@Michael

The puzzle asks us to cook up suitable functors \\(R_1, R_2, R_3\\) and find left adjoints for them, and possibly right adjoints. Currently I can't even cook up the first two functors!

Re the natural isomorphism \\(\mathcal{D}[LX, Y] \cong \mathcal{C}[X, RY]\\), I've found that you don't usually need to worry too much about naturality. The key thing is to look for a bijection between \\(\mathcal{D}\\)-arrows \\(LX \rightarrow Y\\) and \\(\mathcal{C}\\)-arrows \\(X \rightarrow RY\\). Usually something will "jump out" at you, and it will usually turn out to be natural.

What the naturality condition actually says is that if we write \\(\Phi\\) for our bijection, we ought to have

\\[\Phi(Lf \circ x \circ g) = f \circ \Phi(x) \circ Rg\\]

for any arrows \\(f\\) into \\(X\\) and \\(g\\) out of \\(Y\\). But this tends to fall out automatically for any sensible choice of \\(\Phi\\).

The puzzle asks us to cook up suitable functors \\(R_1, R_2, R_3\\) and find left adjoints for them, and possibly right adjoints. Currently I can't even cook up the first two functors!

Re the natural isomorphism \\(\mathcal{D}[LX, Y] \cong \mathcal{C}[X, RY]\\), I've found that you don't usually need to worry too much about naturality. The key thing is to look for a bijection between \\(\mathcal{D}\\)-arrows \\(LX \rightarrow Y\\) and \\(\mathcal{C}\\)-arrows \\(X \rightarrow RY\\). Usually something will "jump out" at you, and it will usually turn out to be natural.

What the naturality condition actually says is that if we write \\(\Phi\\) for our bijection, we ought to have

\\[\Phi(Lf \circ x \circ g) = f \circ \Phi(x) \circ Rg\\]

for any arrows \\(f\\) into \\(X\\) and \\(g\\) out of \\(Y\\). But this tends to fall out automatically for any sensible choice of \\(\Phi\\).