By Dan Laksov

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**Example text**

2) Composite maps. Let L, M, N be A-modules. 1) and t : HomA (∧n L, ∧n M ) ⊗A HomA (∧n M, ∧n L) → HomA (∧n L, ∧n N ). 1) a homomorphism of A-modules w : ΓnA (HomA (L, M )) ⊗A ΓnA (HomA (M, N )) → ΓnA (HomA (L, N )). 3) Lemma. Let L, M, N be free A-modules of finite rank. 2). → Proof. 1) are equal. However, we have that n n n ∧nL,N (γHom (u) ⊗A γHom (v)) = ∧nL,N (γHom (vu)) = ∧n (vu), A (L,M ) A (M,N ) A (L,N ) and n n t(∧nL,M ⊗A ∧nM,N )(γHom (u) ⊗A γHom (v)) A (L,M ) A (M,N ) n n = ∧nM,N γHom (v) ∧nL,M γHom (u) = ∧n v ∧n u.

Let M and N be A-modules, and let B be an A-algebra. 6), so that in particular (tα )α∈I is a family of independent variables over A, and A[t] is the polynomial ring in the variables tα over A. Let (fα )α∈I and (gα )α∈I be elements in B (I) , and let (xα )α∈I and (yα )α∈I be in M (I) , respectively in N (I) . 1) that maps the element α∈I (xα + yα ) ⊗A fα to α∈I (xα ) ⊗A fα + (yα ) ⊗A fα . Its inverse maps ( α∈I xα ⊗A fα ) + ( α∈I yα ⊗A gα ) to α∈I (xα + 0) ⊗ fα + α∈I (0 + yα ) ⊗ gα . Moreover we have a canonical B-bilinear homomorphism (M ⊗A B) × (N ⊗A B) → (M ⊗A B) ⊗B (N ⊗A B).

It follows that every polynomial law U ∈ P n (M, A) is uniquely determined by the elements UA[t] ( α∈J eα ⊗A tα ) for all finite subsets J of I. Moreover, it is clear that we obtain a polynomial law in P n (M, N ) by choosing an arbitrary family (fν )ν∈N(I) of elements fν in A and defining U by UA[t] ( α∈J eα ⊗A tα ) = µ∈N(J ) ,|µ|=n fµ ⊗A tµ . For every ν ∈ N(I) with |ν| = n we let the polynomial law Uν in P n (M, A) be defined by (Uν )A[t] ( α∈J → → f µ ⊗ A tµ = e α ⊗ A tα ) = µ∈N(J ) ,|µ|=n 1 ⊗ A tν 0 when when µ=ν µ=ν for all finite subsets J of I.