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It is well known that, for two functors $F,G : I \to C$ for $I,C$ some $\infty$-categories, the property that a map $\phi: F \to G$ is an equivalence can be checked locally on $I$. Namely, if $\phi(i) : F(i) \to G(i)$ is an equivalence for every $i \in I$, then $\phi$ is an equivalence as a morphism in the $\infty$-category $Fun(I,C)$.

Is the same true for adjointability? Namely:

Let $I$ be an $\infty$-category, and let $F,G : I \to Cat_\infty$ be two functors into the $(\infty,1)$ category of $\infty$-categories. Let $\phi : F \to G$. Suppose that, for each $i \in I$, $\phi(i) : F(i) \to G(i)$ admits left adjoint $L_{\phi(i)}$. Then, there nessecarily exists a map $\psi: G \to F$ restricting to $L_{\phi(i)}$ on every object? Is it essentially unique? does it satisfy some relative version of the property satisfied by left adjoints?

My motivation for believing in it is that the space of choices of the Left adjoint is contractible, so there should be no obstructions to glue them over $I$.

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TL DR: That is not enough. If you let $\psi_i:G(i)\to F(i)$ be the left adjoint of $\phi_i$ you also need the condition that for every $f:i\to j$ the canonical morphism $$\psi_jG(f)\to F(f)\psi_i$$ adjoint to $G(f)\to G(f)\phi_i\psi_i\cong \phi_jF(f)\psi_i$, is an equivalence.

The key word to remember here is relative adjunctions. If we unstraighten your functors $F$ and $G$, we end up with two cartesian fibrations $E\to I^{op}$ and $E'\to I^{op}$ and a map of cartesian fibrations $\phi:E\to E'$ such that for each $i\in I$ the functor $\phi_i:E_i\to E'_i$ is a right adjoint. Then, by theorem 7.3.2.6 in Higher Algebra, there is a relative left adjoint $\psi$. The result you are after would follow if and only if $\psi$ were a map of cartesian fibrations (i.e. iff it sends cartesian edges to cartesian edges). Unwrapping the various definitions, this is exactly the condition I wrote above.

To see a counterexample when the condition is not satisfied, pick two categories $C_0,C_1$ with an initial object and a functor $f:C_0\to C_1$ that does not preserve the initial object (e.g. let $C_0$ be finite pointed sets, $C_1$ be finite sets and $f$ the functor that forgets the pointing). This assembles to a functor $F:\Delta^1\to \mathrm{Cat}_\infty$.

Now let $G:\Delta^1\to \mathrm{Cat}_\infty$ be the constant functor at $*$. There is an obvious natural transformation $\phi:F\to G$ and $\phi_0$ and $\phi_1$ have both left adjoints $\psi_0$ and $\psi_1$ (the inclusion of the initial object). However these do not assemble into a natural transformation $\psi:G\to F$ because $\psi_1$ is not naturally equivalent to $f\psi_0$.

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  • $\begingroup$ oh I see, well, i'm pretty sure that in my situation this condition indeed holds. Thanks! $\endgroup$ – S. carmeli Mar 10 '18 at 14:47
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FWIW, there is nothing "truly $\infty$" about this question; the same question can be asked for 1-categories and the answer is the same. In that case it fits into two abstract frameworks:

1: doctrinal adjunction. For a 2-monad $T$ on a 2-category $K$ and a pseudo $T$-algebra map $(g,\bar{g}):A\to B$ whose underlying morphism $g:A\to B$ in $K$ has a left adjoint $f:B\to A$, there is a canonical induced structure of an oplax $T$-algebra map on $f$ (the mate of the pseudo $T$-morphism structure on $g$), and the whole adjunction lifts to the 2-category of $T$-algebras (and pseudo morphisms) if and only if this oplax structure is in fact a pseudo structure.

Now there is a 2-monad $T$ on $\mathrm{Cat}^{\mathrm{ob}(I)}$ whose algebras are functors $I\to \mathrm{Cat}$. The pseudo $T$-morphisms are pseudo natural transformations, and the lax/oplax $T$-morphisms are lax/oplax natural transformations. When doctrinal adjunction is unraveled in the case of this 2-monad, it amounts to exactly the condition mentioned by Denis.

2: property-like structure. You mentioned that the space of adjoints to a given morphism is contractible (if nonempty), i.e. that "having an adjoint" appears to be a mere property of a morphism (rather than structure on it). From this perspective it may be surprising that the adjoints don't fit together. In fact, though, having an adjoint is something in between a "property" and "structure" called a property-like structure: a structure that is unique on objects when it exists, but is not necessarily preserved by morphisms.

One of the simplest examples of property-like structure is "having an identity element" for a semigroup: a semigroup can have at most one identity element, but a semigroup homomorphism need not preserve identities. A more well-known example is "having colimits" for a category: they are unique (up to unique isomorphism) when they exist, but not every functor preserves them (even up to isomorphism). The latter is an example of a special kind of property-like structure called lax-idempotent, in that it is automatically preserved laxly by every morphism (in this case, the comparison map $\mathrm{colim} \circ F \to F \circ \mathrm{colim}$). More precisely, a 2-monad $T$ on a 2-category $K$ is lax-idempotent if every $K$-morphism between $T$-algebras has a unique structure of lax $T$-morphism.

Now there is a 2-monad $T$ on the 2-category $\mathrm{Cat}^{\mathbf{2}}$ whose algebras are functors equipped with a left adjoint, and this 2-monad is lax-idempotent: the unique lax $T$-morphism structure on a commutative square is, again, its mate under the adjunctions. Now your given natural transformation is a functor $I\to \mathrm{Cat}^{\mathbf{2}}$, and the fact that it has adjunctions "pointwise" means that this functor lifts "objectwise" to $T$-algebras. Lax-idempotence of $T$ therefore implies that the functor lifts to the 2-category of $T$-algebras and lax $T$-morphisms; hence it lifts to pseudo $T$-morphisms if and only if all these mates are isomorphisms.


As far as know, neither of these abstract contexts has yet been worked out in the $\infty$ case. But someone should!

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