I'm interested in the fact that every presheaf on a category $C$ is a colimit of representable functors. The nLab claims that this can be stated as: $$F \simeq \int^{c \in C} Y(c) \otimes F(c)$$ Where $F: C^{op} \to \mathbf{Set}$ and $Y: C \to [C^{op}, \mathbf{Set}]$ is the Yoneda embedding. This can reasonably be called the co-Yoneda Lemma.

The fact that every presheaf can be expressed as a colimit of representables can be proven by shown that: $$F \simeq \mathrm{colim}\left(\left(\int F\right)^{op} \to C \xrightarrow{Y} [C^{op}, \mathbf{Set}] \right)$$ Here $\int F$ is the category of elements. This is proven for example in Mac Lane, Moerdijk - Sheaves in geometry and logic. My question is: Is there an easy proof for $$\int^{c \in C} Y(c) \otimes F(c) \simeq \mathrm{colim}\left(\left(\int F\right)^{op} \to C \xrightarrow{Y} [C^{op}, \mathbf{Set}] \right)$$ The co-Yoneda Lemma and the statement about presheaves seem to be linked by that fact, and the literature suggests (by interchanging the results without mentioning the isomorphism) that this is somehow easy to see, but the only proof I know is by showing the first two identities.


2 Answers 2


Let: $$\Gamma = \operatorname{colim} \left\{ \left(\int F\right)^{op} \to C \to [C^{op}, \mathbf{Set}] \right\}.$$

Then a morphism $u : \Gamma \to Z$ is the same thing as a collection of morphisms: $$u_{c,x} : Y(c) \to Z$$ for all $c \in C$ and $x \in F(c)$, such that if $\varphi : d \to c$ is a morphism in $C$, then the obvious commutative diagram commutes. More precisely, we should have: $$u_{d, F(\varphi)(x)} = u_{c,x} \circ Y(\varphi). \tag{1}$$

This collection is the same thing as a collection $U_c : Y(c) \times F(c) \to Z$ for $c \in C$, defined by: $$U_c(\xi, x) = u_{c,x}(\xi)$$ for all $c \in C$, $x \in F(c)$ and $\xi \in Y(c)$. If you inspect what the equation $(1)$ means for $U$, you will see that it exactly means that $U$ can factor through the coequalizer that usually defines the coend: $$\int^{c \in C} Y(c) \times F(c) = \operatorname{coeq} \left\{ \bigsqcup_{c,d} Y(c) \times F(d) \rightrightarrows \bigsqcup_c Y(c) \times F(c) \right\}$$ And we can check that this defines a bijection on the level of $\hom$-sets (the reverse construction, given some $U$, simply sets $u_{c,x}(\xi) = U_c(\xi,x)$): $$\hom(\Gamma, Z) \cong \hom(\int^c Y(c) \times F(c), Z).$$ I'll let you check that it's natural in $Z$ (simple exercise in definitions), so that finally: $$\Gamma \cong \int^c Y(c) \times F(c).$$

  • $\begingroup$ This probably satisfies the 'easy' requirement. I'm still not sure if I can agree with this isomorphism being used on the nLab without being mentioned (at least I guess that's what they did), but I guess I just have to learn more. Thanks! $\endgroup$ Nov 13, 2014 at 22:15
  • $\begingroup$ should those $f_{c,x}$ be $u_{c,x}$? I can't edit them. $\endgroup$ Jul 28, 2020 at 21:07
  • $\begingroup$ @EmilioMinichiello Yes, thanks. $\endgroup$ Aug 19, 2020 at 9:05

Here is a quick proof of the formula $$F \cong \int^{c \in C} Y(c) \otimes F(c):$$

$$\hom(\int^{c \in C} Y(c) \otimes F(c),G) \cong \int_{c \in C} \hom(Y(c) \otimes F(c),G)$$ $$\cong \int_{c \in C} \hom(F(c),\hom(Y(c),G)) \cong \int_{c \in C} \hom(F(c),G(c))\cong \hom(F,G).$$

  • $\begingroup$ The downvoter may perhaps explain how I can improve my answer. $\endgroup$ Nov 13, 2014 at 16:57
  • 3
    $\begingroup$ You could improve your answer by addressing the question posed by the OP. $\endgroup$
    – user314
    Nov 13, 2014 at 19:30
  • $\begingroup$ This seems fine to me. $\endgroup$ Nov 16, 2014 at 6:05

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