Let $S$ be a small category. Consider its presheaf category $\widehat{S} = [S^{\mathrm{op}},\mathsf{Set}]$. Is there a direct way to see that $\widehat{S}_{\mathrm{fp}}$ is essentially small, and that every object of $\widehat{S}$ is a filtered colimit of objects in $\widehat{S}_{\mathrm{fp}}$? This is true by the various characterizations of locally finitely presentable categories, but I wonder if there is a more direct way. I suspect that $\widehat{S}_{\mathrm{fp}}$ consists of those presheaves which are finite colimits of representable presheaves.

  • $\begingroup$ It's not an answer to your question. Let $\mathcal{G}$ be the full subcategory of finite colimits of representable presheaves. $\mathcal{G} \subset \widehat{S}_{\mathrm{fp}}$. The canonical diagram under $\mathcal{G}$ of a given object $F$ is filtered since $\mathcal{G}$ is closed under finite colimits. And the colimit of this diagram is $F$ since every presheaf is a colimit of representable presheaves. $\endgroup$ – Philippe Gaucher Jun 21 '17 at 11:29

Exercise: If you have finite coproducts and filtered colimits, then you have all coproducts.

Very standard result 1: If you have coequalizers and arbitrary coproducts, then you have all colimits.

Very standard result 2: Presheaves are colimits of representables.

Let $\mathcal{I}$ be a filtered category and $D : \mathcal{I}\to[S^{op},\mathbf{Set}]$. Similarly, let $\mathcal{J}$ be a finite category (i.e. a finite number of arrows) and $D' : \mathcal{J}\to[S^{op},\mathbf{Set}]$. I'll use (co)end syntax for (co)limits because it provides a nice binding syntax, but I'm not actually using any (proper) (co)ends.

  1. Representables are compact: $$\begin{align} \mathsf{Nat}\left(\mathsf{Hom}(-,X),\int^{I\in\mathcal{I}}D(I)\right) & \cong \left(\int^{I\in\mathcal{I}}D(I)\right)\!(X) \\ & \cong\int^{I\in\mathcal{I}}D(I)(X) \\ & \cong\int^{I\in\mathcal{I}}\mathsf{Nat}(\mathsf{Hom}(-,X),D(I)) \end{align}$$
  2. Finite colimits of compact objects are compact: $$\begin{align} \mathsf{Nat}(\int^{J\in\mathcal{J}}D'(J),\int^{I\in\mathcal{I}}D(I)) & \cong \int_{J\in\mathcal{J}}\mathsf{Nat}\left(D'(J),\int^{I\in\mathcal{I}}D(I)\right) \\ & \cong \int_{J\in\mathcal{J}}\int^{I\in\mathcal{I}}\mathsf{Nat}(D'(J),D(I)) \\ & \cong \int^{I\in\mathcal{I}}\int_{J\in\mathcal{J}}\mathsf{Nat}(D'(J),D(I)) \\ & \cong \int^{I\in\mathcal{I}}\mathsf{Nat}(\int^{J\in\mathcal{J}}D'(J),D(I)) \\ \end{align}$$

Thus $\widehat{S}_{fp}$ contains at least the representables and finite colimits of them, and thus coequalizers and finite coproducts. We can get arbitrary colimits, and thus all presheaves, via a filtered colimit of finite coproducts and coequalizers.

  • $\begingroup$ Thank you. We don't have to use coequalizers and coproducts, right? If $X : I \to \mathcal{C}$ is a diagram in a cocomplete category, consider the partial order $\mathcal{F}$ of finite full subcategories of $I$. This is filtered. Then $\mathrm{colim}(X) = \mathrm{colim}_{U \in \mathcal{F}} \,\mathrm{colim}(X|_U)$. $\endgroup$ – HeinrichD Jun 22 '17 at 5:58
  • $\begingroup$ @HeinrichD Yes, that works. The logic is basically the same as showing arbitrary coproducts exist given finite products. It's just a bit more intricate since you aren't just talking about finite subsets of a set. $\endgroup$ – Derek Elkins Jun 22 '17 at 6:15

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