Lemma 76.30.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let
\[ n_{X/Y} : |Y| \to \{ 0, 1, 2, 3, \ldots , \infty \} \]
be the function which associates to $y \in Y$ the number of connected components of $X_ k$ where $\mathop{\mathrm{Spec}}(k) \to Y$ is in the equivalence class of $y$ with $k$ algebraically closed. This is well defined and if $g : Y' \to Y$ is a morphism then
\[ n_{X'/Y'} = n_{X/Y} \circ g \]
where $X' \to Y'$ is the base change of $f$.
Proof.
Suppose that $y' \in Y'$ has image $y \in Y$. Let $\mathop{\mathrm{Spec}}(k') \to Y'$ be in the equivalence class of $y'$ with $k'$ algebraically closed. Then we can choose a commutative diagram
\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[rd] & \mathop{\mathrm{Spec}}(k') \ar[r] & Y' \ar[d] \\ & \mathop{\mathrm{Spec}}(k) \ar[r] & Y } \]
where $K$ is an algebraically closed field. The result follows as the morphisms of schemes
\[ \xymatrix{ X'_{k'} & (X'_{k'})_ K = (X_ k)_ K \ar[l] \ar[r] & X_ k } \]
induce bijections between connected components, see Spaces over Fields, Lemma 72.12.4. To use this to prove the function is well defined take $Y' = Y$.
$\square$
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