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Let $\mathcal{C}$ Be A Category and $S$ a class of morphisms (let us call these weak equivalences) of $\mathcal{C}$. One often defines the localization of $\mathcal{C}$ with respect to $S$ is the category, $S^{-1}\mathcal{C}$ that is initial among all functors that turn the elements of $S$ into isomorphisms. This definition depends on the choice of weak equivalences. What may happen is that their is some other class, $S'$ such that $S^{-1}\mathcal{C}$ and $S'^{-1}\mathcal{C}$ are equivalent categories. Thus, $S$ could be thought of as a presentation of the localization. The question is: is their a presentation-free definition of localization.

More exactly Fill in the blank: A localization of a category $\mathcal{C}$ is a functor, $F:\mathcal{C}\rightarrow\mathcal{D}$ such that ________.

Any partial results are perfectly fine, but not quite what I am looking for.

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  • $\begingroup$ Let me clarify my last sentence. If all their is known are partial results, I would be happy to learn about them. But that is not my hope! $\endgroup$
    – Spice the Bird
    Commented Apr 6, 2012 at 6:22
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    $\begingroup$ $F: \mathcal{A}\to \mathcal{B}$ is a localization if (definition): for any $G: \mathcal{A}\to \mathcal{C}$ such that: $\forall f\in |\mathcal{A}|_1: F(f)$ is isomorphisms $\Rightarrow G(f)$ is a isomorphism, exist uniquely a $H: \mathcal{B}\to \mathcal{C}$ with $G= H\circ F$ $\endgroup$
    – Buschi Sergio
    Commented Apr 6, 2012 at 7:26
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    $\begingroup$ Buschi, That is so obvious (after the fact) and it answers my question. Thank you. $\endgroup$
    – Spice the Bird
    Commented Apr 6, 2012 at 7:52
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    $\begingroup$ I'm quite confused. This is indeed obvious, but I assumed that this is not the answer you wanted. After all it does refer to a class of morphisms $S$ with respect to which you localize. Namely $S = \{ f \mid F(f) \textrm{ is an isomorphism} \}$. $\endgroup$
    – Karol Szumiło
    Commented Apr 6, 2012 at 11:14
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    $\begingroup$ One more thing that may be worth saying explicitly: if $S$ is any class of morphisms and $F : \mathcal{C} \to \mathcal{D}$ a localization with respect to $S$, then $F$ is also a localization with respect to $S_F = \{ f \mid F(f) \textrm{ is an isomorphism} \}$. A class $S$ such that $S = S_F$ is often called saturated. $\endgroup$
    – Karol Szumiło
    Commented Apr 6, 2012 at 19:04

2 Answers 2

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I don't know a general result, but since you also asked for special cases here's one. If $F$ admits a right adjoint, then it is a localization with respect to some class of morphisms if and only if its right adjoint is fully faithful (which is also equivalent to the counit being an isomorphism). The proof is not difficult, it can be found in Proposition I.1.3 of Gabriel, Zisman Calculus of Fractions and Homotopy Theory.

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  • $\begingroup$ Minor remark: This is really a special case; there are localizations which are not left adjoint. On the other hand, it seems to be quite typical. $\endgroup$
    – Martin Brandenburg
    Commented Apr 6, 2012 at 8:55
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It is possible to classify the localizations of locally finitely presentable categories; these are those categories, which are cocomplete, finitely complete, have a small strong generator, such that filtered colimits commute with finite limits. This is even true in the enriched setting. See the paper "Localisation of locally presentable categories" by Day and Street.

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