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I have the following situation: I have a construction that takes an object in a category $\mathcal{C}$ and via some totally noncanonical choices constructs an object in another category $\mathcal{D}$ where different choices result in (noncanonically) isomorphic objects in $\mathcal{D}$.

I also have a second construction where given morphism in $\mathcal{C}$ say $f: X \to Y$, after choosing objects in $\mathcal{D}$ associated to $X$ and $Y$, say $X'$ and $Y'$, I can construct a map $f' : X' \to Y'$.

Is there a way of putting these constructions together and calling it a functor? It seems I would need a new category instead of $\mathcal{D}$ in which the objects are isomorphism classes of objects in $\mathcal{D}$.

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    Make whatever choice you need for each object (you can do this at once if your $C$ is small, and if not you can also do it using universes, global choice, of whatever set-theoretical way of just doing it you choose). This will give you a functor, depending on choices. The functors corresponding to two different choices will be naturally isomorphic. – 2017-01-04
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    (There is no sensible way of constructing a category whose objects are isomorphism classes. You can construct skeletons, though, by picking one object in each isoclass and considering the full subcategory they span; of course, this is not canonical in any sense) – 2017-01-04
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    As Mariano states, yes, you can build a functor though you may need a form of the Axiom of Choice to make all those choices consistently. It may be more natural to present your construction as an [anafunctor](https://ncatlab.org/nlab/show/anafunctor) which was originally motivated as a notion of "functor" that avoids needing the Axiom of Choice in many common situations. – 2017-01-04
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    I think we need to add that this will only work if $f'$ is uniquely determined by $f,X', Y',$ which you don't quite specify. If you're also making further choices to construct $f'$, then there may not be a way to make all these choices simultaneously and compatibly with composition. – 2017-01-05
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    @Mariano: I think you can do it if you put more information into the construction; e.g. make the objects connected groupoids. – 2017-01-05
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    I am curious about your assertion that the isomorphisms are noncanonical. Your second construction, in the case $Y=X$ and $f=id_X$, gives a map $id_X':X'\to X''$ whenever $X'$ and $X''$ are two possible objects associated to $X$. Are these maps isomorphisms? If your second construction respects composition (and identity maps when $X'=Y'$), they ought to be, and so they ought to give "canonical" isomorphisms. And if your second construction doesn't respect composition, I don't see how you could possibly hope to get a functor. – 2017-01-05

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A basic trick here is to "inflate" the category $\mathcal{C}$; e.g.define $\hat{\mathcal{C}}$ to be

  • The objects are pairs $(X, C)$ where $X$ is an object of $\mathcal{C}$ and $C$ represents a particular collection of choices in the construction. (e.g. we might just pick $C = X'$)
  • The morphisms are given by $\hom_\hat{\mathcal{C}}((X,C), (Y,D)) = \hom_\mathcal{C}(X,Y)$

The end result is that the functor $\hat{\mathcal{C}} \to \mathcal{C}$ that forgets the choices is an equivalence functor, and when doing your construction on objects of $\hat{\mathcal{C}}$, you now have canonical choices.

Of course, your construction still may or may not be a functor; e.g. you have to check that it preserves identities and composition.

With a sufficiently strong version of the axiom of choice, equivalence functors have weak inverses, so you also get an equivalence functor $\mathcal{C} \to \hat{\mathcal{C}}$ that makes a choice for each object of $\mathcal{C}$, which can be composed with functors $\hat{\mathcal{C}} \to \mathcal{D}$ to get a functor $\mathcal{C} \to \mathcal{D}$