The Stacks project

Proof. Observe that the map simply sends a triple $(T, g, \mathcal{L})$ as in Examples of Stacks, Section 95.16 to the same triple $(T, g, \mathcal{L})$ but where now we view this as a triple of the kind described in Situation 99.5.1. This works because the invertible $\mathcal{O}_{X_ T}$-module $\mathcal{L}$ is certainly a finitely presented $\mathcal{O}_{X_ T}$-module, it is flat over $T$ because $X_ T \to T$ is flat, and the support is proper over $T$ as $X_ T \to T$ is proper (Morphisms of Spaces, Lemmas 67.30.4 and 67.40.3). Thus the statement makes sense.

Having said this, it is clear that the content of the lemma is the following: given an object $(T, g, \mathcal{F})$ of $\mathcal{C}\! \mathit{oh}_{X/B}$ there is an open subscheme $U \subset T$ such that for a morphism of schemes $T' \to T$ the following are equivalent

1. $T' \to T$ factors through $U$,

2. the pullback $\mathcal{F}_{T'}$ of $\mathcal{F}$ by $X_{T'} \to X_ T$ is invertible.

Let $W \subset |X_ T|$ be the set of points $x \in |X_ T|$ such that $\mathcal{F}$ is locally free in a neighbourhood of $x$. By More on Morphisms of Spaces, Lemma 76.23.8. $W$ is open and formation of $W$ commutes with arbitrary base change. Clearly, if $T' \to T$ satisfies (b), then $|X_{T'}| \to |X_ T|$ maps into $W$. Hence we may replace $T$ by the open $T \setminus f_ T(|X_ T| \setminus W)$ in order to construct $U$. After doing so we reach the situation where $\mathcal{F}$ is finite locally free. In this case we get a disjoint union decomposition $X_ T = X_0 \amalg X_1 \amalg X_2 \amalg \ldots $ into open and closed subspaces such that the restriction of $\mathcal{F}$ is locally free of rank $i$ on $X_ i$. Then clearly

works. (Note that if we assume that $T$ is quasi-compact, then $X_ T$ is quasi-compact hence only a finite number of $X_ i$ are nonempty and so $U$ is indeed open.) $\square$