# Which mappings are functors?

Consider a (somehow naturally) specified mapping from one class of structures into another (or the same) class of structures, e.g. the mapping of a

Is there a somehow standarized way to detect whether such a mapping can be made a functor?

A prerequisite is to equip both sides (classes) of the mapping with morphisms to make them categories. If one has achieved this - usually in one of many possible ways - one has to look for specific mappings of morphisms in the source category to morphisms in the target category that fulfill the functor conditions.

Given such morphisms and a mapping of morphisms one may be able to check whether they fulfill the functor conditions, but - given only the mapping between objects -

how to find appropriate morphisms and their mapping, resp. how to show that you cannot find them?

Only some of the examples above give rise to functors, but I am especially insecure about the last two examples, so I'd like to focus the question:

How do you show that a mapping between structures can not be made a functor?

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An example is in math.stackexchange.com/questions/158438 -- find objects whose morphisms are very restricted – Jack Schmidt Aug 13 '12 at 23:48
@Jack: Thanks for the example, I wouldn't have found it without your help! – Hans Stricker Aug 13 '12 at 23:52
The automorphism group "assignment" is almost never a functor. – Zhen Lin Aug 14 '12 at 2:05
@Zhen Lin: Does that mean, that the relation between a graph and its automorphism group cannot be treated categorically? – Hans Stricker Aug 14 '12 at 8:36
If you delete all non-invertible morphisms in the category you start with, then you get a functor. Otherwise what are you going to do with the morphisms? – Zhen Lin Aug 14 '12 at 11:11

What you specified above is a bunch of mappings from some class to another. Let's denote one such by $f:C\to D$. Now, that mapping $f$ can always be extended to a functor. In fact in two (equally boring) ways. It all depends on how you want to consider the classes $C$ and $D$ as the objects of a category. One way is as the objects of a discrete category: the only arrows are the identity arrows. Another way is as the objects of an indiscrete category: for all objects $x,y$ there is precisely one arrow $x\to y$.
If you equip $C$ and $D$, both, with either the discrete or the indiscrete category structure then $f$ trivially extends (uniquely) to a functor. Of course, things get more interesting when your class of objects comes with a naturally defined notion of morphisms. Then the question whether or not $f$ extends to a functor is less trivial. But there is no hope for an all encompassing answer to that due to the generality of the situation.