A universal property comes from an adjunction. From this point of view, associated graded has no universal property because it is not left or right adjoint.

> Proof. If gr(-) were left (right)
> adjoint, then it would respect
> cokernels (kernels). Consider the
> morphism of filtered vector spaces
> (0⊆0⊆V)→(0⊆V⊆V)
> (the three pieces are the 0-, 1-, and
> 2-filtered parts) which is just the
> identity map on V. It's kernel and
> cokernel are trivial. But the induced
> map
> gr(0⊆0⊆V)→gr(0⊆V⊆V)
> is the zero map from V (in degree 2)
> to V (in degree 1), which has
> non-trivial kernel and cokernel. So
> the associated graded of the
> (co)kernel is not the (co)kernel of
> the associated graded map.

Ben's solution is to write this poorly behaved functor as a composition of two nicer functors. The first functor is Rees:R-filmod→R[t]-grmod (from the category of filtered R-modules to the category of graded R[t]-modules). I think this functor is right adjoint to R[t]/(t-1)⊗-.

The second is R[t]/(t)&otimes;-:R[t]-grmod&rarr;R-grmod, the functor that takes &oplus;N<sub>i</sub> to &oplus;N<sub>i</sub>/N<sub>i-1</sub>. R[t]/(t)&otimes;- is left adjoint to the functor that takes a graded R-module to the same graded module, regarded as an R[t]-module by letting t act by 0.

**Upshot:** associated graded is not an adjoint functor, so it doesn't have a nice universal property by itself, but it is the composition of a right adjoint functor and a left adjoint functor, which do have universal properties.