The Stacks project

59.43 Property (B)

Please see Section 59.41 for the definition of property (B).

Lemma 59.43.1. Let $f : X \to Y$ be a morphism of schemes. Assume (B) holds. Then the functor $f_{small, *} : \mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}) \to \mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale})$ transforms surjections into surjections.

Proof. This follows from Sites, Lemma 7.41.2. $\square$

Lemma 59.43.2. Let $f : X \to Y$ be a morphism of schemes. Suppose

  1. $V \to Y$ is an étale morphism of schemes,

  2. $\{ U_ i \to X \times _ Y V\} $ is an étale covering, and

  3. $v \in V$ is a point.

Assume that for any such data there exists an étale neighbourhood $(V', v') \to (V, v)$, a disjoint union decomposition $X \times _ Y V' = \coprod W'_ i$, and morphisms $W'_ i \to U_ i$ over $X \times _ Y V$. Then property (B) holds.

Proof. Omitted. $\square$

Lemma 59.43.3. Let $f : X \to Y$ be a finite morphism of schemes. Then property (B) holds.

Proof. Consider $V \to Y$ étale, $\{ U_ i \to X \times _ Y V\} $ an étale covering, and $v \in V$. We have to find a $V' \to V$ and decomposition and maps as in Lemma 59.43.2. We may shrink $V$ and $Y$, hence we may assume that $V$ and $Y$ are affine. Since $X$ is finite over $Y$, this also implies that $X$ is affine. During the proof we may (finitely often) replace $(V, v)$ by an étale neighbourhood $(V', v')$ and correspondingly the covering $\{ U_ i \to X \times _ Y V\} $ by $\{ V' \times _ V U_ i \to X \times _ Y V'\} $.

Since $X \times _ Y V \to V$ is finite there exist finitely many (pairwise distinct) points $x_1, \ldots , x_ n \in X \times _ Y V$ mapping to $v$. We may apply More on Morphisms, Lemma 37.40.5 to $X \times _ Y V \to V$ and the points $x_1, \ldots , x_ n$ lying over $v$ and find an étale neighbourhood $(V', v') \to (V, v)$ such that

\[ X \times _ Y V' = R \amalg \coprod T_ a \]

with $T_ a \to V'$ finite with exactly one point $p_ a$ lying over $v'$ and moreover $\kappa (v') \subset \kappa (p_ a)$ purely inseparable, and such that $R \to V'$ has empty fibre over $v'$. Because $X \to Y$ is finite, also $R \to V'$ is finite. Hence after shrinking $V'$ we may assume that $R = \emptyset $. Thus we may assume that $X \times _ Y V = X_1 \amalg \ldots \amalg X_ n$ with exactly one point $x_ l \in X_ l$ lying over $v$ with moreover $\kappa (v) \subset \kappa (x_ l)$ purely inseparable. Note that this property is preserved under refinement of the étale neighbourhood $(V, v)$.

For each $l$ choose an $i_ l$ and a point $u_ l \in U_{i_ l}$ mapping to $x_ l$. Now we apply property (A) for the finite morphism $X \times _ Y V \to V$ and the étale morphisms $U_{i_ l} \to X \times _ Y V$ and the points $u_ l$. This is permissible by Lemma 59.42.3 This gives produces an étale neighbourhood $(V', v') \to (V, v)$ and decompositions

\[ X \times _ Y V' = W_ l \amalg R_ l \]

and $X$-morphisms $a_ l : W_ l \to U_{i_ l}$ whose image contains $u_{i_ l}$. Here is a picture:

\[ \xymatrix{ & & & U_{i_ l} \ar[d] & \\ W_ l \ar[rrru] \ar[r] & W_ l \amalg R_ l \ar@{=}[r] & X \times _ Y V' \ar[r] \ar[d] & X \times _ Y V \ar[r] \ar[d] & X \ar[d] \\ & & V' \ar[r] & V \ar[r] & Y } \]

After replacing $(V, v)$ by $(V', v')$ we conclude that each $x_ l$ is contained in an open and closed neighbourhood $W_ l$ such that the inclusion morphism $W_ l \to X \times _ Y V$ factors through $U_ i \to X \times _ Y V$ for some $i$. Replacing $W_ l$ by $W_ l \cap X_ l$ we see that these open and closed sets are disjoint and moreover that $\{ x_1, \ldots , x_ n\} \subset W_1 \cup \ldots \cup W_ n$. Since $X \times _ Y V \to V$ is finite we may shrink $V$ and assume that $X \times _ Y V = W_1 \amalg \ldots \amalg W_ n$ as desired. $\square$

Lemma 59.43.4. Let $f : X \to Y$ be an integral morphism of schemes. Then property (B) holds.

Proof. Consider $V \to Y$ étale, $\{ U_ i \to X \times _ Y V\} $ an étale covering, and $v \in V$. We have to find a $V' \to V$ and decomposition and maps as in Lemma 59.43.2. We may shrink $V$ and $Y$, hence we may assume that $V$ and $Y$ are affine. Since $X$ is integral over $Y$, this also implies that $X$ and $X \times _ Y V$ are affine. We may refine the covering $\{ U_ i \to X \times _ Y V\} $, and hence we may assume that $\{ U_ i \to X \times _ Y V\} _{i = 1, \ldots , n}$ is a standard étale covering. Write $Y = \mathop{\mathrm{Spec}}(A)$, $X = \mathop{\mathrm{Spec}}(B)$, $V = \mathop{\mathrm{Spec}}(C)$, and $U_ i = \mathop{\mathrm{Spec}}(B_ i)$. Then $A \to B$ is an integral ring map, and $B \otimes _ A C \to B_ i$ are étale ring maps. By Algebra, Lemma 10.143.3 we can find a finite $A$-subalgebra $B' \subset B$ and an étale ring map $B' \otimes _ A C \to B'_ i$ for $i = 1, \ldots , n$ such that $B_ i = B \otimes _{B'} B'_ i$. Thus the question reduces to the étale covering $\{ \mathop{\mathrm{Spec}}(B'_ i) \to X' \times _ Y V\} _{i = 1, \ldots , n}$ with $X' = \mathop{\mathrm{Spec}}(B')$ finite over $Y$. In this case the result follows from Lemma 59.43.3. $\square$

Lemma 59.43.5. Let $f : X \to Y$ be a morphism of schemes. Assume $f$ is integral (for example finite). Then

  1. $f_{small, *}$ transforms surjections into surjections (on sheaves of sets and on abelian sheaves),

  2. $f_{small}^{-1}f_{small, *}\mathcal{F} \to \mathcal{F}$ is surjective for any abelian sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$,

  3. $f_{small, *} : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$ is faithful and reflects injections and surjections, and

  4. $f_{small, *} : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$ is exact.

Proof. Parts (2), (3) we have seen in Lemma 59.42.4. Part (1) follows from Lemmas 59.43.4 and 59.43.1. Part (4) is a consequence of part (1), see Modules on Sites, Lemma 18.15.2. $\square$


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