The Stacks project

Lemma 7.10.6. Any two morphisms $f, g: \mathcal{U} \to \mathcal{V}$ of coverings inducing the same morphism $U \to V$ induce the same map $H^0(\mathcal{V}, \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F})$.

Proof. Let $\mathcal{U} = \{ U_ i \to U\} _{i\in I}$ and $\mathcal{V} = \{ V_ j \to V\} _{j\in J}$. The morphism $f$ consists of a map $U\to V$, a map $\alpha : I \to J$ and maps $f_ i : U_ i \to V_{\alpha (i)}$. Likewise, $g$ determines a map $\beta : I \to J$ and maps $g_ i : U_ i \to V_{\beta (i)}$. As $f$ and $g$ induce the same map $U\to V$, the diagram

\[ \xymatrix{ & V_{\alpha (i)} \ar[dr] \\ U_ i \ar[ur]^{f_ i} \ar[dr]_{g_ i} & & V \\ & V_{\beta (i)} \ar[ur] } \]

is commutative for every $i\in I$. Hence $f$ and $g$ factor through the fibre product

\[ \xymatrix{ & V_{\alpha (i)} \\ U_ i \ar[r]^-\varphi \ar[ur]^{f_ i} \ar[dr]_{g_ i} & V_{\alpha (i)} \times _ V V_{\beta (i)} \ar[u]_{\text{pr}_1} \ar[d]^{\text{pr}_2} \\ & V_{\beta (i)}. } \]

Now let $s = (s_ j)_ j \in H^0(\mathcal{V}, \mathcal{F})$. Then for all $i\in I$:

\[ (f^*s)_ i = f_ i^*(s_{\alpha (i)}) = \varphi ^*\text{pr}_1^*(s_{\alpha (i)}) = \varphi ^*\text{pr}_2^*(s_{\beta (i)}) = g_ i^*(s_{\beta (i)}) = (g^*s)_ i, \]

where the middle equality is given by the definition of $H^0(\mathcal{V}, \mathcal{F})$. This shows that the maps $H^0(\mathcal{V}, \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F})$ induced by $f$ and $g$ are equal. $\square$


Comments (1)

Comment #8571 by Alejandro González Nevado on

SS: Two morphisms between the same pair of covergins of a site inducing (covariantly) the same morphism on the objects of the site covered by these coverings induce (contravariantly) the same map on the zeroth Čech cohomologies of any presheaf of sets on the site with respect to these two coverings.

There are also:

  • 8 comment(s) on Section 7.10: Sheafification

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