Let $X$, $Y\_1$ and $Y\_2$ be a compact Hausdorff spaces and let $\varphi\_i:X\to Y\_i$ be a continuous surjection (and so a quotient map).


Let $\sim$ be the minimal closed equivalence relation on $X$ that includes $(x,y)$ with $\varphi\_1(x)=\varphi\_1(y)$ and $\varphi\_2(x)=\varphi\_2(y)$ (recall that an equivalence relation on $X$ is called *closed* if for every closed $A\subset X$ the union of all classes of elements of $A$ is closed; an equivalence relation induced by $\varphi\_i$ is closed).


One can show that a quotient space $Z$ of $X$ is a compact Hausdorff space. Let $\psi:X\to Z$ be the corresponding quotient map. I want to show that if $\varphi\_i$ are not too wild, then the same is true for $\psi$. Namely:



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> If $\varphi\_i$ are such that $int~\varphi\_i(U)\ne\varnothing$, for every open $U\subset X$, does the same property hold for $\psi$?
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Here is the motivation for this question. I have two closed subalgebras $E\_i$ of $C(X)$ that contain $1$. It is well known that such subalgebras consist of functions of the form $f\circ\varphi\_i$, where $\varphi\_i:X\to Y\_i$ is a surjection onto some compact $Y\_i$, and $f\in C(Y\_i)$.


It is given that $E\_i$ are *regular sublattices* of $C(X)$ (this means that if a net in $E\_i$ decreases to $0$ in $E\_i$, the same is true in $C(X)$). I can show that this property for $E\_i$ translates into the property of $\varphi\_i$ mentioned above and I want to prove that $E\_1\cap E\_2$ also has this property. Perhaps this property has an algebraic equivalent that can be checked for intersection.



Is the union of good equivalence relations on a compact space good?
It seems that the quostion about the skeletal property of $\psi$ has negative answer.


Let us recall that a map $f:X\to Y$ between topological spaces is *skeletal* if for any nonempty open set $U\subseteq X$ the set $\overline{f[U]}$ has non-empty interior in $Y$.


A corresponding counterexample looks as follows. Fix any homeomorphism $h:2^\omega\to 2^\omega\times 2^\omega$ of the Cantor set onto its square.


Let $X=\{0,1\}\times 2^\omega$ and $Y=2^\omega\times 2^\omega$.


For $i\in\{0,1\}$ let $\mathrm{pr}\_i:2^\omega\times 2^\omega\to 2^\omega$ be the projection onto the $i$-th coordinate.


Consider the continuous map $\varphi\_i:X\to 2^\omega$ defined by the formula $\varphi\_i(0,x)=\mathrm{pr}\_i\circ h(x)$ and $\varphi\_i(1,x)=x$ for any $x\in 2^\omega$.


Then the quotient space $Z$ in the qustion of @erz can be identified with the space $Y$ and the quotient map $\psi:X\to Z=Y$ with the diagonal product of the maps $\varphi\_1$ and $\varphi\_2$. It is easy to see that $\psi{\restriction}\{0\}\times 2^\omega$ is open map and the image $\psi[\{1\}\times 2^\omega]$ is nowhere dense in $Z=Y$, being the diagonal of the square $2^\omega\times 2^\omega$. So, $\psi$ is not skeletal in spite of the fact that the maps $\varphi\_1,\varphi\_2$ are open.


