Localization as dual to quotienting? Why does localization look so close to quotienting? Consider the ismorphism theorem:

For a ring map $\phi: R_1 \rightarrow R_2$, let $\ker\phi \equiv \{ r_1\in R_1 : \phi(r_1) = 0 \}$. Then, $\ker \phi$ is an ideal of $R_1$, and there exists an epi-mono factorization of $\phi$ into $R_1 \twoheadrightarrow R_1 /\ker\phi \hookrightarrow R_2$  where $\hookrightarrow$ is injective and $\twoheadrightarrow$ is surjective.

Now consider the similar theorem for localization, where I denote by $R  \hat \times S \equiv S^{-1}R$ for notational suggestivity. Now I write down:

For a ring map $\psi : R_1 \rightarrow R_2$, Let $\operatorname{rek}\psi \equiv \{ r_1 \in R_1 : \psi(r_1) = 1 \}$. Then $\operatorname{rek}\psi$ is a multiplicative subset of $R_1$ and there exists a mono-epi factorization of $\psi$ into $R_1 \hookrightarrow R_1 \hat \times \psi \twoheadrightarrow R_2$

I can build a table:

*

*$\phi \leftrightarrow \psi$.

*$\ker \leftrightarrow\operatorname{rek}$.

*injection $\leftrightarrow$ surjection.

*$/ \leftrightarrow \hat\times$

*ideal $\leftrightarrow$ multiplicative subset.

*$0 \leftrightarrow 1$.

to convert from the quotienting into localization. Is there some "deep" going on here for this duality? This $0 \leftrightarrow 1$ business makes me hopeful that there might be something deeper / categorical lurking in the background.
EDIT: I had only commutative rings in mind when I wrote this. Please feel free to take assumptions on $R$ as required (Commutative, Noetherian, for example), if that allows us to explain this "duality".
 A: The deeper / categorical lurking in the background is the notion of factorization structure.
Let $E$ and $M$ be two classes of morphisms in a category.
Then $(E,M)$ is said to be a factorization structure if:


*

*$E$ and $M$ are closed under composition with isomorphisms;

*every ring homomorphism has a factorization $\mu\circ\varepsilon$ with $\mu\in M$ and $\varepsilon\in E$;

*the unique $(E,M)$-diagonalization property that for every commutative diagram$\require{AMScd}$
$$\begin{CD}
A@>\varepsilon>>B\\
@V\varphi VV @VV\gamma V\\
C@>>\mu> D
\end{CD}$$
with $\varepsilon\in E$ and $\mu\in M$ there exists a unique diagonal $\delta:B\to C$ making the diagram commtative.



In the category of commutative rings we have the following factorization structures:


*

*$E$ is the class of surjective ring homomorphisms and $M$ the class of injective ring homomorphisms;

*$E$ be the class of ring localizations, that's of the form (up to
isomorphism) $A\to S^{-1}A$ where $S\subseteq A$ is a multiplicative
system of $A$ and $M$ be the class of ring homomorphism $\varrho:A\to B$ such that $A^\times=\varrho^{-1}(B^\times)$;

*$E$ is the class of integral ring homomorphisms and $M$ is the class of injective and integrally closed ring homomorphisms.



Consider the second factorization structure in the list above, every homomorphism of commutative rings $\varrho:A\to B$ has as essentially unique $(E,M)$ factorization
$$A\xrightarrow\varepsilon S^{-1}A\xrightarrow\mu B$$
where $S=\psi^{-1}(R_2^\times)$.
Thus, with your notation, we have to take $\operatorname{rek}(\psi)=\psi^{-1}(R_2^\times)=\{ r_1 \in R_1 : \psi(r_1) \in R_2^\times \}$.
A: Fabio gives a very nice answer to your question, but doesn't directly address an important point of confusion in your original post/comments, so I'm adding this answer for posterity. In general, a map $\psi:R_1\rightarrow R_2$  will absolutely not induce an epimorphism $\text{rek}(\psi)^{-1}R_1\twoheadrightarrow R_2$, even if we take the stronger definition $\text{rek}(\psi)=\psi^{-1}(R_2^\times)$ given by Fabio. For instance, if every element of $\text{rek}(\psi)$ is already a unit in $R_1$, then we will just have $\text{rek}(\psi)^{-1}R_1=R_1$, and so using this it is easy to come up with examples where the induced map is not epi.
For instance, take $R_1=\mathbb{Q}$, and $R_2$ any field extension of $\mathbb{Q}$ with a non-trivial automorphism $\alpha$ that fixes $\mathbb{Q}$ pointwise, with $\psi:R_1\hookrightarrow R_2$ the inclusion map. Then $\text{rek}(\psi)=\mathbb{Q}^\times$ , so $\text{rek}(\psi)^{-1}\mathbb{Q}=\mathbb{Q}$ and the induced map to $R_2$ is just $\psi$, which is certainly not an epimorphism. (E.g. $\alpha\circ\psi=\text{id}_{R_2}\circ\psi$ but $\alpha\neq\text{id}_{R_2}$).
Indeed, the polynomial ring example you give in the comments of your post does not hold in general either. If we let $R_1=\mathbb{R}[x]$ and $R_2=\mathbb{R}[x,y]$, with $\psi:R_1\hookrightarrow R_2$ again the inclusion map, then once again $\text{rek}(\psi)=R_1^\times$ but $\psi$ is certainly not epi.
The problem in all of these examples is that $R_2$ can be very big compared to the image of $R_1$; hopefully the above examples clarify that point. (Note however, that – provided $R_2\neq\{0\}$ – the map $R_1\hookrightarrow \text{rek}(\psi)^{-1}R_1$ will still be injective, even if we use Fabio's stronger definition of $\text{rek}(\psi)$, because no element of $\text{rek}(\psi)$ can be a zero-divisor in $R_1$.)
