Can we lift adjunctions to limit stages in an ordinal chain of categories? Assume we have a family of categories indexed by an ordinal $(A_\beta)_{\beta<\alpha}$, with functors (in my mind they're forgetful functors so if it helps, you can assume they're faithful) $U_{\beta, \gamma} : A_\beta \to A_\gamma$ whenever $\gamma > \beta$ that compose as we want them to (essentially we're lookikg at a functor $\alpha^{\operatorname{op}} \to \mathbf{CAT}$).
If we assume that each $U_{\beta,\beta +1}$ as a left-adjoint $F_{\beta,\beta+1}$, then clearly for all finite $n$, $U_{\beta, \beta+n}$ has a left adjoint (composing adjoint functors yields adjoint functors). For instance if $\alpha<\omega$, then we get an adjoint from any $A_\beta$ to $A_0$.
I was wondering if the more general setting (where $\alpha >\omega$ a priori) could yield a similar conclusion : that is, can you lift the chain of adjoints at the limit stages where you can't just say " Lets compose the adjoints" ? 
In other words, assuming we are in the described sutuation, can we conclude that there exists an adjoint to $U_{0,\beta}$ for each $\beta <\alpha$ ?
I tried things assuming the existence of certain limits/colimits in $\mathbf{CAT}$ but it didn't give anything.
 A: After a careful thinking time, it is now apparent to me that the answer is obviously no.
Indeed, taking $\alpha=\omega+1$ and $A_\omega$  to be the empty category (no objects, no arrows), and any $\omega$-chain of adjunctions (for instance $A_n = C$ a fixed category, and the $U$'s between the $C$'s is $\operatorname{id}_C$, and then since $A_\omega$  is initial we don't have a choice so that the appropriate diagrams commute), then clearly we can't always lift the adjunction, because most of the time there won't even be a functor $F: A_0\to A_\omega$ (if $A_0$ is not empty for instance).
So clearly the answer is no, although some deeply burried part of me wanted it to be yes.
A: Let's let $U$ be a continuous functor from $\alpha^{\mathrm{op}}$ to $\mathbf{Cat}$. This means that for each limit ordinal $\beta<\alpha$, $U(\beta)=\lim_{\gamma<\beta} U(\gamma)$. Thinking of the case $\beta=\omega$ may clarify the argument. So suppose $U(\gamma+1\to\gamma)$ has a (say, left) adjoint $F_{\gamma,\gamma+1}$ for each $\gamma<\beta$. Then we want to find a left adjoint for $U(\beta\to 0)$. This will be a functor $F_{0,\beta}:U(0)\to U(\beta)=\lim_{\gamma<\beta} U(\gamma)$. Of course, we'd like to have $F_{0,\beta}$ be induced by a cone whose legs are the $F_{0,\gamma}$. If we can do that, then we'll have $$U(\beta)(F_{0,\beta}x,y)=\lim_{\gamma<\beta}U(\gamma)(F_{0,\gamma}x,U_{\beta,\gamma}y)\cong\lim_{\gamma<\beta}U(0)(x,U_{\gamma 0}U_{\beta,\gamma}y)=U(0)(x,U_{\beta,0}y)$$
as desired. In the first equality I'm using that the limit of a diagram $C_i$ of categories has objects the limit of the objects of the $C_i$ and morphisms $(\lim C_i)((c_i),(c'_i))=\lim (C_i(c_i,c'_i))$. If instead we had a cocontinuous functor out of $\alpha$, things would be trickier since colimits of categories are nasty, but they'd still work out since $\alpha$ is a filtered category.
But we have cheated here, since the cone of $F$s need not be strictly commutative! We have $F_{\gamma,\gamma'}\circ F_{0,\gamma}\cong F_{0,\gamma'}$, since left adjoints are determined up to isomorphism, but there's no reason there should be an equality. Formally, $F$ constitutes a pseudofunctor $\alpha^{\mathrm{op}}\to\mathbf{Cat}$, not a legitimate functor.
We have a few options for getting around this. If we happen to have gotten lucky, so that $F_{\gamma,\gamma'}\circ F_{0,\gamma}=F_{0,\gamma'}$ does hold for every $\gamma,\gamma'<\beta$, then the argument above will go through. We'd like to understand when, in general, we can carefully choose the $F$s to make this work. If $\beta=\omega$, then this is always possible! We can just choose whatever adjoints we want for $F_{\gamma,\gamma+1}$, then defined $F_{\gamma,\gamma'}$ to be an appropriate composition of these. For larger ordinals, I think things also work out, but I'm not completely certain. We could also try to stick with the pseudofunctor version of $F$, which would correspond to requiring that $U$ be pseudocontinuous, but in that case I don't think that $F_{0,\beta}$ is actually guaranteed to be an adjoint anymore.
