How to construct a continuous map such that the preimage of a particular non-open set is not open?

Question

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EDIT

This might be a more clear problem statement:

1. We have an arbitrary topological space $Y$ and an non-open set $B$ in $Y$. How to find a topological space $X$ and a continuous map $F:X\rightarrow Y$ such that $F^{-1}(B)$ in not open in X?

2. We have an arbitrary topological space $Y$ and an non-open set $B$ in $Y$. But now an arbitrary topological space $X$ is given, such that there exists a continuous map $F:X\rightarrow Y$. How to construct a continuous map $G$ such that $G^{-1}(B)$ is not open in $X$?

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Original questions

We have two topological spaces $X$ and $Y$.

If $B$ is not open in $Y$, is there a continuous map $F:X \rightarrow Y$ such that $F^{-1}(B)$ is not open in $X$? How to construct such a map?

Equivelently, if for $\forall$ continuous function $F:X\rightarrow Y$ we have that $F^{-1}(B)$ is open in $X$, is $B$ open in $Y$? Can you give a counter example?

Answer 1

Answer to the original question

Under the discrete topology on a set $X$, any subset of $X$ is open and, for any topological space $Y$, any function $F: X \to Y$ is continuous. For a specific counter-example to both your conjectures, take $X$ to be $\Bbb{R}$ with the discrete topology and $Y$ to be $\Bbb{R}$ with the usual topology and $F : X \to Y$ to be the identity function ($F(x) = x$).

Answer to the revised question

1. Take $X = Y$ and $F : X \to Y$ to be the identity map.

2. There is no such construction that will work for any $X$. First note that the assumption that there exists a continuous function $F : X \to Y$ is not helpful, since it is true for any $X$ and $Y$, because constant functions are continuous. If $X$ is discrete and $G : X \to Y$ is any function, then $G^{-1}(B)$ is open for any $B$.