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Warning: This question is evil.

If $\mathcal{C}$ and $\mathcal{D}$ are categories and $F : \mathcal{C} \rightarrow \mathcal{D}$ is a functor, then $F$ is not necessarily injective on objects. But, I imagine that we can always "make" $F$ injective. To do so, we "inflate" the category $\mathcal{D}$ by adding bunch of objects, thereby obtaining an equivalent category $\mathcal{D}'$ such that the corresponding functor $F' : \mathcal{C} \rightarrow \mathcal{D}'$ is injective-on-objects.

I'm not really sure how to define it though. Ideas, anyone?


Discussion. Here's a couple of examples.

Example 1. Let $\mathcal{C}$ denote an arbitrary category, let $1$ denote a category with precisely one object, and $F : \mathcal{C} \rightarrow 1$ denote the unique such functor. Then $F$ won't be injective unless $\mathcal{C}$ is trivial. But we can let $1' =$ the category whose objects are precisely the objects of $\mathcal{C}$ and whose arrows form a connected equivalence relation. There is an obvious functor $F' : \mathcal{C} \rightarrow 1'$ (just map every object to itself).

Example 2. Lets view the objects of $\mathrm{Grp}$ as ordered pairs $(X,*).$ Then there is a forgetful functor $U : \mathrm{Grp} \rightarrow \mathrm{Set}$ such that $U(X,*) = X.$ However, we can have two distinct groups $(X,*)$ and $(X,\diamond)$ with a common underlying set. So $U$ fails to be injective on objects. However, we should be able to "make" $U$ injective on objects by inflating $\mathrm{Set}.$ In particular, we can just adjoin a new object to $\mathrm{Set}$ for each object of $\mathrm{Grp}.$

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1 Answer 1

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Yes, this can be done. In fact:

Theorem. Every functor admits a factorisation as an injective-on-objects functor followed by a fully faithful surjective-on-objects functor.

Proof. Let $F : \mathcal{C} \to \mathcal{D}$ be a functor. We construct the following category $\mathcal{M}(F)$:

  • The collection of objects is the disjoint union of $\operatorname{ob} \mathcal{C}$ and $\operatorname{ob} \mathcal{D}$.
  • A morphism between two objects $C, C'$ in $\mathcal{C}$ is a morphism $F C \to F C'$ in $\mathcal{D}$; a morphism between two objects in $\mathcal{D}$ is as in $\mathcal{D}$; a morphism from an object $C$ in $\mathcal{C}$ to an object $D'$ in $\mathcal{D}$ is a morphism $F C \to D'$ in $\mathcal{D}$; and a morphism from an object $D$ in $\mathcal{D}$ to an object $C'$ in $\mathcal{C}$ is a morphism $D \to F C'$ in $\mathcal{D}$.
  • Identities and composition are inherited from $\mathcal{D}$.

There is then an obvious injective-on-objects functor $\mathcal{C} \to \mathcal{M} (F)$ and an obvious fully faithful surjective-on-objects functor $\mathcal{M} (F) \to \mathcal{D}$, such that their composite is the original functor $F : \mathcal{C} \to \mathcal{D}$.


There is a different way of replacing a functor with an equivalent injective-on-objects functor, but the statement of this fact is a little bit more complicated.

Definition.

  • An isocofibration is an injective-on-objects functor.
  • An isofibration is a functor $F : \mathcal{C} \to \mathcal{D}$ such that, for all objects $C$ in $\mathcal{C}$ and all isomorphisms $f : F C \to D'$ in $\mathcal{D}$, there exists an object $C'$ in $\mathcal{C}$ and an isomorphism $\tilde{f} : C \to C'$ such that $F \tilde{f} = f$.
  • A trivial isocofibration is an isocofibration that is fully faithful and essentially surjective on objects.
  • A trivial isofibration is a fully faithful functor that is surjective on objects. (Note that any such is automatically an isofibration.)

In this language, the theorem above says that every functor can be factored as an isocofibration followed by a trivial isofibration.

Exercise. Show that every functor can be factored as a trivial isocofibration followed by an isofibration.


Finally, there is also a unique factorisation result:

Exercise. Show that every functor can be factored as a bijective-on-objects functor followed by a fully faithful functor, and that this factorisation is unique up to unique isomorphism.

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Woah thanks for the detailed answer!! Its gonna take me a while to digest all this... –  goblin Aug 30 '13 at 8:38
2  
Notice the similarity to the mapping cone (en.wikipedia.org/wiki/Mapping_cylinder#Interpretation). –  Martin Brandenburg Aug 30 '13 at 8:40

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