It is possible to prove that these two collections generate the same topology on $ \mathbb{X} $? Let $ \mathbb{Y} $ a topological space whose topology $ \tau $ is generated
by a collection of subsets $ \mathcal{B} \subset 2^\mathbb{Y} $.
In other words the topology $ \tau $ of $ \mathbb {Y} $ is the smallest topology
("Smaller" in partial order on $ \subset $ sub-collections in the class of $ 2^\mathbb{Y} $)
containing the collection $ \mathcal{B} $ of subsets of $ \mathbb{Y}$.
Let $ \{\varphi_i\}_{i \in \Lambda} $ a family of functions defined in a set $ \mathbb{X} $ and
taking values ​​in topological space $ \mathbb{Y}$. Put $ \mathcal{A}_{\tau} $ as the collection of all
subsets of $ \mathbb{X}$ in the form
$$
\{\varphi_i^{-1}(V): V \in \tau \mbox{ and } i\in \Lambda \}.
$$
and $ \mathcal{A}_\mathcal{B}$ as
$$
\{\varphi_i^{-1}(B): B\in\mathcal{B}\mbox{ and } i \in \Lambda \}.
$$
It is possible to prove that these two collections generate the same topology on $ \mathbb{X} $?
If not what assumptions are sufficient for this to be true?
I am particularly interested in the case where $ \mathbb{Y} = \mathbb{R} $.
 A: Suppose that $F\subseteq\Lambda$ is finite, $V_i\in\tau$ for each $i\in F$, and $x\in\bigcap_{i\in F}\varphi_i^{-1}[V_i]$. For each $i\in F$ let $y_i=\varphi_i(x)\in V_i$; there is a $B_i\in\mathcal{B}$ such that $y_i\in B_i\subseteq V_i$, so $x\in\bigcap_{i\in F}\varphi_i^{-1}[B_i]\subseteq\bigcap_{i\in F}\varphi_i^{-1}[V_i]$. It follows immediately that $\mathcal{A}_\tau$ and $\mathcal{A}_{\mathcal{B}}$ generate the same topology on $\Bbb X$.
A: If $(S,\mathcal T)$ is a topological space, and $\mathcal C\subset 2^S$, then the topology generated by $\mathcal C$ can be given in the following way: add $S$ and $\emptyset$ to $\mathcal C$ in order to assume WLOG they are in $\mathcal C$, and let $\mathcal F$ the collection of finite intersections of elements of $\mathcal C$. Then the topology generated by $\mathcal C$ consists of arbitrary unions of elements of $\mathcal F$.
Back to the problem. As $\mathcal A_{\tau}$ contains $\mathcal A_{\mathcal B}$, we have to show that each element of $\mathcal A_{\mathcal B}$ is in the topology generated by $\mathcal A_{\tau}$. We just have to show it for elements of the form $\varphi_i^{-1}(V)$ for $V\in\mathcal T$ and $i\in I$. Indeed, if we have done it for these elements, we can do it for arbitrary unions of finite intersections of such elements. Write $V:=\bigcup_{i\in J}\bigcap_{j\in F_i}B_j$, where $J\subset I$ is arbitrary and $F_i$ is finite, $B_j\in\mathcal B.$ 
A: If $\mathcal{B}$ generates $\tau$ then for every $V \in \tau$ there are $(B_{i,j})_{i \in I,1\leq j \leq n_i}$, $B_{i,j} \in \mathcal{B}$ such that $V = \bigcup_{i \in I} \bigcap_{j=1}^{n_i} B_{i,j}$. Since $\varphi^{-1}(\bigcup_{i \in I} \bigcap_{j=1}^{n_i} B_{i,j})$ = $\bigcup_{i \in I} \bigcap_{j=1}^{n_i} \varphi^{-1}(B_{i,j})$ every open set generated by $A_\tau$ is also generated by $A_\mathcal{B}$. The converse is obivous because $\mathcal{B} \subset V$.
$A_\tau$ btw is called the initial topology on $\mathbb{X}$ and is the coarsest (smallest wrt $\subset$) topology on $\mathbb{X}$ which makes all the $\varphi_i$ continuos for a given topology on $\mathbb{Y}$.
A: (I know the OP said to disregard this question, but I discovered some nice things that I'd like to share in this answer.)
