First of all, is it true?

Every presheaf $F:\mathcal{C}^{\text{op}}\to\mathbf{Set}$ can be written as a colimit of representable functors. It can be expressed as a coend, $$\int^{A\in\mathcal{C}} F(A)\times \hom_{\mathcal{C}}(-,A).$$ Can we write a copresheaf $G:\mathcal{D}\to\mathbf{Set}$, similarly, as an end?

  • $\begingroup$ You surely can: see Prop 2.1 here arxiv.org/abs/1501.02503 $\endgroup$ – Fosco Sep 12 '16 at 13:32
  • $\begingroup$ Also, ibid. gives a formal proof of the statement, so that you don't (co)end up (co)vered in dirt having touched a quotient of sets :) $\endgroup$ – Fosco Sep 12 '16 at 13:34
  • $\begingroup$ @FoscoLoregian Thanks! I really like those notes of yours. However, I can not find the notation $Hc^{\mathbf{C}(c,-)}$, appearing inside the end defined anywhere (unless I did not look properly). Do you mean some kind of exponential object? $\endgroup$ – P.P.21 Sep 13 '16 at 8:31
  • $\begingroup$ Oh, I guess you mean a coproduct of the copresheaf $H$ with co-Yoneda, right? $\endgroup$ – P.P.21 Sep 13 '16 at 8:51

The answer to the first question is yes.

You can easily observe that for each object $X \in \mathcal C$ we have the mappings

$$i_1^X \colon \int^{A \in \mathcal C} F(A) \times \hom_{\mathcal C}(X,A) \longrightarrow F(X)$$ $$[(x,\alpha)] \stackrel{i_1^X}{\mapsto} F(\alpha)(x)$$ $$i_2^X \colon F(X) \longrightarrow \int^{A \in \mathcal C} F(A) \times \hom_{\mathcal C}(X,A)$$ $$x \stackrel{i_2^X}{\mapsto} [(x,1_X)]$$ (where by $[(x,\alpha)]$, with $x \in F(A)$ and $\alpha \in \hom_{\mathcal C}(X,A)$, I denote the equivalence class of the pair $(x,\alpha) \in F(A) \times \hom_{\mathcal C}(X,A)$ under the equivalence relation defining the coend).

It is a matter of computation to prove that these mappings are well defined and natural in $X$.

Also a matter computation, but more interesting, these maps are inverse to each other: $$i_2(i_1([x,\alpha]))=i_2(F(\alpha)(x))=[(F(\alpha)x,1_X)]=[x,\alpha]$$ (this equality follows by the properties of coends) $$i_1(i_2(x))=i_1([x,1_X])=F(1_x)(x)=1_{F(X)}(x)=x\ .$$

So the $i_1$ and $i_2$ give natural isomorphisms between the functors $F$ and $\int^{A \in \mathcal C}F(A) \times \hom_{\mathcal C}(-,A)$.

The answer to the second question is yes too. is no, to be more exact what it holds is that $G \cong \int^{A \in \mathcal D}G(A)\times \hom_{\mathcal D}(A,-)$, where the functor on the right is still a coend.

You can regard a functor $G \colon \mathcal D \to \mathbf{Set}$ as a presheaf $G \colon \mathcal {D^{\text{op}}}^\text{op} \to \mathbf {Set}$, then you have that $$G \cong \int^{A \in \mathcal D^\text{op}} G(A) \times \hom_{\mathcal D^\text{op}}(-,A) \cong \int^{A \in \mathcal D}G(A) \times \hom_{\mathcal D}(A,-)$$ where the isomorphisms are natural.

  • $\begingroup$ Ah, thanks! Just out of curiosity, what is the end form of the last integral? (I am struggling a bit to dualize properly) $\endgroup$ – P.P.21 Sep 12 '16 at 9:15
  • $\begingroup$ @P.P.21 is not an end, it is coend. The result is obtained just by duality on the category $\mathcal D$ (we are not dualizing the category $\mathbf{Set}$, that is the reason why we still end up with a coend instead of an end). Apologize for the confusion I misread the question, now it should be correct. $\endgroup$ – Giorgio Mossa Sep 12 '16 at 9:30
  • $\begingroup$ Thanks again. However, I believe that there must be a way to write this as an end - after all a coend is really a colimit, and a colimit is equivalently a limit in the opposite category. I do not mind dualizing - but I can not see how to turn things around properly myself in the context of ends and coends. I am sorry if this does not make sense, I am just trying to familiarize myself with these notions, and I am struggling. $\endgroup$ – P.P.21 Sep 12 '16 at 9:38
  • $\begingroup$ @P.P.21 well of course if you dualize $\mathbf{Set}$, obtaining $\mathbf{Set}^\text{op}$ the coend/colimit above becomes an end/limit. So you get an end of $\mathbf{Set}^\text{op}$-valued functors, the point is that while we are interested in studying $\mathbf{Set}$-valued functors we do not really care about $\mathbf{Set}^\text{op}$-valued functors. $\endgroup$ – Giorgio Mossa Sep 12 '16 at 9:43
  • $\begingroup$ So it would be something like $\int_{A\in \mathcal{C}^{\text{op}}} G^{\text{op}}(A)\coprod\hom_{\mathcal{C}^{\text{op}}}(A,-)$? Or does the $\times$ not turn to a $\coprod$? I am so thankful for your patience... $\endgroup$ – P.P.21 Sep 12 '16 at 9:50

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