Lemma 7.28.7 (0FN2)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 7: Sites and Sheaves
Section 7.28: Localization and morphisms
Lemma 7.28.7 (cite)

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Lemma 7.28.7. Assume given sites $\mathcal{C}', \mathcal{C}, \mathcal{D}', \mathcal{D}$ and functors

\[ \xymatrix{ \mathcal{C}' \ar[r]_{v'} & \mathcal{C} \\ \mathcal{D}' \ar[r]^ v \ar[u]^{u'} & \mathcal{D} \ar[u]_ u } \]

With notation as in Sections 7.14 and 7.21 assume

$u$ and $u'$ are continuous giving rise to morphisms of sites $f$ and $f'$,
$v$ and $v'$ are cocontinuous giving rise to morphisms of topoi $g$ and $g'$,
$u \circ v = v' \circ u'$, and
$v$ and $v'$ are continuous as well as cocontinuous.

Then ¹ $f'_* \circ (g')^{-1} = g^{-1} \circ f_*$ and $g'_! \circ (f')^{-1} = f^{-1} \circ g_!$.

Proof. Namely, we have

\[ f'_*(g')^{-1}\mathcal{F} = (u')^ p((v')^ p\mathcal{F})^\# = (u')^ p(v')^ p\mathcal{F} \]

The first equality by definition and the second by Lemma 7.21.5. We have

\[ g^{-1}f_*\mathcal{F} = (v^ pu^ p\mathcal{F})^\# = ((u')^ p(v')^ p\mathcal{F})^\# = (u')^ p(v')^ p\mathcal{F} \]

The first equality by definition, the second because $u \circ v = v' \circ u'$, the third because we already saw that $(u')^ p(v')^ p\mathcal{F}$ is a sheaf. This proves $f'_* \circ (g')^{-1} = g^{-1} \circ f_*$ and the equality $g'_! \circ (f')^{-1} = f^{-1} \circ g_!$ follows by uniqueness of left adjoints. $\square$

[1] In this generality we don't know $f \circ g'$ is equal to $g \circ f'$ as morphisms of topoi (there is a canonical $2$-arrow from the first to the second which may not be an isomorphism).

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