Using the Yoneda Lemma to construct a left adjoint to the restriction functor $U : C^A \to C^{\operatorname{ob} A}$

Question

I am working through the following exercise of Emily Riehl's Category Theory in Context,

Exercise 5.5.v. Generalizing Exercise 5.5.iv, for any small category $J$ and any cocomplete category $C$ the forgetful functor $C^J \to C^{\operatorname{ob} J}$ admits a left adjoint $\operatorname{Lan}: C^{ \operatorname{ob} J} \to C^J$ that sends a functor $F ∈ C^{\operatorname{ob} J}$ to the functor $\operatorname{Lan}F ∈ C^J$ defined by $$ \operatorname{Lan}F(j) = \coprod_{x \in J}\coprod_{C(x,j)}Fx. $$ (i) Define $\operatorname{Lan}F$ on morphisms in $J$.

(ii) Define $\operatorname{Lan}$ on morphisms in $C^{\operatorname{ob} J}$.

(iii) Use the Yoneda lemma to show that $\operatorname{Lan}$ is left adjoint to the forgetful (restriction) functor $C^J \to C^{\operatorname{ob} J}$.

(iv) Prove that this adjunction is monadic by appealing to the monadicity theorem.

I have managed to do items $(i)$ through $(iii)$, although I haven't appealed to the Yoneda lemma to prove that $\operatorname{Lan}$ is left adjoint to the restriction functor: instead, I have constructed a natural bijection 'by hand'.

How can one prove this via the Yoneda lemma? I would also appreciate a hint for item $(iv)$, as I haven't gathered much intution on split-pairs yet.

Answer 1

I guess it is just the classical application of Yoneda lemma, in the form that adjoints are unique up to canonical iso: call $U$ the forgetful functor; it is just precomposition with the functor $j : Ob(J) \to J$. So, if $U$ has a left adjoint, it must be left extension along $j$.

If you now prove that the functor $L$ Emily defined is such that $$ C^{Ob(J)}(D, UX)\cong C^J(LD,X) $$ (you did it by hand, it's a reasonable way) then you get a natural isomorphism $$ C^J(LD,X) \cong C^J(Lan_jD,X) $$ and now Yoneda lemma entails that $LD\cong Lan_jD$, canonically.

Answer 2

The answer given by fosco actually only seems to use Yoneda to show that, for some $D \in \mathsf{C^{\mathrm{ob}J}}$, Lan$D$ as defined by Riehl is naturally isomorphic to the left Kan extension of $D$ along $j$ - after the natural bijection proving the adjunction $U \vdash$ Lan has already been constructed by hand. The question, however, was to use Yoneda to prove the adjunction itself.

If we wish to find an isomorphism $\mathsf{C^J}(\mathrm{Lan}F, G) \cong \mathsf{C^{\mathrm{ob}J}}(F, UG)$ that is natural in both variables, we can start by fixing $F$ and finding a natural transformation $\mathsf{C^J}(\mathrm{Lan}F, -) \Rightarrow \mathsf{C^{\mathrm{ob}J}}(F, U-)$. But, by Yoneda, the set of all such natural transformations is in bijection with $\mathsf{C^{\mathrm{ob}J}}(F, U\mathrm{Lan}F)$. Thankfully, we have ample choice. The only morphisms in $\mathrm{ob}\mathsf{J}$ are identities, so any collection of morphisms $(Fx \xrightarrow{\alpha_x} U\mathrm{Lan}F(x))_{x \in \mathrm{ob} \mathsf{J}}$ trivially assembles into a natural transformation. An immediately obvious candidate would be to compose the coproduct inclusions leading from $Fx$ to Lan$F(x)$, but since the inner coproduct is indexed by the set $\mathsf{J}(x,x)$ there are potentially many different inclusions we could choose. However, the only element of $\mathsf{J}(x,x)$ that is guaranteed to exist is $1_x$, so we define $\eta_F \in \mathsf{C^{\mathrm{ob}J}}(F, U\mathrm{Lan}F)$ such that $\eta_{Fx} = \iota_x \iota_{1_x}$. (Note the choice of symbol here. This will actually end up defining the unit $\eta : 1_{\mathsf{C^{\mathrm{ob}J}}} \Rightarrow U\mathrm{Lan}$ of the adjunction.) Then the bijection due to Yoneda maps $\eta_F$ to $\Psi(\eta_F) : \mathsf{C^J}(\mathrm{Lan}F, -) \Rightarrow \mathsf{C^{\mathrm{ob}J}}(F, U-)$ defined by $\Psi(\eta_F)_G(\alpha) = \mathsf{C^{\mathrm{ob}J}}(F, U\alpha)(\eta_F) = \alpha\eta_F$ for some $\alpha \in \mathsf{C^J}(\mathrm{Lan}F, G)$. But is this transformation also natural in $F$? $\require{AMScd}$ \begin{CD} \mathsf{C^J}(\mathrm{Lan}F, G) @>\Psi(\eta_F)_G>> \mathsf{C^{\mathrm{ob}J}}(F, UG)\\ @V (\mathrm{Lan}\beta)^* VV @VV \beta^* V\\ \mathsf{C^J}(\mathrm{Lan}F', G) @>>\Psi(\eta_{F'})_G> \mathsf{C^{\mathrm{ob}J}}(F', UG) \end{CD} For some $\beta : F' \Rightarrow F$, that would require $\alpha\eta_F\beta = \alpha\mathrm{Lan}\beta\eta_{F'}$, which should follow from the definition of Lan$\beta$ found in part (ii).

All that remains is to show that our construction is an isomorphism. As far as I can tell, Yoneda is no longer helpful here and one must simply define the inverse. Thus, I will leave this to any interested readers since the original question asker has already done so.