MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

5 added 7 characters in body

(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)

What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
$$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$

Also, if $A\to B_i$ is a finite collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.

Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?

4 corrected typo

(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)

What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.

If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
$$g(s^{-1}a)=g(s)^{-1}g(a)=f(s)^{-1}f(a)=f(s^{-1}a)$$$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$Also, if A\to B_i is a collection of epimorphisms, where the B_i have disjoint support as A-modules, then A\to\prod B_i is an epimorphism. Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose f: A\to B is an epimorphism of rings with no kernel which sends non-units to non-units and such that B has no idempotents. Must f be an isomorphism? 3 inserted "Hausdorff" and converted to LaTeX (Background: In any category, an epimorphism is a morphism f:X→Y f:X\to Y which is "surjective" in the following sense: for any two morphisms g,h:Y→Z, g,h:Y\to Z, if g\circ f=h\circ ff, then g=h. g=h. Roughly, "any two functions on Y Y that agree on the image of X X must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, f f is an epimorphism if its image is dense.) What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal I I in a ring A, A, the quotient map A→A/I A\to A/I is an epimorphism. It's also not hard to see that if S⊂A S\subset A is a multiplicative subset, then the localization A→S-1A A\to S^{-1}A is an epimorphism. Here's a proof to whet your appetite. If g,h:S-1A→B g,h:S^{-1}A\to B are two homomorphisms that agree on A, A, then for any element s-1a∈S-1As^{-1}a\in S^{-1}A, we have g(s-1a) = g(s)-1g(a) = f(s)-1f(a) = f(s-1a)$$g(s^{-1}a)=g(s)^{-1}g(a)=f(s)^{-1}f(a)=f(s^{-1}a)$$Also, if A→Bi$A\to B_i$is a collection of epimorphisms, where the Bi$B_i$have disjoint support as A-modules,$A$-modules, then A→∏Bi$A\to\prod B_i$is an epimorphism. Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose f:A→B$f: A\to B$is an epimorphism of rings with no kernel which sends non-units to non-units and such that B$B$has no idempotents. Must f$f\$ be an isomorphism?