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Suppose $\mathscr{C} \underset{G}{\overset{F}{\rightrightarrows}} \mathsf{Set}$ are such that $\operatorname{El}F$ and $\operatorname{El}G$ have initial objects which lie in fibers over the same object $c \in \operatorname{Ob}\mathscr{C}$. Then $F$ and $G$ are naturally isomorphic.

My Question: What are some other ways that relationships between $\operatorname{El}F$ and $\operatorname{El}G$ imply relationships between $F$ and $G$?


Some Ideas: Ncatlab points out that the functor $$\mathsf{Set}^{\mathscr{C}} \xrightarrow{\text{El}} \mathsf{Cat} \\ F \mapsto \operatorname{El}F$$ has a right adjoint $$\mathsf{Cat} \to \mathsf{Set}^\mathscr{C} \\ C \mapsto \operatorname{Hom}_{\mathsf{Cat}}(J-,C),$$ where $$J:\mathscr{C}^{\mathrm{op}} \to \mathsf{Cat} \\ c \mapsto c \downarrow \mathscr{C},$$ and so $\mathrm{El}$ is cocontinuous. I'm trying to use this adjunction to get some answers to my question.

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  • $\begingroup$ Is the fact you are using true for large categories that are not locally small? $\endgroup$ – Vladimir Sotirov Jul 28 '16 at 16:59
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The category of elements of $\mathcal Set\xrightarrow{F}\mathcal C$ is also a comma category $(\{*\}\downarrow F)$, i.e. a kind of weak pullback of $\mathbf 1\xrightarrow{\{*\}}\mathcal Set\xleftarrow{F}\mathcal C$, in the sense that we have pair of functors $\mathbf 1\leftarrow(\{*\}\downarrow F)\xrightarrow{\Pi_F}\mathcal C$ together with a natural transformation $\{*\}\overset\tau\Rightarrow F\Pi_F\colon (\{*\}\downarrow F)\to\mathcal Set$ so that any other natural transformation $\{*\}G\overset\phi\Rightarrow FH\colon\mathcal W\to\mathcal Set$ factors uniquely as a natural transformation $\{*\}J\overset{\tau J}\Rightarrow F\Pi_FJ$ for a functor $\mathcal W\xrightarrow{J}(\{*\}\downarrow F)$.

$$\require{AMScd}\begin{CD} (\{*\}\downarrow F) @>\Pi_F>>\mathcal C @= \mathcal C\\ @VVV \overset\tau\Rightarrow @VFVV \overset\alpha\Rightarrow @VGVV\\ \mathbf 1 @>\{*\}>>\mathcal Set @= \mathcal Set \end{CD}$$

Now, what is an initial object of $(\{*\}\downarrow F)$? It is literally an element $e\in Fc$ for some objet $c$ of $\mathcal C$ so that for any other element $e'\in Fc'$ there is a unique morphism $c\xrightarrow{f}c'\in\mathcal C$ for which $Ff(e)=e'$. Amazingly, this implies (!) that the a priori classes $\mathcal C(c,d)$ of morphisms $c\to d\in\mathcal C$ for various objects $d\in\mathcal C$ are in bijection with the class of elements of the sets $Fd$. In particular, $F$ is a representable functor, even though the category $\mathcal C$ itself is not necessarily locally small! I think the Yoneda lemma still works for representable functors in this sense, hence the natural isomorphism between two such functors $F$ and $G$ represented by the same object in $\mathcal C$. when $\mathcal C$ is locally small, and by the Yoneda lemma is unique up to isomorphism.


The functoriality in $F$ of the construction $(\{*\}\downarrow F)$ follows from the fact that whiskering is functorial: given a natural transformation $F\overset\alpha\Rightarrow G$, the functor $(\{*\}\downarrow F)\xrightarrow{A}(\{*\}\downarrow G)$ comes from the composite $\{*\}\overset{\tau_F}\Rightarrow F\Pi_F\overset{\alpha\Pi_F}\Rightarrow G\Pi_F$, which must factor uniquely as $\{*\}A\overset{\tau_GA}\Rightarrow G\Pi_GA$.

I don't know that the adjunction you're trying to work out will give something nice; it seems to me like it's one of those "formal" adjunctions that exist purely due to abstract size considerations (e.g. the categories are small so we can construct objects with the universal properties by taking limits over the whole categories).

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    $\begingroup$ this could have been a comment, but the comment box is too narrow. $\endgroup$ – Vladimir Sotirov Jul 21 '16 at 1:03
  • $\begingroup$ I think my question may have been unclear. I understand the quoted theorem, and its relation to the Yoneda Lemma. I am interested in further results along the same lines, i.e. relationships in categories of elements giving relationships between the functors. As for the adjunction, I am comfortable with it: I'm trying to use it to derive other results. $\endgroup$ – Eric Auld Jul 21 '16 at 1:50
  • $\begingroup$ Your question was clear, my comment may have been unclear. I just wanted to provide some interesting context, but the context was too long for a comment. $\endgroup$ – Vladimir Sotirov Jul 21 '16 at 1:56

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