Skip to main content
2 of 3
Made the question more precise by making the dependency on the objects of $\mathcal{D}$ explicit.
Peter Heinig
  • 6.1k
  • 1
  • 27
  • 47

Technical term for representing object of a presheaf determined by a left-adjoint?

If $\mathcal{D}$ is a locally-small category, then a functor $F\colon\mathcal{C}\rightarrow\mathcal{D}$ has a right-adjoint if and only if for each object $d$ of $D$, the presheaf $\mathcal{C}^{\mathrm{op}}\overset{\qquad c\quad\mapsto\quad \mathcal{D}(F(c),d)\qquad }{\longrightarrow}\mathsf{Set}$ is representable.

  • Is there a usual technical term for, given $d$, a representing object of the aforementioned presheaf? Or rather, usual or not, have you ever encountered a specialized term for this, and which do you recommend?
Remarks. Whether any specialized term is needed, is debatable of course, and not the question. There are situations where it is useful to have a term for it, to facilitate discussing the issue, in particular in expositions. Saying ``representing object of $F$'' would be nonsensical. Of course, one can just describe it the way I did, but is there a specialized technical term for this representing object? For want of a standard term, and despite the connotations of "law" and similarity to another, usual notation, I always used to call and denote "the" family

$\mathrm{Ob}(\mathcal{D})\ni d\mapsto $(a representing object of the relevant presheaf) $\in$ $\mathrm{Ob}(\mathcal{C})$

"the" $\mathrm{Law}(F)$, for "left-adjoint witnesses of $F$", since if a family $\mathrm{Law}(F)$ exists, then $F$ is a left-adjoint, and such a family of objects is something of a certificate for its being a left-adjoint.

  • Do you agree that, strictly speaking, it would not be in accordance with the definition of "the" in the to refer to "the law of $F$"?
After all, each "component" of $\mathrm{Law}(F)$ is determined only up to isomorphism in the category $\mathcal{D}$, and the feature of any two laws being determined up to _unique_ isomorphism seems to be totally lacking.
Peter Heinig
  • 6.1k
  • 1
  • 27
  • 47