Up to this point I always just assumed a functor belonged to its domain category. It think this is because, categories are not closed under composition, and functors are basically "the way out" of the category.

But reading about diagrams I come across statements like "a diagram $F$ of type $C$ in $D$", where $F: C \to D$. So in the description of the diagram, it says "in $D$", so it seems $F \in D$.

I know that this is partly because of the viewpoint of having objects and morphisms of $C$ portrayed in $D$, but still.

Reading it up on Wikipedia, I noticed that they never mention it, either.

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    $\begingroup$ Compare with the description of a continuous function $[0,1] \to \mathbb{R}^2$ as a "path in the plane". $\endgroup$ – user14972 Nov 28 '17 at 20:39

Would you say that the function $f: X \to Y$ is an element of the set $X$? No.

Functors, in general, do not "live" anywhere. (I am fully aware that large category proponents disagree with me.) If $C$ is small, then you can make a category out of all functors $F: C \to D$, where the functors are the objects and the natural transformations the morphisms. This is a category because the hom-sets are actual sets due to the smallness of $C$. Some people don't care about this and would consider the same construction a category even when $C$ isn't small.

At any rate, a functor $F: C \to D$ is not a thing in either $C$ or $D$. It is a transformation of the categories, not a resident of either.

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  • $\begingroup$ However, I wouldn't say $g: X \to X$ belonged to $X$ either, because sets, generally, do not contain functions, while categories indeed do contain morphisms. $\endgroup$ – hgiesel Nov 28 '17 at 20:11
  • $\begingroup$ Get concrete if you need to. Take $C$ and $D$ to be the category of groups. Fix a group $A$, and define a functor $F: C \to D$ by $F(X) = A \times X$. Is $F$ a group? $\endgroup$ – Randall Nov 28 '17 at 20:13

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