Let $F$ be a presheaf (of sets) on some topological space $X$, and define the etale space of $F$ as $\widehat{F} = \bigsqcup_{x\in X} F_x$, where $F_x$ is the stalk over $x \in X$. There is a canonical projection $\pi\colon \widehat{F} \to X$, $\pi(F_x) = x$, and the topology on $\widehat{F}$ is supposed to be the coarsest one such that $\pi$ is continuous.
I have some notes that say that then the subsets of $\widehat{F}$ of the form \begin{align*} T(s, U) = \{(x, [s]_x) \mid x \in U\} \end{align*} form a subbase, for $U \subset X$ open, $s \in F(U)$ a section, and $[s]_x$ the germ of $s$ at $x$.
Can that be true, though? If the topology on $\widehat{F}$ is the coarsest one such that $\pi$ is continuous, then a subbase is given by sets of the form \begin{align*} T'(U) = \pi^{-1}(U) = \bigsqcup_{x \in U} F_x \end{align*} for $U$ open.
But if define the sets $T(s, U)$ to be open, then they generate a finer topology than the sets $T'(U)$, since $T(s,U)$ is a (proper) subset of $T'(U)$.
In the answer to this question, the topology is immediately described as the one generated by $T(s,U)$, without any mention of its coarseness.
To be more clear: my question is whether the notes are correct that this is the coarsest topology making $\pi$ continuous and if I am overlooking something?
As per the comments, the usual topolgy is the one generated by the $T(s, U)$.