The Unapologetic Mathematician

Mathematics for the interested outsider

Arrow Categories

One very useful example of a category is the category of arrows of a given category \mathcal{C}.

We start with any category \mathcal{C} with objects {\rm Ob}(\mathcal{C}) and morphisms {\rm Mor}(\mathcal{C}). From this we build a new category called \mathcal{C}^\mathbf{2}, for reasons that I’ll explain later. The objects of \mathcal{C}^\mathbf{2} are just the morphisms of \mathcal{C}. The morphisms of this new category are where things start getting interesting.

Let’s take two objects of \mathcal{C}^\mathbf{2} — that is, two morphisms of \mathcal{C} — and lay them side-by-side:
Now we want something that transforms one into the other. What we’ll do is connect each of the objects on the left to the corresponding object on the right by an arrow:
and require that the resulting square commute: g\circ h=k\circ f as morphisms in \mathcal{C}. This is a morphism from f to g. Sometimes we’ll write (h,k):f\rightarrow g, and sometimes we’ll name the square and write \alpha:f\rightarrow g.

If we have three morphisms f, g, and h in \mathcal{C}, and commuting squares (k_1,k_2):f\rightarrow g and (k_3,k_4):g\rightarrow h then we can get a commuting square (k_3\circ k_1,k_4\circ k_2):f\rightarrow h. We check that this square commutes: h\circ k_3\circ k_1=k_4\circ g\circ k_1=k_4\circ k_2\circ f. This gives a composition of commuting squares. It’s easily checked that this is associative.

Given any morphism f:A\rightarrow B in \mathcal{C} we can just apply the identity arrows to each of A and B to get a commuting square (1_A,1_B) between f and itself. It is clear that this square serves as the identity arrow on the object f in \mathcal{C}^\mathbf{2}, completing our proof that arrows and commuting squares in \mathcal{C} do form a category.

May 23, 2007 - Posted by | Category theory


  1. Thanks for this — nice crisp explanation. Would be even more helpful if it had a couple of simple, finite examples. Posets maybe?

    Comment by Brian Rotman | January 28, 2010 | Reply

  2. Well, these sorts of things show up in all sorts of other categorical constructions I’ve discussed, although I haven’t explicitly linked back here. Look at categories of natural transformations, or of representations, or (coming eventually) bundles.

    Comment by John Armstrong | January 28, 2010 | Reply

  3. How do you show that $hom(f,g)$ and $hom(f’,g’)$ are disjoint?

    Comment by Tony | December 12, 2012 | Reply

    • Tony, this is probably a bit late for you but they are disjoint by construction. We take all the morphisms as objects and add arrows wherever there are commuting squares – at this point it’s just a graph – and as it turns out we can define associative composition of the arrows by the method given above and we get identity arrows out of it too, so it’s a category.

      Comment by Dominic Verdon | November 9, 2013 | Reply

  4. h is used for two different things here – as one of the arrows in the commutative diagram defining the arrow category and as an arrow in the original category (so, an object in the arrow category).

    Comment by hexbienium | September 1, 2014 | Reply

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