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This question is probably trivial but I'm confused anyway. Say I have a function $f$ between two sets, then it is in the definition of a function that for one $x$ in $X$ the value $f(x)$ is unique. Hence a function is not well-defined if $x$ is mapped to two different values. The word "unique" is usually in the definition.
The definitions of a functor $F:C \to D$ between two categories $C$ and $D$ usually just say that an object $X$ in $C$ is assigned to an object $FX$ in $D$ and the same for morphisms. I guess its silly to allow an assignmnt where $FX$ is not unique right? What if $FX$ is unique up to (canonical) isomorphism or something like that?

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up vote 3 down vote accepted

A functor $\mathcal{C} \to \mathcal{D}$ by definition is a pair of functions $\operatorname{ob} \mathcal{C} \to \operatorname{ob} \mathcal{D}$, $\operatorname{mor} \mathcal{C} \to \operatorname{mor} \mathcal{D}$ satisfying various axioms, which I'm sure you know. In particular, if $X$ is an object in $\mathcal{C}$ then there is a unique object $Y$ in $\mathcal{D}$ such that $Y = F X$. There really is no scope for confusion here.

If you want a notion of "functor" where the "value" is only unique up to isomorphism, then perhaps you should look at anafunctors: an anafunctor $\mathcal{C} \to \mathcal{D}$ is a triple $(\mathcal{F}, P, Q)$, where $P : \mathcal{F} \to \mathcal{C}$ is a fully faithful functor that is (strictly) surjective on objects, and $Q : \mathcal{F} \to \mathcal{D}$ is any functor. The "value" of $(\mathcal{F}, P, Q)$ evaluated at an object $X$ in $\mathcal{C}$ is defined to be any object $Y$ in $\mathcal{D}$ for which there exists an object $Z$ in $\mathcal{F}$ such that $P Z = X$ and $Q Z = Y$. Note that if $Z'$ were another such object, then the fact that $P Z = P Z'$ implies there is a unique morphism $e : Z \to Z'$ in $\mathcal{F}$ such that $P e = \textrm{id}_X$; of course, $e$ is an isomorphism whose inverse is the unique morphism $e' : Z' \to Z$ such that $P e = \textrm{id}_X$, so $Q e : Q Z \to Q Z'$ is an isomorphism. Thus the isomorphism type of the "value" is uniquely determined.

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Thanks, I'll look it up. It definitively helped me out. – Michael Scarn Jan 7 '13 at 19:49

No, $FX$ really is unique. A functor $F : \mathcal{C} \to \mathcal{D}$ assigns to each object $X \in \mathcal{C}$ a unique object $FX \in \mathcal{D}$.

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Thanks for the answer. – Michael Scarn Jan 7 '13 at 19:52

The value of a functor $F$, call it $F(X)$, should be uniquely determined. Usually this requirement should not cause any problems.

Sometimes you might find two functors that essentially do the same thing. For example, let the functor $F$ on sets be the identity functor and let $G(X)=\{\{x\}:x \in X\}$. In this case, $G$ is different from $F$, but not in an interesting way. $F$ and $G$ isomorphic functors, which means for all $X$, $F(X) \equiv G(X)$ "naturally".

Here's another situation. Consider the category $Bij$ of finite sets where morphisms are bijections. Let functor $F$ assign to a set $X$ a set of possible linear orderings on $X$ (there are $n!$ ways you can order $X$) and let $G(X)$ be the set of all bijections on $X$ (there are also $n!$ bijections). Even though $F(X) \equiv G(X)$ for all $X$, there is no longer a "canonical" way to pair orders and bijections; $F$ and $G$ are not naturally isomorphic.

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Actually, $F$ and $G$ are not naturally isomorphic more because the automorphisms of $X$ act on $F X$ and $G X$ in very different ways. (The "transport of structure" action on $F X$ is transitive, whereas the conjugation action on $G X$ is not.) – Zhen Lin Jan 7 '13 at 19:33
Thanks for the answer. – Michael Scarn Jan 7 '13 at 19:53

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