Lemma 48.32.6. In Situation 48.16.1 let $f : X \to Y$ and $g : Y \to Z$ be composable morphisms of $\textit{FTS}_ S$. With notation as in Remark 48.32.5 we have $Rg_! \circ Rf_! = R(g \circ f)_!$.

Proof. By the discussion in Categories, Remark 4.22.7 it suffices to show that we obtain the same answer if we compute $\mathop{\mathrm{Hom}}\nolimits$ into $L$ in $D^ b_{\textit{Coh}}(\mathcal{O}_ Z)$. To do this we compute, using the notation in Remark 48.32.5, as follows

\begin{align*} \mathop{\mathrm{Hom}}\nolimits _ Z(Rg_!Rf_!K, L) & = \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _ Z(Rg_!M_ n, L) \\ & = \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _ Y(M_ n, g^!L) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Y(Rf_!K, g^!L) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(K, f^!g^!L) \\ & = \mathop{\mathrm{Hom}}\nolimits _ X(K, (g \circ f)^!L) \\ & = \mathop{\mathrm{Hom}}\nolimits _ Z(R(g \circ f)_!K, L) \end{align*}

The first equality is the definition of $Rg_!Rf_!K$. The second equality is Proposition 48.32.2 for $g$. The third equality is the fact that $Rf_!K$ is given by $(M_ n)$. The fourth equality is Proposition 48.32.2 for $f$. The fifth equality is Lemma 48.16.3. The sixth is Proposition 48.32.2 for $g \circ f$. $\square$

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