Let $\mathcal{C}$ be a category and $\mathcal{W} \subseteq \text{Arr}(\mathcal{C})$ a set (or class) of arrows.

There are (at least) two notions of localization of $\mathcal{C}$ with respect to $\mathcal{W}$, a strict and a weak one. In the strict one, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is a localization iff $\lambda$ inverts all arrows in $\mathcal{W}$ and if for every functor $G: \mathcal{C} \rightarrow \mathcal{E}$ with the same property there exists a unique functor $G':\mathcal{D} \rightarrow \mathcal{E}$ with $$ G = G'\circ{}\lambda $$ It is not hard to see, that such a strict localization functor $\lambda$ must actually be surjective on objects. In the weak version of localization, one merely asks for the existence of a functor $G'$ together with an isomorphism $$ \eta: G \stackrel{\sim}{\rightarrow} G'\circ{}\lambda$$ such that for every other such pair $(G'',\eta')$ there exists a unique isomorphism $\kappa:G' \stackrel{\sim}{\rightarrow} G''$ with $$ (\kappa \circ{} \lambda) \eta = \eta' $$ Now, if such a weak localization functor would always be essentially surjective, then I (think I) can prove that the notions of strict and weak localization are essentially equivalent, in the sense that the existence of one implies that of the other.

So, are all weak localization functors essentially surjective?


For any weak version you have a stong one where $\lambda$ is the identity on objects. Moreover, the target of a (weak) localization functor is essentially unique (up to equivalence). Essential surjectivity is invariant by equivalences, hence the answer to your question is positive.

  • $\begingroup$ That's reassuring to hear! I'm was aware that the two things are equivalent, but how do you prove (either one of) them? I thought I had a really easy proof of the equivalence but I didn't write down the details, and now I'm stuck. $\endgroup$ – Nicolas Schmidt Jun 7 '17 at 11:51
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    $\begingroup$ Well, a strong localization is also a weak one, that's realitively easy. Now, given a weak one $\lambda$, factor $\lambda$ as a composition of two functors, the first one the identity on objects and the second one fully faithful. The first functor is then a strong localization. (I agree, there are details to be filled in, but that's the strategy.) $\endgroup$ – Fernando Muro Jun 7 '17 at 11:54
  • $\begingroup$ That's the exact strategy I had in mind. However I'm stuck at proving that the such factorized functor is again a localization. I had hoped it would be possible to prove that localizations are essentially surjective in an elementary fashion, thus completing the proof that weak and strong localizations are equivalent. Do you have any good reference for localizations (besides Gabriel-Zisman)? $\endgroup$ – Nicolas Schmidt Jun 7 '17 at 13:34
  • $\begingroup$ So what's the problem when proving that it's a localization? The second functor of the composite is an equivalence by construction, hence the first one is a weak localization, which is necessarily strong for it is the identity on objects. $\endgroup$ – Fernando Muro Jun 8 '17 at 13:02
  • $\begingroup$ Why is the second functor an equivalence by construction? It being essentially surjective is the same thing as the whole functor being essentially surjective. I have figured out a way to prove the statement using the explicit description of the localized category by generators and relations, but iirc this description doesn't exist in general. $\endgroup$ – Nicolas Schmidt Jun 8 '17 at 17:02

Here is another argument that doesn't involve mentioning strong localizations at all. A functor is essentially surjective iff it is left orthogonal, in the bicategorical sense, to all fully faithful functors, i.e. a certain square is a bicategorical pullback (see here). This is true for weak localizations essentially because fully faithful functors are conservative (isomorphism-reflecting).

  • $\begingroup$ That's cool! I didn't know there was such a nice categorical characterization of essential surjectivity. Now I only need to sit down and convince myself that it is true. $\endgroup$ – Nicolas Schmidt Jun 8 '17 at 17:23

The question has already been answered by Fernando and Mike, but I think it might be useful to expand on the details and provide another (partial) argument as to why localizations are essentially surjective.

As Fernando suggested, a weak localization functor $$ \lambda: \mathcal{C} \rightarrow \mathcal{D}$$ can be factorized (like any other functor) as the composition $\lambda = \lambda_2 \circ{} \lambda_1$ of an essentially surjective followed by a fully faithful one: $$ \begin{array}{ccccc} \mathcal{C} & & \xrightarrow{\lambda} & & \mathcal{D} \\ & \searrow && \nearrow \\ && \mathcal{C}'\end{array} $$ One simply takes $\mathcal{C}'$ to be the category with the same objects as $\mathcal{C}$ and puts $\text{Hom}_{\mathcal{C'}}(X,Y) := \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$. The functors $\lambda_1$ and $\lambda_2$ are then defined in the obvious way, with $\lambda_1$ being the identity on objects.

