Lemma 92.6.1. Assume (92.6.0.1) induces a quasi-isomorphism $B \otimes _ A^\mathbf {L} A' = B'$. Then, with notation as in (92.6.0.2) and $\mathcal{F}' \in \textit{Ab}(\mathcal{C}')$, we have $L\pi _!(g^{-1}\mathcal{F}') = L\pi '_!(\mathcal{F}')$.
92.6 Functoriality
In this section we consider a commutative square
92.6.0.1
\begin{equation} \label{cotangent-equation-commutative-square} \vcenter { \xymatrix{ B \ar[r] & B' \\ A \ar[u] \ar[r] & A' \ar[u] } } \end{equation}
of ring maps. We claim there is a canonical $B$-linear map of complexes
\[ L_{B/A} \longrightarrow L_{B'/A'} \]
associated to this diagram. Namely, if $P_\bullet \to B$ is the standard resolution of $B$ over $A$ and $P'_\bullet \to B'$ is the standard resolution of $B'$ over $A'$, then there is a canonical map $P_\bullet \to P'_\bullet $ of simplicial $A$-algebras compatible with the augmentations $P_\bullet \to B$ and $P'_\bullet \to B'$. This can be seen in terms of the construction of standard resolutions in Simplicial, Section 14.34 but in the special case at hand it probably suffices to say simply that the maps
\[ P_0 = A[B] \longrightarrow A'[B'] = P'_0,\quad P_1 = A[A[B]] \longrightarrow A'[A'[B']] = P'_1, \]
and so on are given by the given maps $A \to A'$ and $B \to B'$. The desired map $L_{B/A} \to L_{B'/A'}$ then comes from the associated maps $\Omega _{P_ n/A} \to \Omega _{P'_ n/A'}$.
Another description of the functoriality map can be given as follows. Let $\mathcal{C} = \mathcal{C}_{B/A}$ and $\mathcal{C}' = \mathcal{C}_{B'/A}'$ be the categories considered in Section 92.4. There is a functor
\[ u : \mathcal{C} \longrightarrow \mathcal{C}',\quad (P, \alpha ) \longmapsto (P \otimes _ A A', c \circ (\alpha \otimes 1)) \]
where $c : B \otimes _ A A' \to B'$ is the obvious map. As discussed in Cohomology on Sites, Example 21.39.3 we obtain a morphism of topoi $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}')$ and a commutative diagram of maps of ringed topoi
92.6.0.2
\begin{equation} \label{cotangent-equation-double-square} \vcenter { \xymatrix{ (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \underline{B}) \ar[d]_\pi & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \underline{B'}) \ar[d]_\pi \ar[l]^ h & (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \underline{B'}) \ar[d]_{\pi '} \ar[l]^ g \\ (\mathop{\mathit{Sh}}\nolimits (*), B) & (\mathop{\mathit{Sh}}\nolimits (*), B') \ar[l]_ f & (\mathop{\mathit{Sh}}\nolimits (*), B') \ar[l] } } \end{equation}
Here $h$ is the identity on underlying topoi and given by the ring map $B \to B'$ on sheaves of rings. By Cohomology on Sites, Remark 21.38.7 given $\mathcal{F}$ on $\mathcal{C}$ and $\mathcal{F}'$ on $\mathcal{C}'$ and a transformation $t : \mathcal{F} \to g^{-1}\mathcal{F}'$ we obtain a canonical map $L\pi _!(\mathcal{F}) \to L\pi '_!(\mathcal{F}')$. If we apply this to the sheaves
\[ \mathcal{F} : (P, \alpha ) \mapsto \Omega _{P/A} \otimes _ P B,\quad \mathcal{F}' : (P', \alpha ') \mapsto \Omega _{P'/A'} \otimes _{P'} B', \]
and the transformation $t$ given by the canonical maps
\[ \Omega _{P/A} \otimes _ P B \longrightarrow \Omega _{P \otimes _ A A'/A'} \otimes _{P \otimes _ A A'} B' \]
to get a canonical map
\[ L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) \longrightarrow L\pi '_!(\Omega _{\mathcal{O}'/A'} \otimes _{\mathcal{O}'} \underline{B'}) \]
By Lemma 92.4.3 this gives $L_{B/A} \to L_{B'/A'}$. We omit the verification that this map agrees with the map defined above in terms of simplicial resolutions.
Proof. We use the results of Remark 92.5.5 without further mention. We will apply Cohomology on Sites, Lemma 21.39.8. Let $P_\bullet \to B$ be a resolution. If we can show that $u(P_\bullet ) = P_\bullet \otimes _ A A' \to B'$ is a quasi-isomorphism, then we are done. The complex of $A$-modules $s(P_\bullet )$ associated to $P_\bullet $ (viewed as a simplicial $A$-module) is a free $A$-module resolution of $B$. Namely, $P_ n$ is a free $A$-module and $s(P_\bullet ) \to B$ is a quasi-isomorphism. Thus $B \otimes _ A^\mathbf {L} A'$ is computed by $s(P_\bullet ) \otimes _ A A' = s(P_\bullet \otimes _ A A')$. Therefore the assumption of the lemma signifies that $\epsilon ' : P_\bullet \otimes _ A A' \to B'$ is a quasi-isomorphism. $\square$
The following lemma in particular applies when $A \to A'$ is flat and $B' = B \otimes _ A A'$ (flat base change).
