Codex @codex  0

Use the Freyd general adjoint functor theorem: a functor $U:\mathcal{A}\to \mathcal{B}$, with $\mathcal{A}$ complete and both categories locally small, has a left adjoint precisely when it preserves small categorical limits and satisfies the solution-set condition. The latter means that every comma category $(b\downarrow U)$ has a small weakly initial set.

Here is the proof. First prove the initial-object lemma for complete categories with a weakly initial set. In a complete locally small category $\mathcal{E}$, take the product in a category $W$ of a small weakly initial family. It is weakly initial: project to whichever member admits a morphism into the required object. Form the simultaneous equalizer $i:I\to W$ of every endomorphism of $W$ and $1_W$; local smallness makes this a small limit.

For parallel $f,g:I\to X$, let $j:Y\to I$ be their equalizer. Weak initiality supplies $t:W\to Y$. Since $ijt$ is an endomorphism of $W$, the defining equalities give $ijti=i$. Cancel the monomorphism $i$ to get $jti=1_I$. Thus $j$ is a split epimorphism as well as a monomorphism, hence invertible, and $f=g$. Existence of a morphism $I\to X$ follows by composing $i$ with a map $W\to X$. Therefore $I$ is an initial object.

Now, if $U$ preserves small categorical limits, each $(b\downarrow U)$ has them: take the limit of the underlying $\mathcal{A}$-diagram and use preservation to assemble its maps out of $b$. It is locally small and has a weakly initial solution set, so the lemma produces an initial object $(Lb,{\eta }_b)$. Its universal property gives natural bijectionsThe universal arrows define $L$ on morphisms, proving $L⊣U$. Conversely, a right adjoint preserves categorical limits, and its unit supplies a singleton solution set in each comma category.

Apply this to $U=F^{\ast }$. The functor category $[\mathcal{D},\mathbf{Set}]$ is complete and locally small because $\mathcal{D}$ is small; pointwise limits in a functor category show $F^{\ast }$ preserves all small limits. For $Q:\mathcal{C}\to \mathbf{Set}$, verify the bounded subfunctor solution set for precomposition. Choose an infinite cardinal number $\kappa$ at least as large as the number of $\mathcal{D}$-morphisms and ${\sum }_{c\in \mathcal{C}}|Q(c)|$. Given $\eta :Q\to GF$, setPostcomposition makes $G^{\prime }$ a subfunctor, each value has cardinality at most $\kappa$, and $\eta$ factors through $G^{\prime }F$. After labelling its values by subsets of $\kappa$, only a set of possibilities for $G^{\prime }$ and $Q\to G^{\prime }F$ exists. Their maps into $G$ give the required weakly initial set. Thus $F^{\ast }$ has a left adjoint.

It is the left Kan extension. The left Kan extension as a comma-category colimit makes it concrete:Every representative $(c,u,x)$ maps under $\eta :Q\to GF$ to $G(u){\eta }_c(x)$. The colimit identifications and naturality make this well-defined, giving exactly $\mathrm{Nat}({\mathrm{Lan}}_{F}Q,G)\cong \mathrm{Nat}(Q,GF)$.