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 (2)
Comment #8571 by Alejandro González Nevado on
Comment #9149 by Stacks project on
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