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?

  • 4
    $\begingroup$ Tradition. ${}{}$ $\endgroup$ Mar 18, 2016 at 12:46
  • 4
    $\begingroup$ 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). $\endgroup$ Mar 18, 2016 at 12:58
  • 3
    $\begingroup$ 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." $\endgroup$
    – Charlie
    Mar 18, 2016 at 13:00
  • $\begingroup$ Thanks @Charlie . This makes a whole lot of sense. $\endgroup$
    – Cloudscape
    Mar 18, 2016 at 13:23
  • 2
    $\begingroup$ @EricTowers Yet we're happy to talk about coconuts. I've never heard of a contranut. :-) $\endgroup$ Mar 18, 2016 at 21:03

2 Answers 2


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 $\mathsf{C} \to \mathsf{D}$.

Note that some people do use the words "cofunctor" for contravariant functors, I guess it's just a matter of taste as long as every word is defined.

  • $\begingroup$ Interesting answer. $\endgroup$ Mar 18, 2016 at 13:09
  • 1
    $\begingroup$ Worth to add that e.g. a comonad is not the same a monad, despite the underlying functor/co-functor being the same. $\endgroup$ Mar 18, 2016 at 14:24
  • $\begingroup$ Shouldn’t a cofunctor rather go $D^{op}\to C^{op}$? $\endgroup$ Mar 18, 2016 at 16:44
  • $\begingroup$ @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 $C\to D$.) $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.