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

$$\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\Nat}{Nat}$$ 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?)

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).

• compositition symbol $\circ$: \circ, right arrow $\to$: \to or \rightarrow, $\cong$: \cong; please see edits so you can improve your formatting. $\int$: \int – amWhy Feb 27 '20 at 17:56
• @amWhy Ok, thank you. I'll do that in the future but I see no difference on my computer between \cong ($\cong$) and ≅ ($≅$). I imagine there is one on yours. By the way I just put back "counit", you replaced it by "count". – Idéophage Feb 27 '20 at 18:07
• What you pasted looked like $\cong$, so I used the corresponding LaTeX symbol. If you meant to use $\equiv$, then use \equiv. Copy and pastes of text symbols deteriorate more easily, and may not be seen consistently by all users. Also, know that I would not "nitpick" if you hadn't already seemed to exhibit some competence in formatting. I thought you might appreciate the pointers. – amWhy Feb 27 '20 at 18:11
• @amWhy Actually I used unicode characters instead of latex commands intentionally, because I have easy access to them on my keyboard and know mathjax supports them (I don't copy/paste! It would be easier to type the latex.). I found some discussions about this issue on the meta: here and there. But anyway it's not important. – Idéophage Feb 27 '20 at 18:31