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.

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?

share|improve this question

3 Answers 3

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.

share|improve this answer
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 $FX \in \mathcal{D}$.

share|improve this answer
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.

share|improve this answer
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

Your Answer


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.