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 (2)
Comment #8725 by Shiji Lyu on
Comment #9351 by Stacks project on