Lemma 46.4.3. Let A be a ring. Let F be a module valued functor. For every B \in \mathop{\mathrm{Ob}}\nolimits (\textit{Alg}_ A) and B-module N there is a canonical decomposition
F(B[N]) = F(B) \oplus TF(B, N)
characterized by the following properties
TF(B, N) = \mathop{\mathrm{Ker}}(F(B[N]) \to F(B)),
there is a B-module structure TF(B, N) compatible with B[N]-module structure on F(B[N]),
TF is a functor from the category of pairs (B, N),
there are canonical maps N \otimes _ B F(B) \to TF(B, N) inducing a transformation between functors defined on the category of pairs (B, N),
TF(B, 0) = 0 and the map TF(B, N) \to TF(B, N') is zero when N \to N' is the zero map.
Proof.
Since B \to B[N] \to B is the identity we see that F(B) \to F(B[N]) is a direct summand whose complement is TF(N, B) as defined in (1). This construction is functorial in the pair (B, N) simply because given a morphism of pairs (B, N) \to (B', N') we obtain a commutative diagram
\xymatrix{ B' \ar[r] & B'[N'] \ar[r] & B' \\ B \ar[r] \ar[u] & B[N] \ar[r] \ar[u] & B \ar[u] }
in \textit{Alg}_ A. The B-module structure comes from the B[N]-module structure and the ring map B \to B[N]. The map in (4) is the composition
N \otimes _ B F(B) \longrightarrow B[N] \otimes _{B[N]} F(B[N]) \longrightarrow F(B[N])
whose image is contained in TF(B, N). (The first arrow uses the inclusions N \to B[N] and F(B) \to F(B[N]) and the second arrow is the multiplication map.) If N = 0, then B = B[N] hence TF(B, 0) = 0. If N \to N' is zero then it factors as N \to 0 \to N' hence the induced map is zero since TF(B, 0) = 0.
\square
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