Let $X\colon \mathsf{C^{op}}\to \mathsf{Set}$ be a presheaf. It's category of elements, denoted by $\int X$, has pairs $(a,s)$ with $s \in X(a)$ as objects and $f \in \mathrm{Hom}_{\mathsf{D}}(a,b)$ such that $X(f)(t) = s$ as morphisms $(a,s)\to (b,t)$.

This is a theorem from the book Higher Categories and Homotopical Algebra by D.C.Cisinski.

Let $\mathsf{A}$ be a small category, together with a locally small category $\mathsf{C}$ which admits small colimits. For any functor $u\colon \mathsf{A}\to \mathsf{C}$, the functor of evaluation at $u$ $$u^*\colon \mathsf{C}\to\widehat{\mathsf{A}}, Y \mapsto u^*(Y) = (a\mapsto\mathrm{Hom}_{\mathsf{C}}(u(a),Y))$$ has a left adjoint $$u_{!}\colon\widehat{\mathsf{A}}\to\mathsf{C}.$$ Moreover, there is a natural isomorphism $u(a) \cong u_{!}(\mathrm{Hom}_{\mathsf{D}}(-,a)), a \in \mathsf{A}$, such that, for any object $Y$ of $\mathsf{C}$, the induced bijection $\mathrm{Hom}_{\mathsf{C}}(u_{!}(\mathrm{Hom}_{\mathsf{D}}(-,a)),Y) \cong \mathrm{Hom}_{\mathsf{C}}(u(a),Y)$ is the inverse of the composition of the Yoneda bijection $\mathrm{Hom}_{\mathsf{C}}(u(a),Y) = u^*(Y)_a = \mathrm{Hom}_{\widehat{\mathsf{A}}}(\mathrm{Hom}_{\mathsf{D}}(-,a),u^*(Y))$ with the adjunction formula $\mathrm{Hom}_{\widehat{\mathsf{A}}}(\mathrm{Hom}_{\mathsf{D}}(-,a),u^*(Y)) \cong \mathrm{Hom}_{\mathsf{C}}(u_{!}(\mathrm{Hom}_{\mathsf{D}}(-,a)),Y)$.

Cisinski constructs a left adjoint functor $u_!$ by setting, for each presheaf $X$ over $\mathsf{A}$, $u_!(X)$ to be a colimit of the functor $F\colon\int X\to \mathsf{C}$ such that $F(a,s) = u(a)$ (I assume that $F(f\colon (a,s)\to (b,t)) = u(f)$, but the author doesn't explicitly state this, so it is possible I'm mistaken).

It is also stated what is the action of $u_!$ on morphisms of presheaves, and I can't guess it, though it is crucial to the rest of the proof as we need to prove naturality of an adjunction formula. This is my first question.

My second question is why $u(a) \cong u_!(\mathrm{Hom}_{\mathsf{D}}(-,a))$ is unique. Sure, the inverse of a bijection is unique, so the induced bijection $\mathrm{Hom}_{\mathsf{C}}(u_{!}(\mathrm{Hom}_{\mathsf{D}}(-,a)),Y) \cong \mathrm{Hom}_{\mathsf{C}}(u(a),Y)$ is unique, but this doesn't explain why $u(a) \cong u_{!}(\mathrm{Hom}_{\mathsf{D}}(-,a))$ is.


1 Answer 1


Suppose $\psi: X\to Y$ is a morphism of presheaves. Then for $(a,s)\in \int X$, $s\in X(a)$ so that $\psi_a(s) \in Y(a)$. Therefore you get an element $(a,\psi_a(s))\in \int Y$

Now this allows you to define a morphism $\mathrm{colim}_{\int X}F_X \to \mathrm{colim}_{\int Y}F_Y$ (where I write $F_X$ for what you called $F$) by defining $F_X((a,s)) \to F_Y((a,\psi_a(s)))\to \mathrm{colim}_{\int Y}F_Y$ (where the first map is $id_a$ and the second is the inclusion given with the colimit)

That this system of maps is coherent and gives you a map $\mathrm{colim}_{\int X}F_X\to \mathrm{colim}_{\int Y}F_Y$ comes from the fact that if $f: a\to b$ is a morphism, then

$$\require{AMScd}\begin{CD}X(a) @>X(f)>> X(b)\\ @V\psi_aVV @V\psi_bVV\\ Y(a) @>Y(f)>>Y(b) \end{CD}$$

commutes, so that if $X(f)(s) = t$, then $Y(f)(\psi_a(s)) = \psi_b(t)$, so we have a(n obviously commutative) diagram $$\begin{CD} F_X((a,s)) @>>> F_Y((a,\psi_a(s))) @>>> \mathrm{colim}_{\int Y}F_Y \\ @VVV @VVV @VVV\\ F_X((b,t)) @>>> F_Y((b,\psi_b(t))) @>>> \mathrm{colim}_{\int Y}F_Y\end{CD}$$

for each map $f : (a,s)\to (b,t)$ in $\int X$.

This defines the action of $u_!$ on morphisms.

(If you know about co-ends [if you don't, don't read this parenthesis, go below to see the answer to the second question], then $u_!(X) = \int^{a\in A}X(a)\cdot u(a)$ where $X\cdot c = \coprod_X c$ for a set $X$ and an object of $C$ $c$. Then the map $\int^{a\in A}X(a)\cdot u(a) \to \int^{a\in A}Y(a)\cdot u(a)$ is simply induced by $\psi_a\cdot u(a) : X(a)\cdot u(a) \to Y(a)\cdot u(a)$, where one checks similarly as above that this yields a map on co-ends)

For the second question, the Yoneda lemma tells you in particular that $\hom(a,b)\to \hom(\hom(a,-), \hom(b,-))$ is a bijection. So if you have a specific isomorphism $\in \hom(\hom(a,-), \hom(b,-))$ (say "canonical") it gives you a specific isomorphism $\in \hom(a,b)$

Here you have canonical isomorphisms (natural in $a$) $\hom_C(u(a), Y) \cong \hom_C(u_!\hom_D(-,a), Y)$ so, by the Yoneda lemma applied to $C$, they come from specific isomorphisms $u(a) \cong u_!\hom_D(-,a)$ which will also be natural in $a$

  • $\begingroup$ Dear Max, could you expand on why $\eta\colon\mathsf{Hom_D}(a,-)\Rightarrow\mathsf{Hom_D}(b,-)$ being a natural isomorphism implies that $y(\eta) = \eta_a(\mathsf{1}_a)$ is an isomorphism $a\to b$ in $\mathsf{D}$? In the context of my question, we can construct the isomorphism explicitly and then use your approach to prove uniqueness, but it seems you are implying its existence follows directly from Yoneda in a general case. $\endgroup$
    – Jxt921
    Nov 8, 2019 at 20:42
  • 1
    $\begingroup$ It will be an iso $b\to a$, not $a\to b$. But it sinply follows from the Yoneda lemma, it's one of the standard consequences. Indeed, the Yoneda lemma implies in particular that the Yoneda embedding is an embedding, i.e. it is fully faithful; and from this it follows that isomorphisms between the representables come from isomorphisms between the representing objects $\endgroup$ Nov 8, 2019 at 21:03

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