TITLE: Do lifts of maps from extremally disconnected compact Hausdorff spaces into certain colimits exist?
QUESTION [7 upvotes]: Let $T$ be a extremally disconnected compact Hausdorff space. Further let $X$ and $Y$ be two ind-compact Hausdorff spaces, i.e. both are the filtered colimit of compact Hausdorff spaces with injective transition maps. Assume that $f \colon X \to Y$ is a continuous surjective map and $h \colon T \to Y$ is continuous.
My question is wether there exists a continuous lift $g \colon T \to X$ of $h$ such that $h= f \circ g$. I know that this is true if $X$ and $Y$ are compact Hausdorff spaces themselves since extremally disconnected compact Hausdorff spaces are the projective  objects in the category of compact Hausdorff spaces.
I might have to add, that $X$ and $Y$ share the same underlying filtrated category $\mathbb{R}_{>0}$ and whenever $f$ sends an element $x \in X_c$ to an element $f(x) \in Y_d$ then $d \leq c$.

REPLY [11 votes]: In general, the answer is No. The simplest counterexample is the following; I think it also satisfies your further assumptions. Take $Y=\{0,1,2,\ldots,\infty\}$, i.e. the one point compactification of $\mathbb N$, which is compact Hausdorff. Let $X_n=\{0,1,2,\ldots,n,\infty\}$, a finite set, which has a natural inclusion into $Y$; and let $X$ be the colimit of the $X_n$'s, so $X$ is just a discrete infinite set, the disjoint union of $\mathbb N$ and $\infty$. Then $f: X\to Y$ is a continuous surjection, but if $T\to Y$ is a surjection from some extremally disconnected $T$, then it cannot lift to $X$ as otherwise $T\to X$ would be a surjection from a compact space and hence $X$ would have to be compact.
In the positive direction, something useful to know is that if your filtered index category is given by $\mathbb R_{>0}$ (essentially equivalently, the cofinal subcategory $\mathbb N_{>0}$), then for any sequential colimit $Y=\mathrm{colim}_n Y_n$ of compact Hausdorff spaces along closed immersions, any map $T\to Y$ from a compact Hausdorff space $T$ to $Y$ factors over some $Y_n$ -- see, for example, Proposition A.15 (i) here. So if in your situation you further ask that all $X_c\to Y_c$ are surjective, then the answer is Yes.