Definition of final topology

Question

If $X$ is a set, $\{X_i\}$ a family of topological spaces and for each $i\in I$ there is a map $f_i:X_i\rightarrow X$, then one can define a topology on $X$ which is the finest topology for which the maps $f_i$ are all continuous. Sucha topology is the final topology. According to Bourbaki (and also to Wiki) the open sets in $X$ are such that their preimage under each $f_i$ are open sets in $X_i$.

It works fine for the subsets $V\subset Y$ such that

$$\bigcap_{i\in I} f_i(X_i). $$

But I can't understand what happens with the sets $U$ which are not contained in the above intersection. Clearly $f^{-1}_i(U)$ cannot be open in $X_i$, since the operation $f^{-1}_i(U)$ is not defined for at least one index in $I$. A solution would be that $U$ were not open. But then, it may happen that $X$ is not open, which is completely forbidden.

What condition I'm misunderstanding in my above reasoning?

Maybe the subsets $U$ of $X$ such that

$$ U\subseteq \bigcap_{i\in I} f_i(X_i) $$

and $f_i^{-1}(U)$ are open in every space $X_i$ form a basis for the toplogy of $X$ but they are not the whole set of open sets (as for example Bourbaki says)?

Answer 1

There is a problem with the sentence “Clearly $f^{-1}_i(U)$ cannot be open in $X_i$, since the operation $f^{-1}_i(U)$ is not defined for at least one index in $I$.” For each $U\subset Y$ and each $i\in I$, it makes sense to talk about $f_i^{-1}(U)$, since it is defined as $\left\{x\in X\,\middle|\,f(x)\in U\right\}$.

Answer 2

$f^{-1}[U_i]$ is always defined, namely as $\{x\in X_i: f_i(x) \in U_i\}$. This might be empty (or even $X_i$) depending on $U_i$.

Example: suppose $f: X \to Y$ is a constant map with value $p \in Y$. Then

$f^{-1}[O]=\emptyset$ when $p\notin O$ and $f^{-1}[O] = X$ when $p\in O$. So open in $X$ always, regardless of the precise topology on $X$.

So the final topology for $f$, is the discrete topology on $Y$: all sets are open.