Why $F : A \to B$ final is equivalent to $F_! : \hat{A} \to \hat{B} \dashv F^*$ preserving terminal objects?

Question

A functor $F : A \to B $ is final if for every $b \in B$ the under category $ b\downarrow F = \int B(b,F-)$ which is non empty and connected cf ncat

Jean Bénabou in Distributors at work claims that obviously a functor $F$ is final iff the left adjoint $F_! : \hat{A} \to \hat{B}$ (iso to the left extension along $y_A$ of $y_B \circ F$) to the functor $F^* : \hat{B} \to \hat{A}$ of precomposition by $F$ preserves terminal objects.

Is there a short abstract proof of this ?

Answer 1

I don't exactly know if this answers your question since it all depends on what you call "abstract", but here is how one can prove it.

First, show that a category $C$ is connected if and only if the colimit of the diagram $C → \mathrm{Set}$ constant to $1$ is $1$.

The second step is to use the concrete definition of left Kan extension mentioned by Zhen Lin to say that $F_!(1)$ sends $b$ to the colimit of the diagram $b↓F → \mathrm{Set}$ constant at $1$. Thus $F_!(1)(b) = 1$ if and only if $b↓F$ is connected.

We can then argue that $F$ is final if precomposing by $F$ preserves colimits. This is because the universal cocone over a diagram $B→C$ is built by extending the diagram by cocontinuity to the full subcategory of $\hat{B}$ generated by $B$ and the terminal object.

Answer 2

Here is another answer. There are probably others, and it is always interesting to compare them. Thank you for the previous ones, (and for the future ones, if there are some)

I am not sure is this is what Jean Bénabou had in mind, but it might be close to it.

Step 1

At the terminal object $t\in\widehat{A}$ (constantly $1 \in Set$ for all $a \in A$), we have that $$F_!(t) = \operatorname{colim} (y\circ F)$$ The reason being that pulling back along $t: 1 \to \widehat{A}$ is an exact square, and that its pullback object, the category of element of $t$, is isomorphic to $A$ itself $$\int t \simeq A$$

Step 2

Now, the connected bits comes from the fact that $$\operatorname{colim} (y\circ F) : 1 \to \widehat{B}$$ is the transpose of $$\varphi_F \otimes \varphi^! = 1 \xrightarrow{\varphi^!} A \xrightarrow {\varphi_F} B : B^{op}\times 1 \to Set$$ composed as distributors. This composition quantify over the different elements $x\in B(b,Fa)$ which are part of the same connected component.

It is now easy to see that the definition of a final functor maps to this distributor being constantly $1 \in Set$.

This proof is "obvious" inasmuch as the two previous steps are. It seems "intrinsic" at least, in the sense that it only mentions what things are. lower level details (say the $b's\in B$ in the end) play little role. I think it helps to understand what a result means, beside proving it.