Lemma 91.13.3. In the situation above denote $L$ the complex (91.13.0.1). There is a canonical map $L_{B/A} \to L$ in $D(B)$ which induces an isomorphism $\tau _{\geq -2}L_{B/A} \to L$ in $D(B)$.

**Proof.**
Let $P_\bullet \to B$ be a resolution of $B$ over $A$ (Remark 91.5.5). We will identify $L_{B/A}$ with $\Omega _{P_\bullet /A} \otimes B$. To construct the map we make some choices.

Choose an $A$-algebra map $\psi : P_0 \to P$ compatible with the given maps $P_0 \to B$ and $P \to B$.

Write $P_1 = A[S]$ for some set $S$. For $s \in S$ we may write

for some $p_{s, t} \in P$. Think of $F = \bigoplus _{t \in T} P$ as a $(P_1, P_1)$-bimodule via the maps $(\psi \circ d_0, \psi \circ d_1)$. By Lemma 91.13.2 we obtain a unique $A$-biderivation $\lambda : P_1 \to F$ mapping $s$ to the vector with coordinates $p_{s, t}$. By construction the composition

sends $f \in P_1$ to $\psi (d_0(f) - d_1(f))$ because the map $f \mapsto \psi (d_0(f) - d_1(f))$ is an $A$-biderivation agreeing with the composition on generators.

For $g \in P_2$ we claim that $\lambda (d_0(g) - d_1(g) + d_2(g))$ is an element of $Rel$. Namely, by the last remark of the previous paragraph the image of $\lambda (d_0(g) - d_1(g) + d_2(g))$ in $P$ is

which is zero by Simplicial, Section 14.23).

The choice of $\psi $ determines a map

Composing $\lambda $ with the map $F \to F \otimes B$ gives a usual $A$-derivation as the two $P_1$-module structures on $F \otimes B$ agree. Thus $\lambda $ determines a map

Finally, We obtain a $B$-linear map

by mapping $\text{d}g$ to the class of $\lambda (d_0(g) - d_1(g) + d_2(g))$ in the quotient.

The diagram

commutes (calculation omitted) and we obtain the map of the lemma. By Remark 91.11.4 and Lemma 91.11.3 we see that this map induces isomorphisms $H_1(L_{B/A}) \to H_1(L)$ and $H_0(L_{B/A}) \to H_0(L)$.

It remains to see that our map $L_{B/A} \to L$ induces an isomorphism $H_2(L_{B/A}) \to H_2(L)$. Choose a resolution of $B$ over $A$ with $P_0 = P = A[u_ i]$ and then $P_1$ and $P_2$ as in Example 91.5.9. In Remark 91.12.6 we have constructed an exact sequence

where $P_0 = P$ and $J_0 = \mathop{\mathrm{Ker}}(P \to B) = I$. Calculating the Tor group using the short exact sequences $0 \to I \to P \to B \to 0$ and $0 \to Rel \to F \to I \to 0$ we find that $\text{Tor}_2^ P(B, B) = \mathop{\mathrm{Ker}}(Rel \otimes B \to F \otimes B)$. The image of the map $\wedge ^2_ B(I/I^2) \to \text{Tor}_2^ P(B, B)$ under this identification is exactly the image of $TrivRel \otimes B$. Thus we see that $H_2(L_{B/A}) \cong H_2(L)$.

Finally, we have to check that our map $L_{B/A} \to L$ actually induces this isomorphism. We will use the notation and results discussed in Example 91.5.9 and Remarks 91.12.6 and 91.11.5 without further mention. Pick an element $\xi $ of $\text{Tor}_2^{P_0}(B, B) = \mathop{\mathrm{Ker}}(I \otimes _ P I \to I^2)$. Write $\xi = \sum h_{t', t}f_{t'} \otimes f_ t$ for some $h_{t', t} \in P$. Tracing through the exact sequences above we find that $\xi $ corresponds to the image in $Rel \otimes B$ of the element $r \in Rel \subset F = \bigoplus _{t \in T} P$ with $t$th coordinate $r_ t = \sum _{t' \in T} h_{t', t}f_{t'}$. On the other hand, $\xi $ corresponds to the element of $H_2(L_{B/A}) = H_2(\Omega )$ which is the image via $\text{d} : H_2(\mathcal{J}/\mathcal{J}^2) \to H_2(\Omega )$ of the boundary of $\xi $ under the $2$-extension

We compute the successive transgressions of our element. First we have

and next we have

by our choice of the variables $v$ in Example 91.5.9. We may choose our map $\lambda $ above such that $\lambda (u_ i) = 0$ and $\lambda (x_ t) = - e_ t$ where $e_ t \in F$ denotes the basis vector corresponding to $t \in T$. Hence the construction of our map $q$ above sends $\text{d}v_ r$ to

matching the image of $\xi $ in $Rel \otimes B$ (the two minus signs we found above cancel out). This agreement finishes the proof. $\square$

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