Proof.
Choose a bounded above complex $K^\bullet $ of free $R$-modules representing $K$. Then we can choose free $R'$-modules $(K')^ n$ lifting $K^ n$. We can choose $R'$-module maps $(d')^ n_ K : (K')^ n \to (K')^{n + 1}$ lifting the differentials $d^ n_ K : K^ n \to K^{n + 1}$ of $K^\bullet $. Although the compositions
\[ (d')^{n + 1}_ K \circ (d')^ n_ K : (K')^ n \to (K')^{n + 2} \]
may not be zero, they do factor as
\[ (K')^ n \to K^ n \xrightarrow {\omega ^ n_ K} K^{n + 2} \otimes _ R I = I(K')^{n + 2} \to (K')^{n + 2} \]
because $d^{n + 1} \circ d^ n = 0$. A calculation shows that $\omega ^ n_ K$ defines a map of complexes. This map of complexes defines $\omega (K)$.
Let us prove this construction is compatible with a map of complexes $\alpha ^\bullet : K^\bullet \to L^\bullet $ of bounded above free $R$-modules and given choices of lifts $(K')^ n, (L')^ n, (d')^ n_ K, (d')^ n_ L$. Namely, choose $(\alpha ')^ n : (K')^ n \to (L')^ n$ lifting the components $\alpha ^ n : K^ n \to L^ n$. As before we get a factorization
\[ (K')^ n \to K^ n \xrightarrow {h^ n} L^{n + 1} \otimes _ R I = I(L')^{n + 1} \to (L')^{n + 2} \]
of $(d')^ n_ L \circ (\alpha ')^ n - (\alpha ')^{n + 1} \circ (d')_ K^ n$. Then it is an pleasant calculation to show that
\[ \omega ^ n_ L \circ \alpha ^ n = (d_ L^{n + 1} \otimes \text{id}_ I) \circ h^ n + h^{n + 1} \circ d_ K^ n + (\alpha ^{n + 2} \otimes \text{id}_ I) \circ \omega ^ n_ K \]
This proves the commutativity of the diagram in (2) of the lemma in this particular case. Using this for two different choices of bounded above free complexes representing $K$, we find that $\omega (K)$ is well defined! And of course (2) holds in general as well.
If $K$ lifts to $K'$ in $D^-(R')$, then we can represent $K'$ by a bounded above complex of free $R'$-modules and we see immediately that $\omega (K) = 0$. Conversely, going back to our choices $K^\bullet $, $(K')^ n$, $(d')^ n_ K$, if $\omega (K) = 0$, then we can find $g^ n : K^ n \to K^{n + 1} \otimes _ R I$ with
\[ \omega ^ n = (d_ K^{n + 1} \otimes \text{id}_ I) \circ g^ n + g^{n + 1} \circ d_ K^ n \]
This means that with differentials $(d')^ n_ K + g^ n : (K')^ n \to (K')^{n + 1}$ we obtain a complex of free $R'$-modules lifting $K^\bullet $. This proves (1).
Finally, part (3) means the following: Let $R' \to S'$ be a map of rings. Set $S = S' \otimes _{R'} R$ and denote $J = IS' \subset S'$ the square zero kernel of $S' \to S$. Then given $K \in D^-(R)$ the statement is that we get a commutative diagram
\[ \xymatrix{ K \otimes _ R^\mathbf {L} S \ar[d] \ar[rr]_-{\omega (K) \otimes \text{id}} & & (K \otimes ^\mathbf {L}_ R I[2]) \otimes _ R^\mathbf {L} S \ar[d] \\ K \otimes _ R^\mathbf {L} S \ar[rr]^-{\omega (K \otimes _ R^\mathbf {L} S)} & & (K \otimes _ R^\mathbf {L} S) \otimes ^\mathbf {L}_ S J[2] } \]
Here the right vertical arrow comes from
\[ (K \otimes ^\mathbf {L}_ R I[2]) \otimes _ R^\mathbf {L} S = (K \otimes _ R^\mathbf {L} S) \otimes _ S^\mathbf {L} (I \otimes _ R^\mathbf {L} S)[2] \longrightarrow (K \otimes _ R^\mathbf {L} S) \otimes _ S^\mathbf {L} J[2] \]
Choose $K^\bullet $, $(K')^ n$, and $(d')^ n_ K$ as above. Then we can use $K^\bullet \otimes _ R S$, $(K')^ n \otimes _{R'} S'$, and $(d')^ n_ K \otimes \text{id}_{S'}$ for the construction of $\omega (K \otimes _ R^\mathbf {L} S)$. With these choices commutativity is immediately verified on the level of maps of complexes.
$\square$
Comments (1)
Comment #8725 by Shiji Lyu on