Let $X$ and $Y$ be Tychonoff (i.e. completely regular Hausdorff) topological spaces and let $\varphi:X\to Y$ be a continuous surjection that also has a property that $\operatorname{int}\overline{\varphi(U)}\ne\varnothing$, for any open nonempty $U\subset X$ (this property is called skeletal, weakly open, almost open etc in various sources; equivalently, the preimage of any nowhere set is nowhere dense).


Let us consider the topology $\tau$ on $Y$ which is the strongest Tychonoff topology with respect to which $\varphi$ is continuous (this property for $\varphi$ as a map into $(Y,\tau)$ is called $\mathbb{R}$-quotient, which may be viewed as a quotient map in the category of Tychonoff spaces).



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> Is $\varphi$ still skeletal as a map from $X$ into $(Y,\tau)$?
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This is true is $X$ is locally compact, so the counterexamples should be looked for outside of this class.



Is a certain property of a continuous map preserved under a modification of the topology on the target space?
The answer to this question is negative. A suitable counterexample can be constructed as follows.


Let $Y=\mathbb R$ be the real line with the standard Euclidean topology.
Let $\mathbb Q$ be the subspace of rational numbers in $\mathbb R$. Write $\mathbb R\setminus \mathbb Q$ as the union $\bigcup\_{q\in \mathbb Q}X\_q$ of pairwise disjoint dense sets in $\mathbb R$. Let $$X=\mathbb Q\oplus\bigoplus\_{q\in \mathbb Q}(\{q\}\cup X\_q)$$be the topological sum of the spaces $\mathbb Q$ and $\{q\}\cup X\_q$ for $q\in\mathbb Q$.


Let $\varphi:X\to Y$ be the natural projection. It is clear that the spaces $X,Y$ are separable, metrizable (and hence Tychonoff) and the function $\varphi:X\to Y$ is skeletal.


On the other hand, in the $\mathbb R$-quotient topology $\tau$ on $Y=\mathbb R$, the set $\mathbb Q$ is closed and nowhere dense in $(Y,\tau)$, witnessing that the function $\varphi:X\to(Y,\tau)$ is not skeletal.


To show that $\mathbb Q$ is closed in $(Y,\tau)$, choose any point $y\in\mathbb R\setminus \mathbb Q$. Find $q\in\mathbb Q$ such that $y\in X\_q$. Consider the function $f\_q:Y\to\mathbb R$ defined by
$f\_q(x)=|x-q|$ if $x\in X\_q$ and $f\_q(x)=0$, otherwise.


Observe that the composition $f\_q\circ \varphi:X\to\mathbb R$ is continuous, which implies that the set $X\_q=\{x\in Y:f\_q(x)>0\}$ is $\tau$-open, contain $y$, and does not intersect $\mathbb Q$.
This completes the proof of the closedness of $\mathbb Q$.


Assuming that the closed set $\mathbb Q$ is not nowhere dense in $(Y,\tau)$, we can find a nonempty open set $U\subseteq\mathbb Q$. Choose any point $q\in U$ and observe that the preimage $\varphi^{-1}(U)$ is an open set in $X$ containing the point $q\in \{q\}\cup X\_q$. Since $\{q\}$ is nowhere dense in $X\_q$, the set $\varphi^{-1}(U)$ has nonempty intersection with the set $X\_q$ and then $\emptyset \ne \varphi[X\_q\cap \varphi^{-1}(U)]\subseteq U\cap X\_q\subseteq U\setminus\mathbb Q$, which contradicts the choice of $U\subseteq\mathbb Q$.


