I found the following statement in exercise 6.2.24 in Leinster's "Basic category theory" book (freely available for download here: https://arxiv.org/abs/1612.09375):

Given a small category $\mathbf{A}$ and a presheaf $X$ on $\mathbf{A}$, the slice category $\mathbf{PSh}(\mathbf{A})/X=[\mathbf{A}^{\text{op}},\mathbf{Set}]/X$ is equivalent to $\mathbf{PSh}(\mathbf{B})=[\mathbf{B}^{\text{op}}, \mathbf{Set}]$ for some small category $\mathbf{B}$.

I tried to prove it but there is at least one step with which I have some difficulty.

My attempt so far: One can at first assume that $X$ is representable, i.e. $X=H_A$ for some $A\in\mathbf{A}$ (where $H_A(B):=\text{Hom}_{\mathbf{Set}}(B,A)$. One could also write $H(A)$ if $H:\mathbf{A}\rightarrow [\mathbf{A}^{\text{op}},\mathbf{Set}]$ denotes the Yoneda embedding).

According to Proposition 3.2. of this nlab-article about over-categories, there is an equivalence of categories \begin{equation} \begin{split} \mathbf{PSh}(\mathbf{A}/A) \simeq \mathbf{PSh}(\mathbf{A})/H_A. \end{split} \end{equation} To obtain a correspondence between some $\mathbf{PSh}(\mathbf{B})$ and $\mathbf{PSh}(\mathbf{A})/X$, where $X$ is now any presheaf, one could use the fact that every presheaf can be written as a colimit of representables. To this end, let $\mathbf{E}(X)$ be the category of elements of $X$ and $P:\mathbf{E}(X)\to \mathbf{A}$ its projection functor (for the definition cf. Leinster's book linked above, on p. 155, Def. 6.2.16). Then Theorem 6.2.17 on p. 155 of the book reads \begin{equation} \begin{split} X \simeq \lim_{\to\mathbf{E}(X)}H_{P(-)}. \end{split} \end{equation} As a result, \begin{equation} \mathbf{PSh}(\mathbf{A})/X \simeq \mathbf{PSh}(\mathbf{A})/\lim_{\to\mathbf{E}(X)}H_{P(-)}. \end{equation} To finish the proof, one would need to show that the equivalence shown in the nlab-article is stable under limits; in other words that \begin{equation} \text{(eq. (1)) }\qquad \mathbf{PSh}(\mathbf{A})/\lim_{\to\mathbf{E}(X)}H_{P(-)} \simeq \mathbf{PSh}\bigg(\mathbf{A}/\lim_{\to\mathbf{E}(X)}P(-) \bigg). \end{equation} If one could do that, then one could conclude $\mathbf{PSh}(\mathbf{A})/X \simeq \mathbf{PSh}(\mathbf{B})$ for $\mathbf{B}:=\mathbf{A}/\lim_{\to\mathbf{E}(X)}P(-)$.

However, I do not know how to show that (eq. (1)) holds. Many thanks.


1 Answer 1


The trick is that there's no need to take a colimit. The base category can just be taken to be the category of elements.

Here's how it goes:

Suppose I have a map of presheaves on a category $\newcommand\A{\mathcal{A}}\A$, $f: E\to X$.

Then if $e\in E(b)$, $f(e) \in X(b)$, and if $g:a\to b$ is a morphism in $\A$, we know $f(e|_g)=f(e)|_g$. In other words, $f:E\to X$ is equivalent to a presheaf on the category of elements of $X$.

Let me break down how that works. We start with an object in the slice category $f:E\to X$. Let $(a,u)$ be an object in the category of elements of $X$, so $a\in\A$ and $u\in X(a)$. Let $g:a\to b$ be a morphism $(a,u)\to (b,v)$ with $v|_g = u$ in the category of elements of $X$.

Then we define $$E_f(a,u) = f_a^{-1}(\{u\}) = \{e\in E(a) : f(e) = u\}.$$ Then the observation above says that if $g:(a,u)\to (b,v)$, and $e\in E_f(b,v)$ (meaning $f(e)=v$), then $f(e|_g) = f(e)|_g = v|_g=u$, so $e|_g\in E_f(a,u)$. Therefore the presheaf structure on $E$ induces a presheaf structure on $E_f$.

Similarly, a morphism $h:(f:E\to X)\to (f':E'\to X)$ in the slice category will induce a morphism $h : E_f \to E'_{f'}$.

Conversely, to go backwards, if $E$ is a presheaf on the category of elements of $X$, then define $$E(a) = \bigsqcup_{u\in X(a)} E(a,u),$$ and define $f$ by $f(u,e) = u$, for $(u,e)\in E(a)$, $u\in X(a)$, $e\in E(a,u)$.

It's not hard to check that these are inverse equivalence functors.

  • $\begingroup$ Thank you very much for this answer! I was at first a bit confused by the notation $f(e|_g)$ and $v|_g$ but I think I understood now that it means, if $e\in E(b)$, and $g:a\to b$, then $e|_g:=E(g)(e)$. Then $f(e|_g)=f(e)|_g$ more extensively means $f_a(E(g)(e))=X(g)(f_B(e))$, is that correct? In this case, I think, the element $e$ at the very beginning should actually be in $E(b)$ and $f(e)=f_b(e)\in X(b)$ (because $E$ and $X$ are contravariant and if $g:a\to b$, then $E(g):E(b)\to E(a)$), right? (There is also a typo in the line "will induce a morphism $h:E_f\to E_{F'}'$, should be $E_{f'}'$) $\endgroup$
    – exchange
    Commented Apr 22, 2020 at 15:19
  • $\begingroup$ @exchange Yes, sorry for not explaining that. I got the notation from MacLane and Moerdijks' Sheaves in Geometry and Logic. Thanks for catching all my typos, I don't know what I was doing the day I wrote this, wow. $\endgroup$
    – jgon
    Commented Apr 22, 2020 at 17:58
  • $\begingroup$ No problem, thanks for the reference, this is incidentally the next book, I'd like to read and thanks for having written the post. $\endgroup$
    – exchange
    Commented Apr 23, 2020 at 17:13
  • $\begingroup$ I actually found out that it is indeed possible to prove eq. (1) in my question above, using the correspondence with the category of elements of your answer. Basically, $\mathbf{A}/\lim_{\rightarrow\mathbf{E}(X)}P(-)$ is equivalent to the category of elements. Together with your proof, this establishes eq. (1). See also Proposition 2.3 and the remark above of this nlab-article: ncatlab.org/nlab/show/category+of+presheaves $\endgroup$
    – exchange
    Commented Apr 24, 2020 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .