# Why are contravariant functors called contravariant?

I'm just now learning a bit of category theory, and there often seems to be a certain notion, like limits for instance, and if you inverse certain arrows, you obtain a co-object related to that notion (for instance colimits).

Why is it, then, that we speak of contravariant functors instead of co-functors?

• Tradition. ${}{}$ Mar 18, 2016 at 12:46
• Probably at least in part this is by analogy with the terminology of covariant and contravariant tensors in linear algebra, which predates category theory by about a century: The earliest citation in the O.E.D. for contravariant dates to 1853 (from the work of Sylvester, in fact). Mar 18, 2016 at 12:58
• It also makes sense linguistically. Variant as an adjective means "tending to change," so a covariant function is one which "changes with" while a contravariant functor is one which "changes against." Mar 18, 2016 at 13:00
• Thanks @Charlie . This makes a whole lot of sense. Mar 18, 2016 at 13:23
• @EricTowers Yet we're happy to talk about coconuts. I've never heard of a contranut. :-) Mar 18, 2016 at 21:03

The definition of a functor is self-dual. If you reverse all the arrows in the definition of a functor $\mathsf{C} \to \mathsf{D}$, what you get is a functor $\mathsf{C}^\mathrm{op} \to \mathsf{D}^\mathrm{op}$, which is exactly the same thing as a functor $\mathsf{C} \to \mathsf{D}$. So in this sense a cofunctor is just a functor.
Now a contravariant functor is a functor too, but between different categories: a contravariant functor $\mathsf{C} \to \mathsf{D}$ is the same thing as a functor $\mathsf{C}^\mathrm{op} \to \mathsf{D}$, which is exactly the same thing as a functor $\mathsf{C} \to \mathsf{D}^\mathrm{op}$. This is rather unrelated to functors $\mathsf{C} \to \mathsf{D}$.
• Shouldn’t a cofunctor rather go $D^{op}\to C^{op}$? Mar 18, 2016 at 16:44
• @EmilJeřábek no in fact. It's the morphisms of the categories which are turned around when switching to the dual; the arrow between the category is not considered a morphism at that level. (You could of course consider cofunctors on the functor category; but then you'd simply be dealing with morphisms (functors!) $D\to C$... as well as $C\to D$.) Mar 18, 2016 at 19:47
The dual functor of a functor $F: \mathbb{A} \rightarrow \mathbb{B}$ is the functor $F^\mathrm{op}: \mathbb{A}^\mathrm{op}\rightarrow \mathbb{B}^\mathrm{op}$ defined in the obvious way. So, for a duality principle, we would not define co-functors as contravariant functors.