From the construction it's clear that $$ \lambda \text{ ess. surj. } \Leftrightarrow \lambda_2 \text{ ess. surj. } \Leftrightarrow \lambda_2 \text{ is an equivalence} $$ This has nothing to do with $\lambda$ being a localization. However since $\lambda$ is a localization (in the following, "localization" refers to the weak notion), it follows by easy arguments that $$ \lambda_2 \text{ is an equivalence } \Leftrightarrow \lambda_1 \text{ is a localization } \Leftrightarrow \lambda_1 \text{ is a strong localization}$$

The first equivalence follows by observing that $\lambda_1$ already inverts all arrows in $\mathcal{W}$, which implies that for all categories $\mathcal{E}$ the diagram $$ \begin{array}{ccccc} [\mathcal{D},\mathcal{E}] & & \xrightarrow{\lambda^\ast} & & \{ F \in [\mathcal{C},\mathcal{E}] : F(\mathcal{W}) \subseteq \text{Isos}(\mathcal{E}) \} \\ & \searrow && \nearrow \\ && [\mathcal{C}', \mathcal{E}] \end{array} $$ of functor categories is well-defined (and commutative). Because $\lambda$ is a localization, the top arrow is an equivalence. If $\lambda_1$ is a localization, it follows that $\lambda_1^\ast$ and therefore $\lambda_2^\ast$ is an equivalence for all categories $\mathcal{E}$, which implies that $\lambda_2$ itself must be an equivalence. This shows the direction $\Leftarrow$ of the first equivalence, and the direction $\Rightarrow$ is obvious. For the second equivalence the direction $\Leftarrow$ is obvious, and the direction $\Rightarrow$ follows easily because given any weak factorization $$\eta:\widehat{F}\circ{}\lambda_1 \stackrel{\simeq}{\rightarrow} F$$ of a functor $F: \mathcal{C} \rightarrow \mathcal{E}$ inverting the arrows in $\mathcal{W}$, we can simply twist $\widehat{F}$ via $\eta$ to obtain a functor $\widehat{F}'$ giving an on-the-nose-factorization $F = \widehat{F}'\circ{}\lambda_1$: $$ \widehat{F}'(X \stackrel{\phi}{\rightarrow} Y) := \eta(Y) \circ{} \widehat{F}(\phi) \circ{} \eta(X)^{-1} $$

Given these preliminary remarks, let us prove that $\lambda$ is essentially surjective, using

  1. Mike's argument
  2. Fernando's argument
  3. My own (partial) argument

1. Essential surjectivity via left orthogonality to fully faithful functors

According to this nLab entry, a functor $\lambda: \mathcal{C} \rightarrow \mathcal{D}$ is essentially surjective iff it is left orthogonal to fully faithful functors, meaning that for every fully faithful functor $\mu: \mathcal{A} \rightarrow \mathcal{B}$ the induced diagram $$ \require{AMScd} \begin{CD} [\mathcal{D},\mathcal{A}] @>{\mu_\ast}>> [\mathcal{D},\mathcal{B}] \\ @V{\lambda^\ast}VV @V{\lambda^\ast}VV \\ [\mathcal{C},\mathcal{A}] @>{\mu_\ast}>> [\mathcal{C}, \mathcal{B}] \end{CD} $$ is a 2-categorical pullback. Now whatever this means, it should certainly imply the following: given functors $\phi \in [\mathcal{C},\mathcal{A}]$, $\psi \in [\mathcal{D},\mathcal{B}]$ and an isomorphism $\eta: \mu\circ{} \phi \stackrel{\sim}{\rightarrow} \psi\circ{}\lambda$, there exists a functor $\alpha \in [\mathcal{D},\mathcal{A}]$ together with isomorphism $\varepsilon_1: \alpha\circ{}\lambda \stackrel{\sim}{\rightarrow} \phi$ and $\varepsilon_2: \mu\circ{}\alpha \stackrel{\sim}{\rightarrow} \psi$.

This last property implies essential surjectivity of $\lambda$ as follows. Factorize $\lambda = \lambda_2 \circ{} \lambda_1$ as above and consider $\mathcal{A} = \mathcal{C}'$, $\mathcal{B} = \mathcal{C}$, $\mu = \lambda_2$ (fully faithful by construction), $\phi = \lambda_1$, and $\psi = \text{id}$. Because $\psi\circ{}\lambda = \lambda = \lambda_2\circ{}\lambda_1 = \mu \circ{} \phi$ on the nose, we get $\alpha \in [\mathcal{D},\mathcal{C}']$ and isomorphisms $\alpha \circ{} \lambda \simeq \lambda_1$ and $\lambda_2\circ{}\alpha \simeq \text{id}$, showing that $\lambda_2$ is essentially surjective and hence an equivalence, implying that $\lambda = \lambda_2\circ{}\lambda_1$ is essentially surjective (because $\lambda_1$ is by construction).

