Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

To what extent do functors map "elements" of an object to "elements" of another object? They are usually described as just mapping one object to another.

Thanks

share|improve this question
    
What do you mean by 'element'? Are these concrete categories? –  Berci Nov 11 '12 at 13:45
1  
There are categories in practice where objects aren't just "sets with enriched structure." For those categories, there aren't really notions of "elements of an object"... Well, maybe one approach you can take is the functor of points approach: Choose an object $A$ in the category, and for each other object $X$, associate to $X$ the set of maps from $A$ to $X$. –  only Nov 11 '12 at 13:57
3  
Yes, the question is good, but it is not so easy because of the interpretation of 'elements'. Also, there is a concept of 'generalized element' of an object $X$ in a category, and that is none other than an arrow $A\to X$. Of course, in this sense, generalized elements are nicely preserved. –  Berci Nov 11 '12 at 13:59
1  
@ZhenLin: I certainly don't mean to say that I can define a (set-valued) functor on the essential image of a concretization functor. Of course this would depend on remembering how your sets came from objects in each category! All I mean is that if $U_1:\mathcal{C}_1 \rightarrow \mbox{Set}$ and $U_2:\mathcal{C}_2 \rightarrow \mbox{Set}$ are two "naturally-appearing" concrete categories and we have a "naturally-appearing" functor $F:\mathcal{C}_1 \rightarrow \mathcal{C}_2$, then chances are good that we will have functorial-in-$\mathcal{C}_1$ functions $f_X:U_1(X) \rightarrow U_2(FX)$. –  Aaron Mazel-Gee Nov 11 '12 at 16:37
1  
I don't agree. The reason why that claim sounds reasonable is because many functors of interest are part of an adjunction, in which case there is indeed such a natural transformation $U_1 \Rightarrow U_2 F$. But there are also important functors where there is no such natural transformation: consider, for example, the homology functors on the category of chain complexes, or more generally, derived functors... –  Zhen Lin Nov 11 '12 at 17:13

1 Answer 1

In a concrete category, the definition of concrete category tells you the sense. Even though a faithful functor need not be injective on objects, you can view any morphism as a function between sets by passing to the image of the faithful functor to $\mathrm{Set}$.

Even though your comment mentioned concrete categories, it's important to note that if your category is not concrete, then even when you have a decent way of viewing objects as sets, the morphisms may not be describable in terms of mapping elements. The homotopy category of topological spaces has topological spaces as objects (which have underlying sets), but the morphisms are merely homotopy equivalence classes of continuous functions between underlying sets. These morphisms can't be viewed as mapping particular elements of the underlying sets because different members of the equivalence class would do different things to elements.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.