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Extending continuous functions from dense subsets to quasicompacts

$DeclareMathOperatorcl{cl}$I am interested under what assumptions one can always extend continuously a function defined on a dense subset; the range of the function is compact but not necessarily Hausdorff.

That is, I am interested in generalisations of the following theorem [Engelking, General Topology] to non-Hausdorff compact spaces:

3.2.1. THEOREM. Let $A$ be a dense subspace of a topological space $X$ and $f$ a continuous mapping of $A$ to a compact space $Y$. The mapping $f$ has a continuous extension over $X$ if and
only if for every pair $B_1$, $B_2$ of disjoint closed subsets of $Y$ the inverse images $f^{-1}(B_1)$ and
$f^{-1}(B_2)$ have disjoint closures in the space $X$.

I am mostly interested in sufficient conditions.

For example, is the following sufficient?

(i) For each $Z_1, Z_2subset A$, it holds $cl_X(Z_1) cap cl_X( Z_2) = cl_X( cl_A(Z_1)cap cl_A(Z_2))$.

(ii) For each $Zsubset X$ closed, and each pair of closed subsets $Z_1, Z_2subset A$ such that $Zcap A=Z_1cup Z_2$,
there are $Z’_1, Z’_2 subset X$ closed such that $Z=Z’_1cup Z’_2$, and $Z_1=Z’_1cap A$, and $Z_2=Z’_2cap A$, and $Z’_1cap Z’_2=cl_X(Z_1cap Z_2)$.

(iii) $A$ is an open dense subset of $X$.

MathOverflow Asked by user97621 on January 3, 2021

1 Answers

One Answer

$DeclareMathOperatorcl{cl}$Curiously enough, I was looking for a similar result, and I found Blair - Extensions of continuous functions from dense subspaces (and its errata) to be very helpful. Essentially, you define the Lebesgue sets of $fcolon Asubset Xrightarrow Y$ to be $L_{a}(f)={xin A colon ;f(x)leq a}$ and $L^{a}(f)={xin A colon ;f(x)geq a}$, and then you prove that there exists a continuous extension of $f$ from the dense subspace $A$ to $X$ if and only if $$ a<b:rightarrow;;cl_{X}(L_{a}(f)),cap, cl_{X}(L^{b}(f)),=,emptyset,, $$ and $$ bigcap_{n=0}^{infty},cl_{X}(L_{-n}(f),cup,L^{n}(f)),=,emptyset,, $$ where $cl_{X}$ denotes the closure in $X$. Note that both $X$ and $Y$ are generic topological spaces (no compactness is needed). I do not remember the proof, but the paper has it done in full detail.

Answered by Ittiolo on January 3, 2021

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