Lemma 96.3.2. Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Then $(g \circ f)^ p = f^ p \circ g^ p$ and there is a canonical isomorphism ${}_ p(g \circ f) \to {}_ pg \circ {}_ pf$ compatible with adjointness of $(f^ p, {}_ pf)$, $(g^ p, {}_ pg)$, and $((g \circ f)^ p, {}_ p(g \circ f))$.
Proof. Let $\mathcal{H}$ be a presheaf on $\mathcal{Z}$. Then $(g \circ f)^ p\mathcal{H} = f^ p (g^ p\mathcal{H})$ is given by the equalities
\[ (g \circ f)^ p\mathcal{H}(x) = \mathcal{H}((g \circ f)(x)) = \mathcal{H}(g(f(x))) = f^ p (g^ p\mathcal{H})(x). \]
We omit the verification that this is compatible with restriction maps.
Next, we define the transformation ${}_ p(g \circ f) \to {}_ pg \circ {}_ pf$. Let $\mathcal{F}$ be a presheaf on $\mathcal{X}$. If $z$ is an object of $\mathcal{Z}$ then we get a category $\mathcal{J}$ of quadruples $(x, f(x) \to y, y, g(y) \to z)$ and a category $\mathcal{I}$ of pairs $(x, g(f(x)) \to z)$. There is a canonical functor $\mathcal{J} \to \mathcal{I}$ sending the object $(x, \alpha : f(x) \to y, y, \beta : g(y) \to z)$ to $(x, \beta \circ f(\alpha ) : g(f(x)) \to z)$. This gives the arrow in
\begin{align*} ({}_ p(g \circ f)\mathcal{F})(z) & = \mathop{\mathrm{lim}}\nolimits _{g(f(x)) \to z} \mathcal{F}(x) \\ & = \mathop{\mathrm{lim}}\nolimits _\mathcal {I} \mathcal{F} \\ & \to \mathop{\mathrm{lim}}\nolimits _\mathcal {J} \mathcal{F} \\ & = \mathop{\mathrm{lim}}\nolimits _{g(y) \to z} \Big(\mathop{\mathrm{lim}}\nolimits _{f(x) \to y} \mathcal{F}(x)\Big) \\ & = ({}_ pg \circ {}_ pf\mathcal{F})(x) \end{align*}
by Categories, Lemma 4.14.9. We omit the verification that this is compatible with restriction maps. An alternative to this direct construction is to define ${}_ p(g \circ f) \cong {}_ pg \circ {}_ pf$ as the unique map compatible with the adjointness properties. This also has the advantage that one does not need to prove the compatibility.
Compatibility with adjointness of $(f^ p, {}_ pf)$, $(g^ p, {}_ pg)$, and $((g \circ f)^ p, {}_ p(g \circ f))$ means that given presheaves $\mathcal{H}$ and $\mathcal{F}$ as above we have a commutative diagram
\[ \xymatrix{ \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{X})}(f^ pg^ p\mathcal{H}, \mathcal{F}) \ar@{=}[r] \ar@{=}[d] & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{Y})}(g^ p\mathcal{H}, {}_ pf\mathcal{F}) \ar@{=}[r] & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{Y})}(\mathcal{H}, {}_ pg{}_ pf\mathcal{F}) \\ \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{X})}((g \circ f)^ p\mathcal{G}, \mathcal{F}) \ar@{=}[rr] & & \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{Y})}(\mathcal{G}, {}_ p(g \circ f)\mathcal{F}) \ar[u] } \]
Proof omitted. $\square$
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