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1. • CommentRowNumber1.
   • CommentAuthorAli Caglayan
   • CommentTimeMar 18th 2019
   • (edited Mar 19th 2019)
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   Author: Ali Caglayan
   Format: MarkdownItexIn 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.

   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.

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMar 18th 2019
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   Author: Mike Shulman
   Format: MarkdownItexThe answer is yes. But in order to make it precise one probably has to choose a particular model.

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

3. • CommentRowNumber3.
   • CommentAuthorDmitri Pavlov
   • CommentTimeMar 18th 2019
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   Author: Dmitri Pavlov
   Format: MarkdownItexThe 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.

   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.

4. • CommentRowNumber4.
   • CommentAuthorAli Caglayan
   • CommentTimeMay 26th 2019
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   Author: Ali Caglayan
   Format: MarkdownItexSo 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](http://www.math.jhu.edu/~eriehl/elements.pdf).

   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.

5. • CommentRowNumber5.
   • CommentAuthorAli Caglayan
   • CommentTimeMay 26th 2019
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   Author: Ali Caglayan
   Format: MarkdownItexAm I correct in thinking that because we are considering oo-initial objects, we typically don't even need to consider higher coherences?

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

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeMay 28th 2019
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   Author: Mike Shulman
   Format: MarkdownItexThat probably depends on your definition of $\infty$-initial...

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