This is probably wrong or I'm talking funny but I'll try **Puzzle 163** the best I can in hopes of someone giving a solution.

So using John's tip in [comment 27](https://forum.azimuthproject.org/discussion/comment/19814/#Comment_19814) I will define the functors first.

So say we have a category \\(\mathcal{C}, \mathcal{D} \in Ob(\mathbf{Cat})\\) and a functor \\(F : \mathcal{C} \rightarrow \mathcal{D}\\). Also \\(x,y \in Ob(\mathcal{C})\\). We also need \\(S,T \in \mathbf{Set}\\) and a function \\(f : S \rightarrow T\\).

**\\(\mathbf{Cat} \stackrel{R_1}{\longrightarrow} \mathbf{Preorder}\\)**

Let \\([\leq]\\) be the isomorphism class on the morphisms in \\(homset(x,y)\\) so that
$$homset(x,y) \iff x \; [\leq] \; y$$

Define \\(R_1(C)\\) to be the preorder formed by applying **Reflexivity** and **Transivity** using \\([\leq]\\).

Define \\(R_1(F)\\) to be the monotone function such that \\(x \; [\leq] \; y \iff m_1(x) \; [\leq] \; m_1(y)\\) where \\(m_1\\) is a monotone function in \\(\mathbf{Preorder}\\).

**\\(\mathbf{Preorder} \stackrel{R_2}{\longrightarrow} \mathbf{Poset}\\)**

Let \\([x]\\) be the isomorphism class on the morphisms in \\(x \in Ob(C)\\) so that
$$x \; [\leq] \; y \; and \; y \; [\leq] \; x \iff [x] \; = \; [y]$$

Define \\(R_2R_1(C)\\) to be the preorder formed by applying **Reflexivity** and **Transivity** using \\([\leq]\\) and \\([x]\\).

Define \\(R_2R_1(F)\\) to be the monotone function such that \\([x] \; [\leq] \; [y] \iff m_2([x]) \; [\leq] \; m_2([y])\\) where \\(m_2\\) is a monotone function in \\(\mathbf{Poset}\\).

**\\(\mathbf{Poset} \stackrel{R_3}{\longrightarrow} \mathbf{Set}\\)**

Define \\(R_3R_2R_1(C)\\) to be the set \\(Ob(\mathcal{C})\\) such that \\([x] \; [\leq] \; [y] \iff [x],[y] \in Ob(\mathcal{C})\\).

Define \\(R_3R_2R_1(F)\\) to be the function \\(m_3\\) such that \\([x] \rightarrow m_2([x])\\) for all \\([x] \in Ob(\mathcal{C})\\)

Then we can daisy chain the naturality square Cheuk Man used and proved the naturality for in [comment 39](https://forum.azimuthproject.org/discussion/comment/19830/#Comment_19830) to define the left adjoints.

\[
\begin{matrix}
\mathbf{Cat}(L_1L_2L_3(S), \mathcal{C}) & \overset{\mathbf{Cat}(L_1L_2L_3(\mathbf{f}), \mathbf{F})}\longrightarrow & \mathbf{Cat}(L_1L_2L_3(T), \mathcal{D}) \\\\
\alpha_{L_2L_3(S),\mathcal{C}} \downarrow & & \alpha_{L_2L_3(T),\mathcal{D}} \downarrow \\\\
\mathbf{Pre}(L_2L_3(S), R_1(\mathcal{C})) & \underset{\mathbf{Pre}(L_2L_3(\mathbf{f}), R_1(\mathbf{F}))}\longrightarrow & \mathbf{Pre}(L_2L_3T, R_1(\mathcal{D}))\\\\
\alpha_{L_3(S),R_1(\mathcal{C})} \downarrow & & \alpha_{L_3(T),R_1(\mathcal{D})} \downarrow \\\\
\mathbf{Pos}(L_3(S), R_2R_1(\mathcal{C})) & \underset{\mathbf{Pos}(L_3(\mathbf{f}), R_2R_1(\mathbf{F}))}\longrightarrow & \mathbf{Pos}(L_3(T), R_2R_1(\mathcal{D}))\\\\
\alpha_{S,R_2R_1(\mathcal{C})} \downarrow & & \alpha_{T,R_2R_1(\mathcal{D})} \downarrow \\\\
\mathbf{Set}(S, R_3R_2R_1(\mathcal{C})) & \underset{\mathbf{Set}(\mathbf{f}, R_3R_2R_1(\mathbf{F}))}\longrightarrow & \mathbf{Set}(T, R_3R_2R_1(\mathcal{D}))\\\\
\end{matrix}
\]