Remark 91.11.4. We can make the comparison map of Lemma 91.11.3 explicit in the following way. Let $P_\bullet $ be the standard resolution of $B$ over $A$. Let $I = \mathop{\mathrm{Ker}}(A[B] \to B)$. Recall that $P_0 = A[B]$. The map of the lemma is given by the commutative diagram

We construct the downward arrow with target $I/I^2$ by sending $\text{d}f \otimes b$ to the class of $(d_0(f) - d_1(f))b$ in $I/I^2$. Here $d_ i : P_1 \to P_0$, $i = 0, 1$ are the two face maps of the simplicial structure. This makes sense as $d_0 - d_1$ maps $P_1$ into $I = \mathop{\mathrm{Ker}}(P_0 \to B)$. We omit the verification that this rule is well defined. Our map is compatible with the differential $\Omega _{P_1/A} \otimes _{P_1} B \to \Omega _{P_0/A} \otimes _{P_0} B$ as this differential maps $\text{d}f \otimes b$ to $\text{d}(d_0(f) - d_1(f)) \otimes b$. Moreover, the differential $\Omega _{P_2/A} \otimes _{P_2} B \to \Omega _{P_1/A} \otimes _{P_1} B$ maps $\text{d}f \otimes b$ to $\text{d}(d_0(f) - d_1(f) + d_2(f)) \otimes b$ which are annihilated by our downward arrow. Hence a map of complexes. We omit the verification that this is the same as the map of Lemma 91.11.3.

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