It's well known that a category $\mathcal{C}$ having (conical) limits/colimits of shape $\mathcal{D}$ is equivalent to the diagonal functor $\Delta^\mathcal{D}_\mathcal{C}:\mathcal{C}\to\mathcal{C}^\mathcal{D}$ having a left/right adjoint respectively.

Is there a similar characterization of weighted limits/colimits as left/right adjoints, maybe to a diagonal functor into $\mathcal{C}^{(\mathcal{V}^\mathcal{D})}$ for $\mathcal{V}$ a closed symmetric monoidal category or some such?

Going through Awodey's introductory category theory text there was no mention of weighted (co)limits, but I need to step up to $2$-category theory for Borceux/Janelidze's (non-Galoisian) categorical Galois theory and weighted (co)limits give the correct fully weak notion of $2$-categorical (co)limits.

I've seen it mentioned elsewhere that weighted (co)limits are equivalent to conical (co)limits and cotensoring over the walking arrow category, but cotensoring isn't mentioned in Awodey either so this characterization is somewhat opaque to me. The adjoint characterization of (co)limits is very clear, so hopefully an adjoint characterization of weighted (co)limits can demystify their workings as well.

If they aren't special cases of ($1$-)adjoints that would be interesting too, since the moral of $1$-category theory (as I understand it) is that 'every ubiquitous concept in regular mathematics is an adjunction or an adjoint functor'. It is noted in the answer here that weighted limits are a fundamentally $2$-categorical notion and considering them on a $1$-categorical level necessarily misses some of the subtleties involved, so perhaps they are a particular case of a $2$-adjunction?

  • $\begingroup$ Is there a reason for the downvote? If this is too simple and a better fit at MSE I'll gladly shift it over, but the linked question there seemed like it received relatively little attention and might have been a better fit here. $\endgroup$ – Alec Rhea Mar 8 '19 at 21:57

Here's one way in which weighted limits are adjoints. Let $D$ be a $V$-category and $W:D\to V$ a weight, and $C$ a $V$-category. Suppose that $C$ has copowers, so for $k\in V$ and $x,y\in C$ we have $C(k\odot x,y) \cong V(k,C(x,y))$. Then there is a $V$-functor $\Delta^W:C \to C^D$ defined by $\Delta^W(c)(d) = W(d) \odot c$, and a right adjoint to this functor sends $F\in C^D$ to an object $l$ such that $$C(c,l) \cong C^D(W(-)\odot c,F) \cong V^D(W,C(c,F))$$ which is the definition of a weighted limit $\lim^W F$.

You can get rid of the assumption of copowers if you take $W$ as the argument of the functor rather than $F$: there's a contravariant functor $\nabla^F : C^{op} \to V^D$ defined by $\nabla^F(c)(d) = C(c,F(d))$, and a mutual right adjoint to this functor sends $W\in V^D$ to an object $l$ such that $$C(c,l) \cong V^D(W,C(c,F(d)))$$ once again the definition of a weighted limit $\lim^W F$.

  • $\begingroup$ Much appreciated Mike, this is exactly what I was looking for. Is there any non-enriched adjunction that gives rise to the notion of a weighted limit, or is enriching over the weighting category necessary to capture the correct notion? $\endgroup$ – Alec Rhea Mar 9 '19 at 1:08
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    $\begingroup$ @AlecRhea An unenriched adjunction will only give you an unenriched universal property, whereas a weighted limit proper must have an enriched universal property. So that can only happen if you're in a case where unenriched universal properties imply enriched ones, but that happens (for limits) if $C$ has copowers, since $V_0(k,C(c,l)) \cong C_0(k\odot c,l)$. $\endgroup$ – Mike Shulman Mar 9 '19 at 3:46

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