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Is there any one (or more) categories that doesn't have a functor? Functors go between categories, so is there any category that only has an identity functor but no other functor that maps it to another category?

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You can always define the opposite functor into the opposite category. If the category is symmetric then the opposite category can be identified with it in a natural way, but whether or not this makes it the same category depends on what you want. – Alex Becker May 20 '13 at 1:59
@Alex. I don't get it. What is your opposite functor? In general there is no interesting functor (by which I always mean covariant, since contravariant functors are in fact covariant functors in disguise) $C \to C^{op}$. – Martin Brandenburg May 20 '13 at 9:33
up vote 2 down vote accepted

Every category has at least 2 outgoing functors: the identity functor plus at least another one. Actually every category has infinitely many outgoing functors. You can easily see this by considering the cases of:

  • the empty category which goes to every other one (it is the initial category, after all)
  • the non empty categories which go trivially or non trivially to every other non empty and/or non terminal category

However the empty category has only one ingoing functor. The identity functor. Every other category has more than one ingoing functor

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A very strong no! For every two categories $C,D$ and an object $d\in ob(D)$ there is a functor $F:C\to D$ mapping every thing in $C$ to $d$ and $id_d$.

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mapping every thing in C to d or to D? – Ellen May 20 '13 at 2:09
every object to $d$ and every arrow to $id_d$. – Ittay Weiss May 20 '13 at 2:11
Of course, if $D$ is empty... – Zhen Lin May 20 '13 at 7:33
Who assumes that? :) For me, sets are discrete categories, including the empty set. – Berci May 20 '13 at 10:22
@Berci I'm just saying that above I said "... $D$ and an object $d\in ob(D)$". Of course I'm not claiming that the empty category does not exist. And as a side note, for me sets are just indiscrete categories. – Ittay Weiss May 20 '13 at 10:26

For every category $C$ there is a functor from $C$ to the category $1$ consisting of exactly one object and one arrow.

So you can map every category $C$ to the category $1$.

Conversely, $1$ can be embedded into any other category.

So every category "has a functor" other than identity functor.

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Of course, for one category this is the identity. – Alex Becker May 20 '13 at 2:03
Your answer is false if taken literally: $1$ does not embed into the empty category, and the category $1$ only has the identity endofunctor. – Zhen Lin May 20 '13 at 7:34

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