5
$\begingroup$

It is well-documented that an erasable delta-functor $(F^n,\delta)$ is universal. However, in 'small' abelian categories (in the technical sense or otherwise), there aren't always enough objects to erase all of $F^n(A)$ simultaneously. So consider the following variant:

Definition. Let $\mathscr A$ be an abelian category and $R$ a ring. Then an additive functor $F \colon \mathscr A \to \operatorname{Mod}_R$ is locally erasable if for every $A \in \mathscr A$ and every $x \in F(A)$, there exists a monomorphism $A \hookrightarrow B$ such that $x$ maps to $0$ in $F(B)$.

For instance, Yoneda's $\operatorname{Ext}^i(-,-)$ will not always be erasable in each variable, but it is always locally erasable. (I only care about situations where it is actually a set.)

It is not so hard to prove that locally erasable delta-functors are still universal, by a slight variation of the usual proof. This surely must have come up before, so:

Question. Is there a reference for this result?

$\endgroup$

1 Answer 1

10
$\begingroup$

This is Proposition 4.2 in Buchsbaum’s Satellites and universal functors, Annals of Mathematics 71(2), pp. 199–209 (1960).

Well, to be precise, that is the dual result (for contravariant functors). But replacing $\mathscr{A}$ with $\mathscr{A}^\text{op}$ gives the version in the question.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for finding this; this is exactly what I was looking for! Even in 1960, it wasn't worth checking all the details :) (saves me a page of embarrassingly trivial maths, although exposition is also a valuable thing I suppose). $\endgroup$ Commented Sep 2 at 21:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .