[proofplan] For each object $c \in \mathcal C$, we study the comma category $(c \downarrow G)$. The solution-set condition gives a weakly initial set in this comma category, while completeness of $\mathcal D$ and preservation of limits by $G$ imply that $(c \downarrow G)$ is complete. We then prove, inside the argument, the initial-object criterion that a complete locally small category with a weakly initial set has an initial object. Applying this criterion to every $(c \downarrow G)$ produces universal arrows $c \to GLc$, and these universal arrows assemble uniquely into a left adjoint $L: \mathcal C \to \mathcal D$. [/proofplan]

[step:Show that each comma category $(c \downarrow G)$ is complete] Fix an object $c \in \mathcal C$. Let $\mathcal A_c := (c \downarrow G)$ denote the comma category whose objects are pairs $(d,\alpha)$, where $d \in \mathcal D$ and $\alpha: c \to Gd$ is a morphism in $\mathcal C$, and whose morphisms \begin{align*} u: (d,\alpha) \to (d',\alpha') \end{align*} are morphisms $u: d \to d'$ in $\mathcal D$ satisfying \begin{align*} G(u) \circ \alpha = \alpha'. \end{align*} Let $J$ be a small category, and let \begin{align*} F: J \to \mathcal A_c \end{align*} be a diagram. Write \begin{align*} F(j) = (d_j,\alpha_j) \end{align*} for each object $j \in J$, and write $u_a: d_j \to d_k$ for the morphism in $\mathcal D$ underlying $F(a)$ whenever $a: j \to k$ is a morphism in $J$. Since $\mathcal D$ is complete, the diagram \begin{align*} D: J \to \mathcal D, \qquad j \mapsto d_j \end{align*} has a limit. Let $d \in \mathcal D$ be this limit, with projection morphisms \begin{align*} \pi_j: d \to d_j \end{align*} for $j \in J$. Because $G$ preserves small limits, the cone \begin{align*} G(\pi_j): Gd \to Gd_j \end{align*} is a limit cone for the diagram $G \circ D: J \to \mathcal C$. The morphisms $\alpha_j: c \to Gd_j$ form a cone from $c$ to $G \circ D$, since for every morphism $a: j \to k$ in $J$, \begin{align*} G(u_a) \circ \alpha_j = \alpha_k \end{align*} by the definition of morphisms in $\mathcal A_c$. Therefore the universal property of the limit $Gd$ gives a unique morphism \begin{align*} \alpha: c \to Gd \end{align*} such that \begin{align*} G(\pi_j) \circ \alpha = \alpha_j \end{align*} for every $j \in J$. The pair $(d,\alpha)$, together with the morphisms \begin{align*} \pi_j: (d,\alpha) \to (d_j,\alpha_j), \end{align*} is a cone in $\mathcal A_c$. To verify its universal property, let $(e,\beta)$ be any object of $\mathcal A_c$ equipped with a cone \begin{align*} \rho_j: (e,\beta) \to (d_j,\alpha_j) \end{align*} over $F$. The underlying morphisms $\rho_j: e \to d_j$ form a cone over $D$, so there is a unique morphism $\rho: e \to d$ in $\mathcal D$ satisfying \begin{align*} \pi_j \circ \rho = \rho_j \end{align*} for every $j \in J$. Applying $G$ and composing with $\beta$ gives \begin{align*} G(\pi_j) \circ G(\rho) \circ \beta = G(\rho_j) \circ \beta = \alpha_j = G(\pi_j) \circ \alpha \end{align*} for every $j \in J$. Since the cone $(G(\pi_j))_{j \in J}$ is limiting, it is jointly monic, and hence \begin{align*} G(\rho) \circ \beta = \alpha. \end{align*} Thus $\rho$ is a morphism $(e,\beta) \to (d,\alpha)$ in $\mathcal A_c$, and its uniqueness follows from uniqueness in $\mathcal D$. Hence $\mathcal A_c$ is complete. [/step]

[step:Use the solution-set condition to obtain a weakly initial set] By the solution-set condition at $c$, there exist a set $I_c$, objects $d_i \in \mathcal D$, and morphisms $\eta_i: c \to Gd_i$ for $i \in I_c$ such that every morphism $f: c \to Gd$ factors as \begin{align*} f = G(h) \circ \eta_i \end{align*} for some $i \in I_c$ and some morphism $h: d_i \to d$ in $\mathcal D$. Define \begin{align*} S_c := \{(d_i,\eta_i) : i \in I_c\}. \end{align*} This is a set of objects of $\mathcal A_c$. If $(d,f)$ is any object of $\mathcal A_c$, then the factorization condition gives an object $(d_i,\eta_i) \in S_c$ and a morphism $h: d_i \to d$ in $\mathcal D$ such that \begin{align*} G(h) \circ \eta_i = f. \end{align*} Thus $h$ is a morphism \begin{align*} h: (d_i,\eta_i) \to (d,f) \end{align*} in $\mathcal A_c$. Therefore $S_c$ is a weakly initial set in $\mathcal A_c$. [/step]

[step:Construct an initial object from a weakly initial set in a complete locally small category] [claim:Every complete locally small category with a weakly initial set has an initial object] Let $\mathcal A$ be a complete locally small category, and suppose that $S$ is a weakly initial set of objects of $\mathcal A$. Then $\mathcal A$ has an initial object. [/claim] [proof] Since $\mathcal A$ is complete and $S$ is a set, the product \begin{align*} W := \prod_{s \in S} s \end{align*} exists in $\mathcal A$. For every object $A \in \mathcal A$, choose an object $s \in S$ and a morphism $r_s: s \to A$, which exists because $S$ is weakly initial. Let \begin{align*} p_s: W \to s \end{align*} be the product projection. Then $r_s \circ p_s: W \to A$ is a morphism, so $W$ is weakly initial. Because $\mathcal A$ is locally small, the collection \begin{align*} \operatorname{End}_{\mathcal A}(W) := \operatorname{Hom}_{\mathcal A}(W,W) \end{align*} is a set. Since $\mathcal A$ is complete, the product \begin{align*} P := \prod_{a \in \operatorname{End}_{\mathcal A}(W)} W \end{align*} exists. Define two morphisms \begin{align*} \Delta: W \to P, \qquad A: W \to P \end{align*} as follows: the $a$-component of $\Delta$ is $\operatorname{id}_W: W \to W$, and the $a$-component of $A$ is $a: W \to W$. Let \begin{align*} e: E \to W \end{align*} be the equalizer of $\Delta$ and $A$. Equivalently, for every endomorphism $a: W \to W$, \begin{align*} a \circ e = e. \end{align*} We prove that $E$ is initial. First let $A_0$ be any object of $\mathcal A$. Since $W$ is weakly initial, there exists a morphism $r: W \to A_0$. Hence \begin{align*} r \circ e: E \to A_0 \end{align*} is a morphism, so at least one morphism $E \to A_0$ exists. It remains to prove uniqueness. Let \begin{align*} f,g: E \to A_0 \end{align*} be morphisms. Since $\mathcal A$ is complete, let \begin{align*} m: B \to E \end{align*} be the equalizer of $f$ and $g$. Since $W$ is weakly initial, there exists a morphism \begin{align*} \ell: W \to B. \end{align*} The composite \begin{align*} a := e \circ m \circ \ell: W \to W \end{align*} is an endomorphism of $W$. By the defining property of $e$, we have \begin{align*} a \circ e = e. \end{align*} Substituting the definition of $a$ gives \begin{align*} e \circ m \circ \ell \circ e = e. \end{align*} Since equalizers are monomorphisms, $e$ is monic, and therefore \begin{align*} m \circ \ell \circ e = \operatorname{id}_E. \end{align*} Because $m$ equalizes $f$ and $g$, we obtain \begin{align*} f = f \circ \operatorname{id}_E = f \circ m \circ \ell \circ e = g \circ m \circ \ell \circ e = g \circ \operatorname{id}_E = g. \end{align*} Thus there is exactly one morphism from $E$ to every object $A_0 \in \mathcal A$, so $E$ is initial. [/proof] Apply the claim to $\mathcal A_c = (c \downarrow G)$. This category is locally small because $\mathcal C$ and $\mathcal D$ are locally small and \begin{align*} \operatorname{Hom}_{\mathcal A_c}((d,\alpha),(d',\alpha')) \subset \operatorname{Hom}_{\mathcal D}(d,d') \end{align*} is a subset defined by the equation $G(u)\circ \alpha=\alpha'$. By the previous two steps, $\mathcal A_c$ is complete and has a weakly initial set. Hence $\mathcal A_c$ has an initial object. [/step]

