# Are the categories, $\widehat{\mathcal{A}}\downarrow X$ and $\widehat{\mathcal{A}\downarrow X}$ equivalent.

The title is fairly self explanatory. Let $\mathcal{A}$ be a small category , $X$ be an object of $\widehat{\mathcal{A}}$. Then the question is, are the categories, $\widehat{\mathcal{A}\downarrow X}$, and $\widehat{A}\downarrow X$ equivalent? I am 99% sure that this is true. Furthermore, if their is a reference for this, that would be helpful.

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And what does $\hat{A}$ mean? – David Carchedi Apr 8 '11 at 2:11
I like this result, because it seems so trivial that one should expect an "abstract nonsense" proof, but there is none. You have to carefully construct the equivalence ... – Martin Brandenburg Apr 8 '11 at 8:32
... therfore certainly a good exercise for starters in category theory. – Martin Brandenburg Apr 8 '11 at 8:33
I appreciate this comment, I was looking for such a proof, now I know I should stop looking. – Spice the Bird Apr 11 '11 at 5:06
Usually, if an abstract-nonsense fact doesn't seem to have an abstract-nonsense proof, it means you haven't gotten abstract enough or nonsensical enough. (-: See my answer below. – Mike Shulman Apr 14 '11 at 20:35

By $\widehat{\mathcal{A}}$ you must mean the category of presheaves on $\mathcal{A}$. I take it that $X$ denotes a presheaf, and that $\mathcal{A}/X$ is the comma category $y_{\mathcal{A}} \downarrow X$ where $y_{\mathcal{A}}$ is the Yoneda embedding.

Another reference for this fact is Tom Leinster's book Higher Operads, Higher Categories, p. 394. I'm not sure who first discovered it, but it's ancient folklore.

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The hat does refer to the presheaves, the down arrow is the comma category. – Spice the Bird Apr 8 '11 at 2:32
In my arXiv copy of the book, the fact is Prop. 1.1.7 on page 9, and there is no p. 394. – Mike Shulman Apr 14 '11 at 20:13
Thanks, Mike. I got that page number from a quick googling which pointed to Tom's book on Google Books, p. 394 (this must be the bound CUP version). It mentions the result in passing and refers back to 1.1.7 as you say. – Todd Trimble Apr 14 '11 at 21:30

Actually, there is an abstract-nonsense proof (although that doesn't make constructing the equivalence by hand any less of a good exercise). It goes through the equivalence of presheaves on a category A with discrete fibrations over A, so $\widehat{A} \simeq \mathrm{DFib}(A) \subset (\mathrm{Cat}\downarrow A)$. Note that DFib(A) is a full subcategory of $\mathrm{Cat}\downarrow A$: any map between discrete fibrations is a map of fibrations.

Furthermore, a composite of discrete fibrations is a discrete fibration, and conversely any map between discrete fibrations is itself a discrete fibration. Thus, for any discrete fibration B → A, we have $\mathrm{DFib}(A)\downarrow B \simeq \mathrm{DFib}(B)$. Thus, if $B = A\downarrow X$ is the category of elements of a presheaf $X\colon A^{op} \to \mathrm{Set}$, then we have

$(\widehat{A}\downarrow X) \simeq (\mathrm{DFib}(A)\downarrow B) \simeq \mathrm{DFib}(B) \simeq \widehat{B} = \widehat{A \downarrow X}$

I'm guessing $\widehat{A}$ denotes the presheaf category? If so, see Johstone's "Sketches of an Elephant" page 8.