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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
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1 Answer

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.

Regardless, most often you start with categories and consider naturally arising functors. It is not so often that you concentrate only on the objects and just conjure some mapping and casually wonder 'can I extend it to a functor?'.

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