Lemma 22.34.1. Let R be a ring. Let (A, \text{d}), (B, \text{d}), and (C, \text{d}) be differential graded R-algebras. Let N be a differential graded (A, B)-bimodule. Let N' be a differential graded (B, C)-module. Assume (22.34.0.1) is an isomorphism. Then the composition
\xymatrix{ D(A, \text{d}) \ar[rr]^{- \otimes _ A^\mathbf {L} N} & & D(B, \text{d}) \ar[rr]^{- \otimes _ B^\mathbf {L} N'} & & D(C, \text{d}) }
is isomorphic to - \otimes _ A^\mathbf {L} N'' with N'' = N \otimes _ B N' viewed as (A, C)-bimodule.
Proof.
Let us define a transformation of functors
(- \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} N' \longrightarrow - \otimes _ A^\mathbf {L} N''
To do this, let M be a differential graded A-module with property (P). According to the construction of the functor - \otimes _ A^\mathbf {L} N'' of the proof of Lemma 22.33.2 the plain tensor product M \otimes _ A N'' represents M \otimes _ A^\mathbf {L} N'' in D(C, \text{d}). Then we write
M \otimes _ A N'' = M \otimes _ A (N \otimes _ B N') = (M \otimes _ A N) \otimes _ B N'
The module M \otimes _ A N represents M \otimes _ A^\mathbf {L} N in D(B, \text{d}). Choose a quasi-isomorphism Q \to M \otimes _ A N where Q is a differential graded B-module with property (P). Then Q \otimes _ B N' represents (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} N' in D(C, \text{d}). Thus we can define our map via
(M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} N' = Q \otimes _ B N' \to M \otimes _ A N \otimes _ B N' = M \otimes _ A^\mathbf {L} N''
The construction of this map is functorial in M and compatible with distinguished triangles and direct sums; we omit the details. Consider the property T of objects M of D(A, \text{d}) expressing that this map is an isomorphism. Then
if T holds for M_ i then T holds for \bigoplus M_ i,
if T holds for 2-out-of-3 in a distinguished triangle, then it holds for the third, and
T holds for A[k] because here we obtain a shift of the map (22.34.0.1) which we have assumed is an isomorphism.
Thus by Remark 22.22.5 property T always holds and the proof is complete.
\square
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