MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a slightly pedantic question about the "2-categorical" nature of localization. Recall the definition:

Definition. A localization of a category $\cal C$ with respect to a class of morphisms $W$ is a category ${\cal C}[W^{-1}]$ together with a functor $q:{\cal C}\to {\cal C}[W^{-1}]$ such that

  1. $q$ sends the morphisms of $W$ to isomorphisms;
  2. for any functor $F:{\cal C}\to {\cal D}$ which sends the morphisms of $W$ to isomorphisms, there is a unique functor $G:{\cal C}[W^{-1}]\to {\cal D}$ such that $G\circ q=F$.

Note that I'm using the "strict" version of the definition, which characterizes the category ${\cal C}[W^{-1}]$, if it exists, up to a canonical isomorphism of categories (as opposed to a weaker version that only requires $G \circ q$ to be naturally isomorphic to $F$ and which only characterizes ${\cal C}[W^{-1}]$ up to equivalence).

The above very familiar definition can be rephrased as follows: condition (1) is equivalent to saying that for any category ${\cal D}$, the functor $-\circ q : {\cal D}^{{\cal C}[W^{-1}]} \to {\cal D}^{\cal C}$ factors through the full subcategory ${\cal D}^{({\cal C},W)}$ of ${\cal D}^{\cal C}$ consisting of those functors $F:{\cal C}\to {\cal D}$ which send the morphisms in $W$ to isomorphisms, and the universal property (2) states that the functor $-\circ q : {\cal D}^{{\cal C}[W^{-1}]} \to {\cal D}^{({\cal C},W)}$ is bijective on objects.

But what about natural transformations? It should follow from the above definition that localization is actually "2-categorical" in the sense that the functor $-\circ q : {\cal D}^{{\cal C}[W^{-1}]}\to {\cal D}^{\cal C}$ is fully faithful, so that $-\circ q : {\cal D}^{{\cal C}[W^{-1}]} \to {\cal D}^{({\cal C},W)}$ is an isomorphism of categories, not just a bijection between objects.

This would be immediate provided that $q$ is bijective on objects. It is probably obvious that the definition of localization forces this, but I can't see it. It is clear to me that a localization $q$ must be surjective on objects but I'm missing why it must be injective on objects. I'm sure there is a very simple reason, flying straight from the definition, that I'm missing.

Aside: one reason why someone might care about this detail comes from the theory of derivators. To show, for example, how derived categories give rise to derivators one needs to induce natural transformations between functors between derived categories.

share|cite|improve this question
It doesn't seem like bijection on objects (for q) would be forced by the definition... For example, we could replace the usual construction of C[W^{-1}] with its skeleton... this should have strictly fewer objects in a lot of cases it seems (e.g. inverting rational equivalences in the homotopy category, or something). – Dylan Wilson Sep 9 '11 at 1:09
@Dylan - 'bijection on objects' relates to the induced functor between hom-categories, not the localisation functor. – David Roberts Sep 9 '11 at 1:41
@David: In the second last paragraph, I also mention bijection on objects for the canonical functor q as a means of establishing that the induced functor between hom-categories is fully faithful. – Beren Sanders Sep 9 '11 at 1:51
@Dylan: But it seems to me that the skeleton won't necessarily satisfy the strict universal property. – Beren Sanders Sep 9 '11 at 1:52
+1 for giving strict higher categories some love. – Harry Gindi Sep 26 '11 at 13:05
up vote 6 down vote accepted

Let $\widetilde{\mathcal{C}}$ be the category with same objects as $\mathcal{C}$ and exactly one morphism between each pair of objects. Then there exists a unique functor $F:\mathcal{C} \to \widetilde{\mathcal{C}}$ that is the identity on objects. It sends all morphisms in $W$ to isomorphisms (since all morphisms of $\widetilde{\mathcal{C}}$ are isomorphisms). Thus it factors as $F = G q$ for a unique $G : \mathcal{C}[W^{-1}]\to\widetilde{\mathcal{C}}$. Since $F$ is the identity on objects, it follows that $q$ is injective on objects.

share|cite|improve this answer
Thank you, this is what I was missing! – Beren Sanders Sep 9 '11 at 12:18

You can also see that the 2-dimensional universal property of q, that $-\circ q$ is fully faithful, follows directly from the 1-dimensional universal property you describe in your definition. More generally any cocone in a 2-category (Cat here) which satisfies the 1-dimensional universal property of the colimit (coinverter here) also satisfies the 2-dimensional universal property so long as the 2-category you are working in has cotensors with 2, the category with two objects $0$ and $1$ and a single arrow $0 \to 1$. Concretely this works in the above case as follows. Let $q$ satisfies the 1-dimensional universal property and suppose we have a pair of functors $F,G:{\cal C}\to {\cal D}$ which both invert the W's and a natural transformation $\alpha:F \to G$. Such a triple $(F,\alpha,G)$ uniquely correspond to a functor $\hat{\alpha}:C \to Ar(D)$ where Ar(D) is the arrow category of D, whose objects are arrows in D, and morphisms commuting squares (this is the cotensor of D with 2 in Cat). Because isomorphisms in Ar(D), equally the functor category from 2 to D, are those nat. transformations which are pointwise isos it is clear the $\hat{\alpha}$ inverts the W's since both F and G do. Thus you obtain a unique functor $\beta:{\cal C}[W^{-1}] \to D$ such that $\beta q = \hat{\alpha}$ and now postcomposing $\beta$ with the evident, and universal, two functor projections and natural transformation from $Ar(D)$ to $D$ gives the two functors and natural transformation from ${\cal C}[W^{-1}]$ to $D$ that you are after. Apologies if the tex is madness as I can't see it on this computer.

share|cite|improve this answer

The localisation of a category (ignoring size issues), when defined in the 'up to equivalence' way, is a coinverter in $Cat$. This is an example of a weighted colimit in the $Cat$-enriched category $Cat$. If one wants the 'up to isomorphism' version, then what you are dealing with is a strict coinverter. This obviously satisfies a slightly different universal property.

share|cite|improve this answer

Considering the groupoids $Grpd({C})$ this is maked by the free category generate by the morphisms of $\mathcal{C}$, their formal inverse and formal composition, and quotient this free category by the relations maked from the composition of $\mathcal{C}$. Then considering the natural functor $F: \mathcal{C}\to Grp({C})$ that is the identity on objects, factorizing it follow that your functor $q$ is a section on object part is a section, then injective. Of course $q$ is surjective , this follow from the (strict) universal property (2), and considering the full category generated by the image of $q$.

share|cite|improve this answer

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.