Notations. Let me start by introducing a slightly different notation, writing $D^\#$ instead of $\mathcal{A}_D$: $$ D^\# = \{ V,f : V \in D : f^{-1}[V] \} $$
Read this as "the set which, for each $V$ and $f$ such that $V \in D$, contains $f^{-1}[V]$".  Here, and in the rest of the post, we let $f$ range implicitly over $\{\varphi_i\}_{i \in \Lambda}$.
We use $D^\cup$ and $D^\cap$ for the set of all arbitrary unions and of finite intersections of elements of $D$, respectively:
$$
\begin{align}
& D^\cup = \left\{ \mathscr{F} : \mathscr{F} \subseteq D : \bigcup \mathscr{F} \right\} \\
& D^\cap = \left\{ F,G : F,G \in D : F \cap G \right\} \\
\end{align}
$$
(See also another answer of mine.)  This means that we can write "the topology generated by $B$" as $B^{\cap\cup}$.
Problem statement. Using these notations, we are asked whether, for all $B$ and $\tau$ with $\tau = B^{\cap\cup}$, we have $\tau^{\#\cap\cup} = B^{\#\cap\cup}$; or equivalently, whether for all $B$
$$
(0) \;\;\; B^{\cap\cup\#\cap\cup} = B^{\#\cap\cup}
$$
Proof.  The only thing we know about $^\#$, $^\cap$, and $^\cup$ are their definitions: these apply $f^{-1}[\cdot]$, $\cap$, and $\cup$ many times.  And looking at the latter operators, we know that the order of $f^{-1}[\cdot]$ and $\cap$ doesn't matter, and the same for $f^{-1}[\cdot]$ and $\cup$.  (I will not prove those distribution properties here.)  Therefore it seems likely that for all $D$
$$
(1) \;\;\; D^{\cup\#} = D^{\#\cup} \\
(2) \;\;\; D^{\cap\#} = D^{\#\cap} \\
$$
If we would be able to prove these, then $(0)$ directly follows from a simple calculation:
$$
\begin{align}
& B^{\cap\cup\#\cap\cup} \\
= & \;\;\;\;\;\text{"using $(1)$ with $D := B^\cap$"} \\
& B^{\cap\#\cup\cap\cup} \\
= & \;\;\;\;\;\text{"using $(2)$ with $D := B$"} \\
& B^{\#\cap\cup\cap\cup} \\
= & \;\;\;\;\;\text{"$^{\cap\cup}$ is idempotent"} \\
& B^{\#\cap\cup} \\
\end{align}
$$
(I won't prove here that $^{\cap\cup}$ is idempotent.)
It turns out we can prove $(1)$, as follows:
$$
\begin{align}
& D^{\cup\#} \\
= & \;\;\;\;\;\text{"definition of $^\#$"} \\
& \{ V,f : V \in D^\cup : f^{-1}[V] \} \\
= & \;\;\;\;\;\text{"write $V$ as a union using the definition of $^\cup$"} \\
& \{ \mathscr{F},f : \mathscr{F} \subseteq D : f^{-1}\left[\bigcup \mathscr{F} \right] \} \\
= & \;\;\;\;\;\text{"distribution property of $^{-1}$"} \\
& \{ \mathscr{F},f : \mathscr{F} \subseteq D : \bigcup \{ V : V \in \mathscr{F} : f^{-1}[V]\} \} \\
= & \;\;\;\;\;\text{"rewrite using the definition of $^{\cup}$"} \\
& \{ V,f : V \in D : f^{-1}[V] \}^{\cup} \\
= & \;\;\;\;\;\text{"definition of $^\#$"} \\
& D^{\#\cup} \\
\end{align}
$$
The proof for $(2)$ is very similar.
Observations. As a first observation, I quite like how the $^\#$ notation works out, and interacts with the $^\cap$ and $^\cup$ notations I'd invented earlier: $(0)$ brings out the structure of the problem in a way that is much clearer than writing $(\mathcal{A}_{B^{\cap\cup}})^{\cap\cup} = (\mathcal{A}_B)^{\cap\cup}$: after making the latter explicit, I suddenly realized that $\mathcal{A}$ is essentially a unary operator.
Also, I like how we've been able to avoid $\subseteq$, and instead have an equality proof.  That is not what I expected when I started to work on this proof.
Finally, I dislike the set comprehension manipulation: the two steps above "using the definition of $^\cup$" are not really easy to follow.  I don't yet see how to fix that.