Therefore it only remains to see that a localization functor has the property above. But given $\mu$, $\phi$, $\psi$, and an isomorphism $\eta: \mu\circ \phi \stackrel{\sim}{\rightarrow} \psi\circ \lambda$ as in the statement of the property, because $\lambda$ inverts the arrows in $\mathcal{W}$, so must $\mu \circ{} \phi$; since $\mu$ is fully faitful and therefore detects isomorphisms, $\phi$ itself must already $\mathcal{W}$. The universal property of the localization then gives a functor $\alpha \in [\mathcal{D},\mathcal{A}]$ and an isomorphism $\varepsilon_1: \alpha \circ{} \lambda \stackrel{\sim}{\rightarrow} \phi$ out-of-the-box. Moreover the universal property also states that $\lambda^\ast$ defines a bijection $$ \text{Nat}(\mu\circ \alpha, \psi) \stackrel{\sim}{\rightarrow} \text{Nat}(\mu\circ \alpha \circ \lambda, \psi \circ \lambda)$$ which we can use to lift the isomorphism given by the composition $$ \mu\circ\alpha\circ\lambda \stackrel{\mu\varepsilon_1}{\rightarrow} \mu\circ\phi \stackrel{\eta}{\rightarrow} \psi\circ\lambda$$ This lift is again an isomorphism (because of the fully faithfulness of $\lambda^\ast$), giving the desired second isomorphism $\varepsilon_2$.

2. Constructing a quasi-inverse to $\lambda_2$

Because $\lambda_1$ inverts the arrows in $\mathcal{W}$ as we've seen already, we can apply the universal property of $\lambda$ to get a weak factorization of $\lambda_1$: $$\eta: \mu \circ\lambda \stackrel{\sim}{\rightarrow} \lambda_1$$ From this we deduce an isomorphism $$\lambda_2\eta: \lambda_2\circ\mu\circ\lambda \stackrel{\sim}{\rightarrow} \lambda_2 \circ \lambda_1 = \lambda$$ Applying the second part of the universal property of $\lambda$, the fully faithfulness of $\lambda^\ast$: $$\text{Nat}(\lambda_2\circ\mu , \text{id}) \stackrel{\sim}{\rightarrow} \text{Nat}(\lambda_2\circ\mu\circ\lambda , \lambda)$$ we deduce a natural transformation $\kappa: \lambda_2\circ\mu \rightarrow \text{id}$ that must be an isomorphism for the same argument used in the previous section.

3. Using the description of morphisms in localized categories

One can verify directly that $\lambda_1$ is a localization, therefore proving the essential surjectivity of $\lambda_2$, by simply writing down the objects asked for by the universal property. The verification that these objects have the required properties is not obvious, but follows easily if one knows that all morphisms between objects $\lambda(X)$, $\lambda(Y)$ in the localized category are given by images of morphisms in $\mathcal{C}$ and inverses of morphisms in $\mathcal{W}$.

Let $F: \mathcal{C} \rightarrow \mathcal{E}$ be a functor inverting $\mathcal{W}$. By the universal property of $\lambda$, we get a weak factorization $$ \eta: \widehat{F}\circ\lambda \stackrel{\sim}{\rightarrow} F$$ which we can also read as an equivalence $\widehat{F}'\circ\lambda_1 \simeq F$ if we put $\widehat{F}' := \widehat{F}\circ\lambda_2$, proving one half of the universal property for $\lambda_1$. To show the other half, we must prove that for any two functors $F,G: \mathcal{C}' \rightarrow \mathcal{E}$ we get a bijection $$\lambda_1^\ast: \text{Nat}(F,G) \stackrel{\sim}{\longrightarrow} \text{Nat}(F\circ\lambda_1 , G\circ\lambda_1 )$$ Since $\lambda_1$ is the identity on objects, this map is obviously injective. To see surjectivity, we would have to know that if for an $\eta:F\circ\lambda_1 \rightarrow G\circ\lambda_1$ the diagrams $$ \require{AMScd} \begin{CD} F(X) @>\eta(X)>> G(X) \\ @VF(\phi)VV @VG(\phi)VV \\ F(Y) @>\eta(Y)>> G(Y)\end{CD} $$ commuted for all $\phi \in \lambda(\text{Hom}_{\mathcal{C}}(X,Y)) \subseteq \text{Hom}_{\mathcal{C}'}(X,Y) = \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$, then they commuted for all $\phi \in \text{Hom}_{\mathcal{D}}(\lambda(X),\lambda(Y))$. But given a description of the morphisms in $\mathcal{D}$ as compositions of images of morphisms in $\mathcal{C}$ and inverses of images of morphisms in $\mathcal{W}$, this would be immediate.

Such as description would follow for instance from the explicit construction of the localized category via generators and relations (which may not exist due to size issues).


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