11
$\begingroup$

It is a fairly special property for the Yoneda embedding $A \hookrightarrow \mathcal{P}A$ of a category to have a left adjoint defined everywhere (this happens just when $A$ is total). However, a partially defined such left adjoint can express the existence of any colimits in $A$. For instance, if $F\colon I\to A$ is any functor, then colimits of $F$ are the same as values of such a partially defined left adjoint at the presheaf $\mathrm{colim}_i\; A(-,F(i))$. A similar idea works for weighted limits in an enriched setting. In fact, $A$ is small-cocomplete just when its Yoneda embedding has a left adjoint defined at all small presheaves (presheaves that are small colimits of representables).

My question is, given a presheaf $X\in \mathcal{P}A$, what is a name for the value of a (partially defined) left adjoint to the Yoneda embedding at $X$?

It is tempting to want to call it the "colimit" of $X$, except that $X$ itself is a functor $A^{op} \to \mathrm{Set}$, and we are certainly not talking about the colimit of that functor. If it helps, note that the object in question is equivalently the colimit of the identity functor of $A$ weighted by $X$.

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Perhaps you could call it the "realisation" of $X$? After all, a presheaf on $A$ is, in a some sense, a formal colimit. $\endgroup$
    – Zhen Lin
    May 17, 2012 at 17:44
  • $\begingroup$ This is the (may be partial definited) LEft Kan extension of Yoneda embedding $Y: \mathcal {A}\to \mathcal {A}^>$ respect to $1_{\mathcal{A}}$. In my notes (personal, Italian) I like call it Yoneda-Kan extensions, is usual in topics as completions of categories, shape theory (categorical aspect) , kinds of "geometrical realization". $\endgroup$
    – Buschi Sergio
    May 17, 2012 at 17:57
  • 2
    $\begingroup$ I guess "realization" isn't bad. $\endgroup$
    – Mike Shulman
    May 17, 2012 at 21:33
  • 1
    $\begingroup$ Personally, I like to think of presheaves as a kind of external logic for the category, thus a name that I like is "internalization", in the sense of "internal logic" of the category. $\endgroup$
    – Ivan Di Liberti
    Aug 6, 2020 at 18:12
  • 1
    $\begingroup$ @IvanDiLiberti But this left adjoint rarely preserves very much "logic". At most one can expect it to preserve finite limits. $\endgroup$
    – Mike Shulman
    Aug 7, 2020 at 15:43

1 Answer 1

Reset to default
5
$\begingroup$

The question is: Given a functor $F : A^{op} \to \mathsf{Set}$, how do we call an object $?(F)$ in $A$ satisfying the universal property

$\hom(?(F),X) \cong \hom(F,\hom(-,X))$

for all $X \in A$? Some people call it a corepresenting object of $F$. The reason is that a representing object of $F$ is some object $!(F)$ satisfying $\hom(X,!(F)) \cong \hom(\hom(-,X),F)$, since the left hand side simplifies to $F(X)$ by the Yoneda Lemma. Remark that every representing object is also a corepresenting object.

If $F$ is a moduli problem in algebraic geometry, then $?(F)$ with some additional assumptions is usually also called a coarse moduli space (whereas $!(F)$ is the fine moduli space). One of the many references is Definition 2.1. (2) in Adrian Langer's "Moduli Spaces Of Sheaves On Higher Dimensional Varieties", as well as Definition 2.2.1 in "The Geometry of Moduli Spaces of Sheaves" by Huybrecht and Lehn. Perhaps someone can add the original reference.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Hmm... to me a "corepresenting object" is a representing object for a covariant functor. $\endgroup$
    – Mike Shulman
    May 17, 2012 at 21:32
  • 3
    $\begingroup$ Yes this terminology also appears. But the distinction between covariant and contravariant functors is artificial because we can use dual categories. The notions are just the same, as well as their representability. But in the above definition of corepresentabilty, something new happens and deserves to be named "co". We don't understand maps $\hom(-,X) \to F$, but rather maps $F \to \hom(-,X)$. $\endgroup$
    – Martin Brandenburg
    May 18, 2012 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.