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Tag 03PQ

Chapter 50: Étale Cohomology > Section 50.29: Neighborhoods, stalks and points

Lemma 50.29.4. Let $S$ be a scheme, and let $\overline{s}$ be a geometric point of $S$. The category of étale neighborhoods is cofiltered. More precisely:

  1. Let $(U_i, \overline{u}_i)_{i = 1, 2}$ be two étale neighborhoods of $\overline{s}$ in $S$. Then there exists a third étale neighborhood $(U, \overline{u})$ and morphisms $(U, \overline{u}) \to (U_i, \overline{u}_i)$, $i = 1, 2$.
  2. Let $h_1, h_2: (U, \overline{u}) \to (U', \overline{u}')$ be two morphisms between étale neighborhoods of $\overline{s}$. Then there exist an étale neighborhood $(U'', \overline{u}'')$ and a morphism $h : (U'', \overline{u}'') \to (U, \overline{u})$ which equalizes $h_1$ and $h_2$, i.e., such that $h_1 \circ h = h_2 \circ h$.

Proof. For part (1), consider the fibre product $U = U_1 \times_S U_2$. It is étale over both $U_1$ and $U_2$ because étale morphisms are preserved under base change, see Proposition 50.26.2. The map $\overline{s} \to U$ defined by $(\overline{u}_1, \overline{u}_2)$ gives it the structure of an étale neighborhood mapping to both $U_1$ and $U_2$. For part (2), define $U''$ as the fibre product $$ \xymatrix{ U'' \ar[r] \ar[d] & U \ar[d]^{(h_1, h_2)} \\ U' \ar[r]^-\Delta & U' \times_S U'. } $$ Since $\overline{u}$ and $\overline{u}'$ agree over $S$ with $\overline{s}$, we see that $\overline{u}'' = (\overline{u}, \overline{u}')$ is a geometric point of $U''$. In particular $U'' \not = \emptyset$. Moreover, since $U'$ is étale over $S$, so is the fibre product $U'\times_S U'$ (see Proposition 50.26.2). Hence the vertical arrow $(h_1, h_2)$ is étale by Remark 50.29.2 above. Therefore $U''$ is étale over $U'$ by base change, and hence also étale over $S$ (because compositions of étale morphisms are étale). Thus $(U'', \overline{u}'')$ is a solution to the problem. $\square$

    The code snippet corresponding to this tag is a part of the file etale-cohomology.tex and is located in lines 3226–3242 (see updates for more information).

    \begin{lemma}
    \label{lemma-cofinal-etale}
    Let $S$ be a scheme, and let $\overline{s}$ be a geometric point of $S$.
    The category of \'etale neighborhoods is cofiltered. More precisely:
    \begin{enumerate}
    \item Let $(U_i, \overline{u}_i)_{i = 1, 2}$ be two \'etale neighborhoods of
    $\overline{s}$ in $S$. Then there exists a third \'etale neighborhood
    $(U, \overline{u})$ and morphisms
    $(U, \overline{u}) \to (U_i, \overline{u}_i)$, $i = 1, 2$.
    \item Let $h_1, h_2: (U, \overline{u}) \to (U', \overline{u}')$ be two
    morphisms between \'etale neighborhoods of $\overline{s}$. Then there exist an
    \'etale neighborhood $(U'', \overline{u}'')$ and a morphism
    $h : (U'', \overline{u}'') \to (U, \overline{u})$
    which equalizes $h_1$ and $h_2$, i.e., such that
    $h_1 \circ h = h_2 \circ h$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    For part (1), consider the fibre product $U = U_1 \times_S U_2$.
    It is \'etale over both $U_1$ and $U_2$ because \'etale morphisms are
    preserved under base change, see
    Proposition \ref{proposition-etale-morphisms}.
    The map $\overline{s} \to U$ defined by $(\overline{u}_1, \overline{u}_2)$
    gives it the structure of an \'etale neighborhood mapping to both
    $U_1$ and $U_2$. For part (2), define $U''$ as the fibre product
    $$
    \xymatrix{
    U'' \ar[r] \ar[d] & U \ar[d]^{(h_1, h_2)} \\
    U' \ar[r]^-\Delta & U' \times_S U'.
    }
    $$
    Since $\overline{u}$ and $\overline{u}'$ agree over $S$ with $\overline{s}$,
    we see that $\overline{u}'' = (\overline{u}, \overline{u}')$ is a geometric
    point of $U''$. In particular $U'' \not = \emptyset$.
    Moreover, since $U'$ is \'etale over $S$, so is the fibre product
    $U'\times_S U'$ (see
    Proposition \ref{proposition-etale-morphisms}).
    Hence the vertical arrow $(h_1, h_2)$ is \'etale by
    Remark \ref{remark-etale-between-etale} above.
    Therefore $U''$ is \'etale over $U'$ by base change, and hence also
    \'etale over $S$ (because compositions of \'etale morphisms are \'etale).
    Thus $(U'', \overline{u}'')$ is a solution to the problem.
    \end{proof}

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