# Every slice of a presheaf category is again a presheaf category

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 $$$$\begin{split} \mathbf{PSh}(\mathbf{A}/A) \simeq \mathbf{PSh}(\mathbf{A})/H_A. \end{split}$$$$ To obtain a correspondence between some $$\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}$$ its projection functor (for the definition cf. Leinster's book linked above, on p. 155, Def. 6.2.16). Then Theorem 6.2.17 on p. 155 of the book reads $$$$\begin{split} X \simeq \lim_{\to\mathbf{E}(X)}H_{P(-)}. \end{split}$$$$ As a result, $$$$\mathbf{PSh}(\mathbf{A})/X \simeq \mathbf{PSh}(\mathbf{A})/\lim_{\to\mathbf{E}(X)}H_{P(-)}.$$$$ To finish the proof, one would need to show that the equivalence shown in the nlab-article is stable under limits; in other words that $$$$\text{(eq. (1)) }\qquad \mathbf{PSh}(\mathbf{A})/\lim_{\to\mathbf{E}(X)}H_{P(-)} \simeq \mathbf{PSh}\bigg(\mathbf{A}/\lim_{\to\mathbf{E}(X)}P(-) \bigg).$$$$ If one could do that, then one could conclude $$\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.

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.

• 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 $E_{f'}'$) Commented Apr 22, 2020 at 15:19
• @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.
– jgon
Commented Apr 22, 2020 at 17:58
• No problem, thanks for the reference, this is incidentally the next book, I'd like to read and thanks for having written the post. Commented Apr 23, 2020 at 17:13
• 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 Commented Apr 24, 2020 at 15:39