Kan extension and representable functors Consider some category $\mathcal{C}$ with some full subcategory $\mathcal{G}$ such that $\mathcal{G}$ generates $\mathcal{C}$ (the example I have in mind the $Alg_A^{free}\subset Alg_A$). A functor $F:\mathcal{G}\rightarrow Sets$ can be extended by Kan extension to a functor $F':\mathcal{C}\rightarrow Sets$. I'm curious about the interplay of the representability of these two functors. For instance, if we assume that $F'$ is representable, then it commutes with limits, and since the inclusion functor $\mathcal{G}\hookrightarrow \mathcal{C}$ also commutes with limits, it is necessary for $F$ to respect limits.
If $F$ is representable, is $F'$ representable? I've tried using the explicit (co-)limit representation of Kan extensions, but didn't get anywhere.
 A: Yes, $F'$ will always be corepresentable if $F$ is.
Let's say $i:\mathcal G\hookrightarrow\mathcal C$ is the fully faithful dense inclusion you mention, and $F:\mathcal G\to\mathbf{Set}$ some functor, then your functor $F'$ (as you mention) is the left Kan extension $F'=\operatorname{Lan}_iF$, which (as per the $n$Lab) has the formula
$$
F'(c) = \int^{g\in\mathcal G}\mathcal C(i(g),c)\odot F(g)
$$
(where the copower in $\mathbf{Set}$ is just given by the cartesian product). Now, suppose $F'=\mathcal G(a,-)$ is corepresentable, then the above formula reduces to
$$
F'(c) = \int^{g\in\mathcal G}\mathcal C(g,c)\times\mathcal C(a,g)
$$
using that $\mathcal G$ is a full subcategory so that $\mathcal G(a,g)=\mathcal C(a,g)$. Since composition $\circ:\mathcal C(g,c)\times\mathcal C(a,g)\to\mathcal C(a,c)$ is extranatural in $g$, we get a canonical map $F'(c)\to\mathcal C(a,c)$.
Explicitly, we $F'(c)$ is the set of pairs $(\varphi,\psi)$ of morphisms $\varphi:g\to c$ and $\psi:a\to g$ (where $g$ ranges over $\mathcal G$) modulo the equivalence induced by identifying $(\lambda\circ\gamma,\rho)$ and $(\lambda,\gamma\circ\rho)$ for $\gamma:g\to g'$, $\lambda:g'\to c$, and $\rho:a\to g$.
The canonical map is then just composing pairs; that is, $(\varphi,\psi)\mapsto\varphi\circ\psi$.
However, the equivalence allows us to identify any pair $(\varphi,\psi)=(\varphi,\psi\circ\operatorname{id}_a)$ with $(\varphi\circ\psi,\operatorname{id}_a)$, so every element of $F'(c)$ is canonically of the form $(\xi,\operatorname{id}_a)$ for $\xi:a\to c$. Moreover, under this identification, any equivalent $(\lambda\circ\gamma,\rho)$ and $(\lambda,\gamma\circ\rho)$ get sent to $(\lambda\circ\gamma\circ\rho,\operatorname{id}_a)$, which shows that this representative is unique. Moreover, the canonical map $F'(c)\to\mathcal C(a,c)$ reduces to just projection $(\xi,\operatorname{id}_a)\mapsto\xi$. Therefore, it turns out that $F'(c)=\mathcal C(a,c)$, showing that $F'$ is indeed corepresentable, and by the same corepresenting object as $F$.
This is an explicit way of seeing it, but this readily generalises a lot. If we take $\mathcal V$ to be any (Bénabou) cosmos, $\mathcal C$ a $\mathcal V$-enriched category, and $\mathcal G$ a full sub-$\mathcal V$-category of $\mathcal C$, then any corepresentable functor $F=\mathcal G(a,-):\mathcal G\to\mathcal V$ admits a left Kan extension as before (this time the copowering being given by the tensor product of $\mathcal V$) with
$$
F'(c) := \operatorname{Lan}_{\mathcal G\hookrightarrow\mathcal C}F(c) = \int^{g\in\mathcal G}\mathcal C(g,c)\otimes\mathcal C(a,g)
$$
and the map $F'(c)\to\mathcal C(a,c)$ induced by composition as before will be an isomorphism in $\mathcal V$ natural in $c$, so $F'$ is again corepresented by $a$.