Lemma 92.6.2. If (92.6.0.1) induces a quasi-isomorphism $B \otimes _ A^\mathbf {L} A' = B'$, then the functoriality map induces an isomorphism
\[ L_{B/A} \otimes _ B^\mathbf {L} B' \longrightarrow L_{B'/A'} \]
Proof. We will use the notation introduced in Equation (92.6.0.2). We have
\[ L_{B/A} \otimes _ B^\mathbf {L} B' = L\pi _!(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) \otimes _ B^\mathbf {L} B' = L\pi _!(Lh^*(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B})) \]
the first equality by Lemma 92.4.3 and the second by Cohomology on Sites, Lemma 21.39.6. Since $\Omega _{\mathcal{O}/A}$ is a flat $\mathcal{O}$-module, we see that $\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}$ is a flat $\underline{B}$-module. Thus $Lh^*(\Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B}) = \Omega _{\mathcal{O}/A} \otimes _\mathcal {O} \underline{B'}$ which is equal to $g^{-1}(\Omega _{\mathcal{O}'/A'} \otimes _{\mathcal{O}'} \underline{B'})$ by inspection. we conclude by Lemma 92.6.1 and the fact that $L_{B'/A'}$ is computed by $L\pi '_!(\Omega _{\mathcal{O}'/A'} \otimes _{\mathcal{O}'} \underline{B'})$. $\square$
Remark 92.6.3. Suppose that we are given a square (92.6.0.1) such that there exists an arrow $\kappa : B \to A'$ making the diagram commute: In this case we claim the functoriality map $P_\bullet \to P'_\bullet $ is homotopic to the composition $P_\bullet \to B \to A' \to P'_\bullet $. Namely, using $\kappa $ the functoriality map factors as where $P_{A'/A', \bullet }$ is the standard resolution of $A'$ over $A'$. Since $A'$ is the polynomial algebra on the empty set over $A'$ we see from Simplicial, Lemma 14.34.3 that the augmentation $\epsilon _{A'/A'} : P_{A'/A', \bullet } \to A'$ is a homotopy equivalence of simplicial rings. Observe that the homotopy inverse map $c : A' \to P_{A'/A', \bullet }$ constructed in the proof of that lemma is just the structure morphism, hence we conclude what we want because the two compositions are the two maps discussed above and these are homotopic (Simplicial, Remark 14.26.5). Since the second map $P_\bullet \to P'_\bullet $ induces the zero map $\Omega _{P_\bullet /A} \to \Omega _{P'_\bullet /A'}$ we conclude that the functoriality map $L_{B/A} \to L_{B'/A'}$ is homotopic to zero in this case.
\[ \xymatrix{ B \ar[r]_\beta \ar[rd]_\kappa & B' \\ A \ar[u] \ar[r]^\alpha & A' \ar[u] } \]
\[ P_\bullet \to P_{A'/A', \bullet } \to P'_\bullet \]
\[ \xymatrix{ P_\bullet \ar[r] & P_{A'/A', \bullet } \ar@<1ex>[rr]^{\text{id}} \ar@<-1ex>[rr]_{c \circ \epsilon _{A'/A'}} & & P_{A'/A', \bullet } \ar[r] & P'_\bullet } \]
Lemma 92.6.4. Let $A \to B$ and $A \to C$ be ring maps. Then the map $L_{B \times C/A} \to L_{B/A} \oplus L_{C/A}$ is an isomorphism in $D(B \times C)$.
Proof. Although this lemma can be deduced from the fundamental triangle we will give a direct and elementary proof of this now. Factor the ring map $A \to B \times C$ as $A \to A[x] \to B \times C$ where $x \mapsto (1, 0)$. By Lemma 92.5.8 we have a distinguished triangle
\[ L_{A[x]/A} \otimes _{A[x]}^\mathbf {L} (B \times C) \to L_{B \times C/A} \to L_{B \times C/A[x]} \to L_{A[x]/A} \otimes _{A[x]}^\mathbf {L} (B \times C)[1] \]
in $D(B \times C)$. Similarly we have the distinguished triangles
\[ \begin{matrix} L_{A[x]/A} \otimes _{A[x]}^\mathbf {L} B \to L_{B/A} \to L_{B/A[x]} \to L_{A[x]/A} \otimes _{A[x]}^\mathbf {L} B[1]
\\ L_{A[x]/A} \otimes _{A[x]}^\mathbf {L} C \to L_{C/A} \to L_{C/A[x]} \to L_{A[x]/A} \otimes _{A[x]}^\mathbf {L} C[1]
\end{matrix} \]
Thus it suffices to prove the result for $B \times C$ over $A[x]$. Note that $A[x] \to A[x, x^{-1}]$ is flat, that $(B \times C) \otimes _{A[x]} A[x, x^{-1}] = B \otimes _{A[x]} A[x, x^{-1}]$, and that $C \otimes _{A[x]} A[x, x^{-1}] = 0$. Thus by base change (Lemma 92.6.2) the map $L_{B \times C/A[x]} \to L_{B/A[x]} \oplus L_{C/A[x]}$ becomes an isomorphism after inverting $x$. In the same way one shows that the map becomes an isomorphism after inverting $x - 1$. This proves the lemma. $\square$
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