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Let $\mathsf{C}$ be a category and $S$ be a collection of morphisms in $\mathsf{C}$. In this generality, we can construct the localization $S^{-1}\mathsf{C}$ by posing its objects to be the same as those in $\mathsf{C}$ and a morphism $M\to N$ to be a path

where the $L_i$ are objects of $\mathsf{C}$, the arrows to the right are morphisms of $\mathsf{C}$, and the arrows to the left are elements of $S$, up to equivalence. This equivalence relation, denoted $\sim$, says basically that compositions should behave well

that we may ignore identities

and that arrows to the left correspond to inverses

Usually, $S$ satisfies the axioms of a (left) multiplicative system which implies that every path is equivalent to a path of the form

which we call a roof. We wish to see if two roofs define the same morphism without ever dealing with longer paths, so we say that two roofs are $\sim_L$ equivalent if there exists an object $L$ in $\mathsf{C}$ and morphisms $p_1:L_1\to L$, $p_2:L_2\to L$ making the diagram

commute, and such that $p_2\circ s_2=p_1\circ s_1$ is in $S$.

In order for this to be useful, of course both notions should coincide. That is, there should exist a dashed isomorphism making the diagram

commute. I can verify that $\sim_L$ is indeed an equivalence relation (as is $\sim$), that such a map exists and that it is surjective. How can I prove that it's also injective?

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    $\begingroup$ The way I generally see this proved is by directly showing that the "Roof" description satisfies the universal property of the localisation, which implies it is equivalent to the other definition of the localisation rather than directly proving they are equivalent. $\endgroup$
    – Simon Henry
    Commented Nov 18, 2021 at 16:02
  • $\begingroup$ Dear @SimonHenry, I also know multiples references that do this. But I have to say that it feels very unsatisfying to me; basically the "raison d'être" of the axioms for a multiplicative system is the need to simplify paths, obtaining roofs. (This allows us to add morphisms when $\mathsf{C}$ is preadditive, for example.) $\endgroup$
    – Gabriel
    Commented Nov 18, 2021 at 16:04
  • $\begingroup$ That is, it seems unsatisfying to me to ignore that we have a clear inclusion of the set of roofs in the set of paths. $\endgroup$
    – Gabriel
    Commented Nov 18, 2021 at 16:06
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    $\begingroup$ I understand you want a more direct proof, and I don't know one in the literature. But you can directly translate the proofs I'm mentioning in what you are after, it is just going to be a little more complicated. It is important to note that the approach I'm mentioning do not "ìgnore" that roof are a special case of paths : the functor you get from the universal property send a roof to what the corresponding path is sent to. So you'll construct a functor from path to roof (up to equivalence) and show it is inverse to the obvious inclusions. $\endgroup$
    – Simon Henry
    Commented Nov 18, 2021 at 22:33
  • $\begingroup$ I guess a direct proof that it's injective should look like "if two roofs become equivalent, translate every use of one of the rules making paths equivalent to a use of one of the rules making roofs equivalent, using the fact that $S$ is a multiplicative system". This would be some sort of rewriting proof. It's tedious but it'll get you there $\endgroup$
    – Maxime Ramzi
    Commented Feb 2, 2022 at 19:39

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I did a subtle viewpoint shift, which made everything way simpler. Instead of proving that our map is injective (which amounts to verifying that two $\sim$-equivalent roofs are $\sim_L$-equivalent, and is very messy), we define a left inverse. In case this may help someone in the future, I'll leave here my proof that this works.

(To be honest, this isn't as messy as I thought this would be. Verifying that the category whose morphisms are $\sim_L$-equivalence classes of roofs is indeed a category and satisfies the universal property of localisation in all details is actually worse.)

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