Lemma 62.4.12. Let $f : X \to Y$ and $g : Y \to Z$ be composable locally quasi-finite morphisms of schemes. Then there is a canonical isomorphism of functors

$(g \circ f)_! \longrightarrow g_! \circ f_!$

These isomorphisms satisfy the following properties:

1. If $f$ and $g$ are separated, then the isomorphism agrees with Lemma 62.3.13.

2. If $g$ is separated, then the isomorphism agrees with Lemma 62.4.11.

3. For a geometric point $\overline{z} : \mathop{\mathrm{Spec}}(k) \to Z$ the diagram

$\xymatrix{ ((g \circ f)_!\mathcal{F})_{\overline{z}} \ar[d] \ar[rr] & & \bigoplus \nolimits _{g(f(\overline{x})) = \overline{z}} \mathcal{F}_{\overline{x}} \ar@{=}[d] \\ (g_!f_!\mathcal{F})_{\overline{z}} \ar[r] & \bigoplus \nolimits _{g(\overline{y}) = \overline{z}} (f_!\mathcal{F})_{\overline{y}} \ar[r] & \bigoplus \nolimits _{g(f(\overline{x})) = \overline{z}} \mathcal{F}_{\overline{x}} }$

is commutative where the horizontal arrows are given by Lemma 62.4.5.

4. Let $h : Z \to T$ be a third locally quasi-finite morphism of schemes. Then the diagram

$\xymatrix{ (h \circ g \circ f)_! \ar[r] \ar[d] & (h \circ g)_! \circ f_! \ar[d] \\ h_! \circ (g \circ f)_! \ar[r] & h_! \circ g_! \circ f_! }$

commutes.

5. Suppose that we have a diagram of schemes

$\xymatrix{ X' \ar[d]_{f'} \ar[r]_ c & X \ar[d]^ f \\ Y' \ar[d]_{g'} \ar[r]_ b & Y \ar[d]^ g \\ Z' \ar[r]^ a & Z }$

with both squares cartesian and $f$ and $g$ locally quasi-finite. Then the diagram

$\xymatrix{ a^{-1} \circ (g \circ f)_! \ar[d] \ar[rr] & & (g' \circ f')_! \circ c^{-1} \ar[d] \\ a^{-1} \circ g_! \circ f_! \ar[r] & g'_! \circ b^{-1} \circ f_! \ar[r] & g'_! \circ f'_! \circ c^{-1} }$

commutes where the horizontal arrows are those of Lemma 62.4.10.

Proof. If $f$ and $g$ are separated, then this is a special case of Lemma 62.3.13. If $g$ is separated, then this is a special case of Lemma 62.4.11 which moreover agrees with the case where $f$ and $g$ are separated.

Construction in the general case. Choose an open covering $Y = \bigcup Y_ i$ such that the restriction $g_ i : Y_ i \to Z$ of $g$ is separated. Set $X_ i = f^{-1}(Y_ i)$ and denote $f_ i : X_ i \to Y_ i$ the restriction of $f$. Also denote $h = g \circ f$ and $h_ i : X_ i \to Z$ the restriction of $h$. Consider the following diagram

$\xymatrix{ \bigoplus \nolimits _{i_0, i_1} h_{i_0i_1, !}\mathcal{F}|_{X_{i_0i_1}} \ar[r] \ar[d] & \bigoplus \nolimits _{i_0} h_{i_0, !}\mathcal{F}|_{X_{i_0}} \ar[r] \ar[d] & h_!\mathcal{F} \ar[r] \ar@{..>}[dd] & 0 \\ \bigoplus \nolimits _{i_0, i_1} g_{i_0i_1, !} f_{i_0i_1, !}\mathcal{F}|_{X_{i_0i_1}} \ar[r] \ar[d] & \bigoplus \nolimits _{i_0} g_{i_0, !} f_{i_0, !}\mathcal{F}|_{X_{i_0}} \ar[d] \\ \bigoplus \nolimits _{i_0, i_1} g_{i_0i_1, !} (f_!\mathcal{F})|_{Y_{i_0i_1}} \ar[r] & \bigoplus \nolimits _{i_0} g_{i_0, !} (f_!\mathcal{F})|_{Y_{i_0}} \ar[r] & g_!f_!\mathcal{F} \ar[r] & 0 }$

By Lemma 62.4.7 the top and bottom row in the diagram are exact. By Lemma 62.4.11 the top left square commutes. The vertical arrows in the lower left square come about because $(f_!\mathcal{F})|_{Y_{i_0i_1}} = f_{i_0i_1, !}\mathcal{F}|_{X_{i_0i_1}}$ and $(f_!\mathcal{F})|_{Y_{i_0}} = f_{i_0, !}\mathcal{F}|_{X_{i_0}}$ as the construction of $f_!$ is local on the base. Moreover, these equalities are (of course) compatible with the identifications $((f_!\mathcal{F})|_{Y_{i_0}})|_{Y_{i_0i_1}} = (f_!\mathcal{F})|_{Y_{i_0i_1}}$ and $(f_{i_0, !}\mathcal{F}|_{X_{i_0}})|_{Y_{i_0i_1}} = f_{i_0i_1, !}\mathcal{F}|_{X_{i_0i_1}}$ which are used (together with the covariance for open embeddings for $Y_{i_0i_1} \subset Y_{i_0}$) to define the horizontal maps of the lower left square. Thus this square commutes as well. In this way we conclude there is a unique dotted arrow as indicated in the diagram and moreover this arrow is an isomorphism.

Proof of properties (1) – (5). Fix the open covering $Y = \bigcup Y_ i$. Observe that if $Y \to Z$ happens to be separated, then we get a dotted arrow fitting into the huge diagram above by using the map of Lemma 62.4.11 (by the very properties of that lemma). This proves (2) and hence also (1) by the compatibility of the maps of Lemma 62.4.11 and Lemma 62.3.13. Next, for any scheme $Z'$ over $Z$, we obtain the compatibility in (5) for the map $(g' \circ f')_! \to g'_! \circ f'_!$ constructed using the open covering $Y' = \bigcup b^{-1}(Y_ i)$. This is clear from the corresponding compatibility of the maps constructed in Lemma 62.4.11. In particular, we can consider a geometric point $\overline{z} : \mathop{\mathrm{Spec}}(k) \to Z$. Since $X_{\overline{z}} \to Y_{\overline{z}} \to \mathop{\mathrm{Spec}}(k)$ are separated maps, we find that the base change of $(g \circ f)_!\mathcal{F} \to g_! f_! \mathcal{F}$ by $\overline{z}$ is equal to the map of Lemma 62.3.13. The reader then immediately sees that we obtain property (3). Of course, property (3) guarantees that our transformation of functors $(g \circ f)_! \to g_! \circ f_!$ constructed using the open covering $Y = \bigcup Y_ i$ doesn't depend on the choice of this open covering. Finally, property (4) follows by looking at what happens on stalks using the already proven property (3). $\square$

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