# Is a certain property of a continuous map preserved under a modification of the topology on the target space?

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).

Is $$\varphi$$ still skeletal as a map from $$X$$ into $$(Y,\tau)$$?

This is true is $$X$$ is locally compact, so the counterexamples should be looked for outside of this class.

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 $$U\subseteq\mathbb Q$$.

• Thank you! After the formula the second $\mathbb{Q}$ should be multiplied with $0$, right?
– erz
Jan 29, 2022 at 21:44
• Actually, why is $X$ Tychonoff?
– erz
Jan 29, 2022 at 21:44
• @erz Ups! Indeed, $X$ is not even regular. So, you accepted my answer too fast. I should think a bit more. Sorry for this gap. Jan 29, 2022 at 22:39
• @erz Now I have corrected the example, so it is became Tychonoff. Jan 30, 2022 at 13:09
• I don't think I understand your construction of $X$ and $\varphi$. If $X$ is a disjoint union of subsets of $\mathbb{R}$ which add up to $\mathbb{R}$, then $\varphi$ is the identity map, and so $(Y,\tau)=X$. So it is probably something else but i cannot surmise it from the solution
– erz
Jan 31, 2022 at 5:39