[step:Interpret the initial object as a universal arrow] For each object $c \in \mathcal C$, choose an initial object of $\mathcal A_c = (c \downarrow G)$ and denote it by \begin{align*} (Lc,\eta_c), \end{align*} where $Lc \in \mathcal D$ and $\eta_c: c \to G(Lc)$ is a morphism in $\mathcal C$. Initiality means that for every object $d \in \mathcal D$ and every morphism \begin{align*} f: c \to Gd \end{align*} in $\mathcal C$, there exists a unique morphism \begin{align*} \bar f: Lc \to d \end{align*} in $\mathcal D$ such that \begin{align*} G(\bar f) \circ \eta_c = f. \end{align*} Thus $\eta_c: c \to G(Lc)$ is a universal arrow from $c$ to $G$. [/step]

[step:Assemble the universal arrows into a left adjoint] We define a functor \begin{align*} L: \mathcal C \to \mathcal D \end{align*} on objects by the assignment $c \mapsto Lc$ chosen above. For a morphism \begin{align*} \alpha: c \to c' \end{align*} in $\mathcal C$, define \begin{align*} L\alpha: Lc \to Lc' \end{align*} to be the unique morphism in $\mathcal D$ satisfying \begin{align*} G(L\alpha) \circ \eta_c = \eta_{c'} \circ \alpha. \end{align*} This morphism exists and is unique by the universal property of $\eta_c$, applied to the morphism $\eta_{c'} \circ \alpha: c \to G(Lc')$. We verify functoriality. For the identity morphism $\operatorname{id}_c: c \to c$, both $L(\operatorname{id}_c)$ and $\operatorname{id}_{Lc}$ satisfy \begin{align*} G(-) \circ \eta_c = \eta_c. \end{align*} By uniqueness in the universal property of $\eta_c$, \begin{align*} L(\operatorname{id}_c) = \operatorname{id}_{Lc}. \end{align*} If $\alpha: c \to c'$ and $\beta: c' \to c''$ are morphisms in $\mathcal C$, then \begin{align*} G(L\beta \circ L\alpha) \circ \eta_c &= G(L\beta) \circ G(L\alpha) \circ \eta_c \\ &= G(L\beta) \circ \eta_{c'} \circ \alpha \\ &= \eta_{c''} \circ \beta \circ \alpha. \end{align*} The morphism $L(\beta \circ \alpha): Lc \to Lc''$ is uniquely characterized by the same equation, so \begin{align*} L(\beta \circ \alpha)=L\beta \circ L\alpha. \end{align*} Thus $L$ is a functor. [/step]

[step:Identify the adjunction bijection] For each $c \in \mathcal C$ and $d \in \mathcal D$, define \begin{align*} \Phi_{c,d}: \operatorname{Hom}_{\mathcal D}(Lc,d) \to \operatorname{Hom}_{\mathcal C}(c,Gd) \end{align*} by \begin{align*} \Phi_{c,d}(h) := G(h) \circ \eta_c. \end{align*} The universal property of $\eta_c$ says exactly that $\Phi_{c,d}$ is a bijection for every pair $(c,d)$. It remains only to record naturality. Let $\alpha: c' \to c$ be a morphism in $\mathcal C$, let $k: d \to d'$ be a morphism in $\mathcal D$, and let $h: Lc \to d$ be a morphism in $\mathcal D$. Then \begin{align*} \Phi_{c',d'}(k \circ h \circ L\alpha) &= G(k \circ h \circ L\alpha) \circ \eta_{c'} \\ &= G(k) \circ G(h) \circ G(L\alpha) \circ \eta_{c'} \\ &= G(k) \circ G(h) \circ \eta_c \circ \alpha \\ &= G(k) \circ \Phi_{c,d}(h) \circ \alpha. \end{align*} This is precisely the naturality condition for the family of bijections \begin{align*} \operatorname{Hom}_{\mathcal D}(Lc,d) \cong \operatorname{Hom}_{\mathcal C}(c,Gd). \end{align*} Therefore $L$ is left adjoint to $G$. Hence $G$ has a left adjoint, as claimed. [/step]