Consider a family of maps $(f_t)_{t\in\mathbb{T}} : S\to S'$. I would like to define a topology making this family continuous. To do so, I consider a topology $\mathcal{U'}$ on $S'$ and I define the topology $\mathcal{U}$ of $S$ which consists of all the preimage of $f_t$ by an element of $\mathcal{U'}$. Clearly we have :
- $\emptyset$ and $S$ are in $\mathcal{U}$
- If $A_i$ is a finite family of $\mathcal{U}$, then $\cup_{i=1}^{n}A_i$ : For each $A_i$ there exists $A_i^{'}\in\mathcal{U'}$ such that $f^{-1}(A_i^{'}) = A_i$ and we conclude by the property of the preimage which allows to invert union and function.
- If $A_i$ is a finite family of $\mathcal{U}$, then $\cap_{i=1}^{n}A_i$ : For each $A_i$ there exists $A_i^{'}\in\mathcal{U'}$ such that $f^{-1}(A_i^{'}) = A_i$ and we conclude by the property of the preimage which allows to invert intersection and function.
Now I would like to find a subbase of $\mathcal{U}$. My attempt is as follows :
Consider the family
$$ \mathcal{A} = \{ f_{t}^{-1}(A') : t\in\mathbb{T} , A'\in\mathcal{U'}\} $$
Consider now $B\in\mathcal{U}$, we know that there exists $B'\in\mathcal{U'}$ and $t\in\mathbb{T}$ such that :
$$ f_{t}^{-1}(B')\cap S = B $$
Since this holds for all elements of $\mathcal{U}$, we have shown that every element of $\mathcal{U}$ can be written as a finite intersection of elements of $\mathcal{A}$.
I would like to know if my attempt is correct please ? And if you have another way to find a subbase.
Thank you a lot for your help !