Modules over a category ring
Let
Proof
Let
and for a morphism
and extend this definition linearly first for nonhomogenous vectors and then general
Now suppose
defines an
so by linearity
Conversely, suppose
for ; for and .
Now suppose
which is well-defined since
so the following diagram commutes
whence
It is not difficult to see the natural equivalences required to make this an equivalence.
#state/tidy | #lang/en | #SemBr
Footnotes
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assuming the Axiom of Choice. ↩