**Proof.**
Suppose to begin with that $\mathcal{T}$ is a scheme $T$, let $U \to \mathcal{X}$ be a smooth surjective morphism whose source is a scheme, and let $T' = T \times _{\mathcal{X}} U$. Let $f': T' \to U$ be the pull-back of $f$ over $U$, and let $g: T' \to T$ be the projection.

Lemma 105.5.9 shows that $\dim _{t'}(T'_{f'(t')}) = \dim _{g(t')}(T_{f(g(t'))}),$ for $t' \in T'$, while, since $g$ is smooth and surjective (being the base-change of a smooth surjective morphism) the map induced by $g$ on underlying topological spaces is continuous and open (by Properties of Spaces, Lemma 64.4.6), and surjective. Thus it suffices to note that part (1) for the morphism $f'$ follows from Morphisms of Spaces, Lemma 65.34.4, and part (2) from either of Morphisms, Lemma 29.29.4 or Morphisms, Lemma 29.34.12 (each of which gives the result for schemes, from which the analogous results for algebraic spaces can be deduced exactly as in Morphisms of Spaces, Lemma 65.34.4.

Now return to the general case, and choose a smooth surjective morphism $h:V \to \mathcal{T}$ whose source is a scheme. If $v \in V$, then, essentially by definition, we have

\[ \dim _{h(v)}(\mathcal{T}_{f(h(v))}) = \dim _{v}(V_{f(h(v))}) - \dim _{v}(V_{h(v)}). \]

Since $V$ is a scheme, we have proved that the first of the terms on the right hand side of this equality is upper semi-continuous (and even locally constant if $f$ is smooth), while the second term is in fact locally constant. Thus their difference is upper semi-continuous (and locally constant if $f$ is smooth), and hence the function $\dim _{h(v)}(\mathcal{T}_{f(h(v))})$ is upper semi-continuous on $|V|$ (and locally constant if $f$ is smooth). Since the morphism $|V| \to |\mathcal{T}|$ is open and surjective, the lemma follows.
$\square$

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