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I'm trying to get my head around what a cograph of an n-functor is. We (some n-Lab people) are discussing it here. As a start, I'd be happy to understand what the cograph of a 0-functor, i.e. function between sets, is.


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2 Answers 2

In formal terms, the graph of a map $f: X \to Y$ is just the map $$(1, f): X \to X \times Y.$$ This makes sense in any category with products, denoted $\times$. Of course, one has a certain mental picture, at least in the category of sets: I'm thinking of an $X$-axis, and a $Y$-axis, and the graph as a curve in the absolutely ordinary, school-mathematics way. I'm also visualizing the way in which each point on the $X$-axis has assigned to it a point on the curve, namely, the one directly above it. This assignment is the map $(1, f)$ itself. The image of this map is the curve.

In formal terms, the cograph of a map $f: X \to Y$ is just the map $$[f, 1]: X + Y \to Y.$$ This makes sense in any category with sums (coproducts), denoted $+$. (Usually one would write a column vector instead of $[f, 1]$, but I don't want to figure out how to typeset that, and I hope you know what I mean.) This corresponds to a different mental picture of a map (again sticking to maps of sets): it's the one where you draw the set $X$ on the left, as a bunch of dots in a circle, the set $Y$ on the right, as another bunch of dots in a circle, and arrows going from the various points of $X$ to their images in $Y$.

This is discussed in Lawvere and Rosebrugh's book Sets for Mathematics. If you follow the link and download the sample chapter, you'll find a picture of the kind I mean, and the word "cograph", on page 2.

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Is this `picture of a map' also the category of elements if we view $X\to Y$ as a functor $[1]\to Sets$? –  Ma Ming Mar 24 '14 at 23:26
Yes, that's exactly what it is. –  Tom Leinster Mar 26 '14 at 0:48

I believe the usual notion of cograph for a function f : A → B between sets is the quotient of A ∐ B by the relation f(a) ~ b, together with its inclusions from A and B. From a homotopy theoretic point of view, you could also call it the mapping cylinder of f. As discussed in your link, rather than identify f(a) and b (or equivalently glue in a 1-cell between f(a) and b) you could insert a non-invertible 1-morphism to obtain a category. So there are constructions which yield each of a 0-category and a 1-category. However, in the n-categorical (or (∞,n)-categorical) case it seems that you always just get a n-category, never an (n+1)-category.

I think what's going on is the fact that you're looking at a diagram of sets, or more generally n-categories, indexed by a 1-category {• → •}. The result is a max(n,1)-category. If you consider a natural transformation between two functors between 1-categories, maybe you can define a cograph which is naturally a 2-category.

(And by the way, I have no idea what is meant on cograph of a functor by "The cograph 2-pullback [diagram] is computed by the ordinary pullback [diagram]", although the latter diagram makes sense and the rest of the page is fine. Oh I see, "2-pullback" should be "comma object".)

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