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• CommentRowNumber1.
• CommentAuthorAli Caglayan
• CommentTimeMar 18th 2019
• (edited Mar 18th 2019)

In category theory, a functor $U: C \to D$ can have the property of having a left adjoint iff $\forall X \in D$, $\exists FX \in C$ and $\exists \eta_X : X \to U(F X)$ such that, $\forall A \in C$ and forall $f : X \to U(A)$, $\exists ! g : F X \to A$ such that the triangle commutes: $f = U(g) \circ \eta_X$.

This definition is nice because it is very minimalistic, and it can be canonically extended to construct a left adjoint functor when it holds. I suppose another way to say this is that these are the necessary and sufficient conditions for a left adjoint defined any other way.

My question is: Do we have a similar minimal sort of definition for having an oo-left adjoint?

I don’t have any particular model of oo-cat in mind.

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeMar 18th 2019

The answer is yes. But in order to make it precise one probably has to choose a particular model.

• CommentRowNumber3.
• CommentAuthorDmitri Pavlov
• CommentTimeMar 18th 2019

The first functor should probably be called U, not F. Also A∈C, not A∈X.

The condition after “such that” can be reformulated by saying that the canonical functor FX/C→X/U induced by U and precomposition with η_X is an equivalence of categories.

Presumably, replacing equivalences with ∞-equivalences should yield the desired outcome.

• CommentRowNumber4.
• CommentAuthorAli Caglayan
• CommentTimeMay 26th 2019

So I had a chat with Emily and it turns out there is a way to make this precise:

A functor $U : C \to D$ has a left adjoint if for all $X$ in $D$, the comma category $X / U$ has an initial object $(FX, \eta_X)$.

Then you should be able to stick an oo infront of everything.

This is proposition 4.1.5 in her new book.

• CommentRowNumber5.
• CommentAuthorAli Caglayan
• CommentTimeMay 26th 2019

Am I correct in thinking that because we are considering oo-initial objects, we typically don’t even need to consider higher coherences?

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeMay 28th 2019

That probably depends on your definition of $\infty$-initial…