When is $\mathrm{Hom}(i(x), y)\to\mathrm{Hom}(f(x),\mathrm{Lan}_i(f)(y))$ an isomorphism?

Question

$\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Nat}{Nat}$ Let $i : A\to B$ and $f : A\to C$ be functors.

In the case where $A=C$ and $f$ is the identity functor, if the left Kan extension of $f$ along $i$ is pointwise, then it is a right adjoint to $i$. In general, we can build a transformation $$\mathrm{Hom}(i(x), y)\to \mathrm{Hom}(f(x),\mathrm{Lan}_i(f)(y))$$ natural in $x$ and $y$ that specializes to the adjunction in the case of the previous sentence. In general, this transformation only have an initial property (it is a translation of the initial property of $\mathrm{Lan}_i(f)$) and is not an isomorphism.

An other case where it is an isomorphism arises when we consider the nerve of some functor $F : A→B$. It is the left Kan extension of the Yoneda embedding $A\to [A^{\mathrm{op}},\mathrm{Set}]$ along $F$. The Yoneda lemma says that the morphism above is an isomorphism.

Question: Is there something interesting to say about when this morphism is an isomorphism? Some necessary or sufficient condition? (Maybe nothing?)

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I add more details below.

Construction of the morphism. The Yoneda lemma imply that for any functor $g : B→C$ (and as above $i:A\to B$ and $f:A\to C$), the set of natural transformations $f\to g\circ i$ is in natural bijection with the natural transformations $$\mathrm{Hom}(i(x), y)\to \mathrm{Hom}(f(x),g(y))\text{.}$$ We can use the end/coend calculus for this: $$\begin{align*} \Nat(f,g\circ i) &\cong \int_x \Hom(f(x), g(i(x)))\\ &\cong \int_x \Hom\left[f(x), ∫_y g(y)^{\Hom(i(x),y)}\right]\\ &\cong ∫_{x,y} \Hom[\Hom(i(x),y), \Hom(f(x),g(y))] \end{align*}$$

Thus a Kan extension can equally be defined using the morphism this question is about.

There is no implication between "$f\to g\circ i$ is an isomorphism" and the property that the corresponding morphism on the other side is an isomorphism (when $g = \mathrm{Lan}_i(f)$). Indeed, in the case of the nerve functor, $f\to g\circ i$ is not an isomorphism. And in the case of the realization functor, $f\to g\circ i$ is an isomorphism but the corresponding transformation is not.

I would enjoy an answer involving profunctors to explain how the "bending" works and what corresponds to what (a bit like we can do with the units/counit of an adjunction to make correspond the zig-zag identities with the property that the two natural transformations are inverses).

Answer 1

This post on MO seems to answer my question: https://mathoverflow.net/questions/252383/the-kan-construction-profunctors-and-kan-extensions. It takes into account both the situations in the OP.

I yet have to read it and will edit this answer when I do if nobody posts an other answer meanwhile.