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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 $f^{-l}(B_1)$ and $f^{-l}(B_2)$ have disjoint closures in the space $X$.

I am mostly interested in sufficient conditions.

For example, is the following sufficient?

(i) For each $Z_1, Z_2\subset A$, 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 $Z\subset X$ closed, each closed subsets $Z_1, Z_2\subset A$ such that $Z\cap A=Z_1\cup Z_2$, there are $Z'_1, Z'_2 \subset X$ closed such that $Z=Z'_1\cup Z'_2$, and $Z_1=Z'_1\cap A$, and $Z_2=Z'_2\cap A$, and $Z'_1\cap Z'_2=cl_X(Z_1\cap Z_2)$.

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

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Curiously enough, I was looking for a similar result, and I found this paper (and its errata) to be very helpful. Essentially, you define the Lebesgue sets of $f\colon A\subset X\rightarrow Y$ to be $L_{a}(f)=\{x\in A \colon \;f(x)\leq a\}$ and $L^{a}(f)=\{x\in A \colon \;f(x)\geq a\}$, 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.

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