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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?

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    $\begingroup$ Tradition. ${}{}$ $\endgroup$ Mar 18, 2016 at 12:46
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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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.

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  • $\begingroup$ Interesting answer. $\endgroup$ Mar 18, 2016 at 13:09
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    $\begingroup$ 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
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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.

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