order topology). It is well-known that $$\ddot{\mathbb I}$$ is a compact Hausdorff space admitting a continuous surjective map $$\ddot{\mathbb I}\to [0,1]$$. Then the square $$K=\ddot{\mathbb I}\times\ddot{\mathbb I}$$ admits a continuous surjective map $$f:K\to[0,1]^2$$ such that for every point $$z\in[0,1]^2$$ the preimage $$f^{-1}(z)$$ contains at most 4 points. Using the answer (and comments) to this question, it can be shown that $$f$$ has the desired property: it has a Borel section if and only if CH holds. More details on the proof can be found in this paper. $$\quad\square$$
Problem. Is there a 2-to-1 map $$f:K\to M$$ from a compact (Rosenthal) space onto a compact metrizable space $$M$$, which has no Borel selections?