0P5Z
Proposition 4.3.9. There is a differential functor given by
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There is a closed isomorphism of functors
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If and are invertible, then defines a morphism
of -representations .
0P60
Proof. Let be an object of . Let , an element of .
We have
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It follows that
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hence is an object of . We put
.
Let . We have a
commutative diagram
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and it follows that .
We put . This makes into a
differential functor .
We have
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and
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We have
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hence
defines
a closed isomorphism . The naturality
of and implies immediately that of .
We have for . Together with (4.3.4),
it follows that
, hence
defines a morphism of -representations.
∎