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.
$\begingroup$
$\endgroup$
5
-
$\begingroup$ And what does $\hat{A}$ mean? $\endgroup$– David CarchediCommented Apr 8, 2011 at 2:11
-
1$\begingroup$ 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 ... $\endgroup$– Martin BrandenburgCommented Apr 8, 2011 at 8:32
-
$\begingroup$ ... therfore certainly a good exercise for starters in category theory. $\endgroup$– Martin BrandenburgCommented Apr 8, 2011 at 8:33
-
$\begingroup$ I appreciate this comment, I was looking for such a proof, now I know I should stop looking. $\endgroup$– Spice the BirdCommented Apr 11, 2011 at 5:06
-
1$\begingroup$ 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. $\endgroup$– Mike ShulmanCommented Apr 14, 2011 at 20:35
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
3
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.
answered Apr 8, 2011 at 2:21
-
$\begingroup$ The hat does refer to the presheaves, the down arrow is the comma category. $\endgroup$ Commented Apr 8, 2011 at 2:32
-
1$\begingroup$ In my arXiv copy of the book, the fact is Prop. 1.1.7 on page 9, and there is no p. 394. $\endgroup$ Commented Apr 14, 2011 at 20:13
-
$\begingroup$ 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. $\endgroup$ Commented Apr 14, 2011 at 21:30
$\begingroup$
$\endgroup$
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
$(\widehat{A}\downarrow X) \simeq (\mathrm{DFib}(A)\downarrow B) \simeq \mathrm{DFib}(B) \simeq \widehat{B} = \widehat{A \downarrow X}$
See also this question.
answered Apr 14, 2011 at 20:27
$\begingroup$
$\endgroup$
1
I'm guessing $\widehat{A}$ denotes the presheaf category? If so, see Johstone's "Sketches of an Elephant" page 8.
-
$\begingroup$ Thank you both for your answers, I will have a look at both references. $\endgroup$ Commented Apr 8, 2011 at 2:42