Lemma 6.30.3. With notation as above. For each $U \in \mathcal{B}$, let $C(U) \subset \text{Cov}_\mathcal {B}(U)$ be a cofinal system. For each $U \in \mathcal{B}$, and each $\mathcal{U} : U = \bigcup U_ i$ in $C(U)$, let coverings $\mathcal{U}_{ij} : U_ i \cap U_ j = \bigcup U_{ijk}$, $U_{ijk} \in \mathcal{B}$ be given. Let $\mathcal{F}$ be a presheaf of sets on $\mathcal{B}$. The following are equivalent

1. The presheaf $\mathcal{F}$ is a sheaf on $\mathcal{B}$.

2. For every $U \in \mathcal{B}$ and every covering $\mathcal{U} : U = \bigcup U_ i$ in $C(U)$ the sheaf condition $(**)$ holds (for the given coverings $\mathcal{U}_{ij}$).

Proof. We have to show that (2) implies (1). Suppose that $U \in \mathcal{B}$, and that $\mathcal{U} : U = \bigcup _{i\in I} U_ i$ is an arbitrary covering by elements of $\mathcal{B}$. Because the system $C(U)$ is cofinal we can find an element $\mathcal{V} : U = \bigcup _{j \in J} V_ j$ in $C(U)$ which refines $\mathcal{U}$. This means there exists a map $\alpha : J \to I$ such that $V_ j \subset U_{\alpha (j)}$.

Note that if $s, s' \in \mathcal{F}(U)$ are sections such that $s|_{U_ i} = s'|_{U_ i}$, then

$s|_{V_ j} = (s|_{U_{\alpha (j)}})|_{V_ j} = (s'|_{U_{\alpha (j)}})|_{V_ j} = s'|_{V_ j}$

for all $j$. Hence by the uniqueness in $(**)$ for the covering $\mathcal{V}$ we conclude that $s = s'$. Thus we have proved the uniqueness part of $(**)$ for our arbitrary covering $\mathcal{U}$.

Suppose furthermore that $U_ i \cap U_{i'} = \bigcup _{k \in I_{ii'}} U_{ii'k}$ are arbitrary coverings by $U_{ii'k} \in \mathcal{B}$. Let us try to prove the existence part of $(**)$ for the system $(\mathcal{U}, \mathcal{U}_{ij})$. Thus let $s_ i \in \mathcal{F}(U_ i)$ and suppose we have

$s_ i|_{U_{ii'k}} = s_{i'}|_{U_{ii'k}}$

for all $i, i', k$. Set $t_ j = s_{\alpha (j)}|_{V_ j}$, where $\mathcal{V}$ and $\alpha$ are as above.

There is one small kink in the argument here. Namely, let $\mathcal{V}_{jj'} : V_ j \cap V_{j'} = \bigcup _{l \in J_{jj'}} V_{jj'l}$ be the covering given to us by the statement of the lemma. It is not a priori clear that

$t_ j|_{V_{jj'l}} = t_{j'}|_{V_{jj'l}}$

for all $j, j', l$. To see this, note that we do have

$t_ j|_ W = t_{j'}|_ W \text{ for all } W \in \mathcal{B}, W \subset V_{jj'l} \cap U_{\alpha (j)\alpha (j')k}$

for all $k \in I_{\alpha (j)\alpha (j')}$, by our assumption on the family of elements $s_ i$. And since $V_ j \cap V_{j'} \subset U_{\alpha (j)} \cap U_{\alpha (j')}$ we see that $t_ j|_{V_{jj'l}}$ and $t_{j'}|_{V_{jj'l}}$ agree on the members of a covering of $V_{jj'l}$ by elements of $\mathcal{B}$. Hence by the uniqueness part proved above we finally deduce the desired equality of $t_ j|_{V_{jj'l}}$ and $t_{j'}|_{V_{jj'l}}$. Then we get the existence of an element $t \in \mathcal{F}(U)$ by property $(**)$ for $(\mathcal{V}, \mathcal{V}_{jj'})$.

Again there is a small snag. We know that $t$ restricts to $t_ j$ on $V_ j$ but we do not yet know that $t$ restricts to $s_ i$ on $U_ i$. To conclude this note that the sets $U_ i \cap V_ j$, $j \in J$ cover $U_ i$. Hence also the sets $U_{i \alpha (j) k} \cap V_ j$, $j\in J$, $k \in I_{i\alpha (j)}$ cover $U_ i$. We leave it to the reader to see that $t$ and $s_ i$ restrict to the same section of $\mathcal{F}$ on any $W \in \mathcal{B}$ which is contained in one of the open sets $U_{i \alpha (j) k} \cap V_ j$, $j\in J$, $k \in I_{i\alpha (j)}$. Hence by the uniqueness part seen above we win. $\square$

Comment #1804 by Keenan Kidwell on

In the final line of the first text block of the proof, the inclusion $V_j\subseteq U_{\alpha(i)}$ should be $V_j\subseteq U_{\alpha(j)}$.

Comment #6519 by Wet Lee on

ln the third paragraph of the proof of Lemma 6.30.3, should be ?

There are also:

• 6 comment(s) on Section 6.30: Bases and sheaves

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).