I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres Trees (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.

so I define $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).

the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, only using the universal property? I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. a major step would be the case of an integral domain.

EDIT: A new improved version of this question can be found here.

I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. A very nice example for this is given in Serres Trees (normal form for elements in amalgamated sums of subgroups). Up to now, it works in all examples I've came across. Even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. But I'm stuck with localizations of rings (or monoids, or modules). Rings and monoids are here assumed to be commutative.

So I define $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. Here $S$ is a submonoid of a ring $A$. It can be shown with rather general facts that $S^{-1} A$ exists, in several ways. But that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).

The definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). Clearly $a/s=b/t$ holds when $uta=usb$ for some $u \in S$. But how can we prove the converse, only using the universal property? I hope my aim is clear. In particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

Here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. Again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. To prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. Then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. This maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

As already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. For we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if $a$ is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. A major step would be the case of an integral domain.

EDIT: A new improved version of this question can be found here.

I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres Trees (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.

so I define $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).

the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, only using the universal property? I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. a major step would be the case of an integral domain.

EDIT: A new improved version of this question can be found here.

I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres Trees (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.

so I define $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).

the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, only using the universal property? I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. a major step would be the case of an integral domain.

EDIT: A new improved version of this question can be found here.

I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres Trees (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.

so I define $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).

the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, only using the universal property? I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. a major step would be the case of an integral domain.

EDIT: now I offer a bounty. it might be a silly question to some of you, but for me it's important ;-).

I've been studying universal objects of universal algebra in a quite general setting and try to exhibit the structure of their elements just using the universal property. a very nice example for this is given in Serres Trees (normal form for elements in amalgamated sums of subgroups). up to know, it works in all examples I've came across. even tensor products, see: Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs. but I'm stuck with localizations of rings (or monoids, or modules). rings and monoids are here assumed to be commutative.

so I define $S^{-1} A$ to be a ring which represents the subfunctor of $\hom(A,-)$, which maps elements of $S$ to units. here $S$ is a submonoid of a ring $A$. it can be shown with rather general facts that $S^{-1} A$ exists, in several ways. but that's not the point: I want to avoid explicit constructions (I might elaborate the reasons later).

the definition implies that there is a natural homomorphism $A \to S^{-1} A$, which is denoted simply by $a \mapsto a$, and that every element of $S^{-1} A$ has the form $a/s$ ($a \in A, s \in S$). clearly $a/s=b/t$ holds, when $uta=usb$ for some $u \in S$. but how can we prove the converse, only using the universal property? I hope my aim is clear. in particular, it would be cheeting applying the universal property to another explicit constructed model of $S^{-1} A$.

here is an example how elements might be described without using any construction: we want to show that in the category of abelian groups, elements of the coproduct $A+B$ (provided it exists) have a unique representation $a+b$, where $a \in A$ and $b \in B$. again we have an abuse of notation here, $a$ also means the image of $a$ in $A+B$. to prove this, observe that $\{a+b : a \in A, b \in B\}$ is a subgroup of $A+B$ which also satisfies the universal property. then it follows that every element has the form $a+b$. now define $A+B \to A$ by extending $id : A \to A$ and $0 : B \to A$. this maps $a+b \mapsto a$. hence $a$ is unique, and similar also $b$.

as already said, this also works in other situations, but it get's more complicated. conclusion: we don't have to invent other objects to study universal objects. for we may apply the universal property to themselves! I hope that this also works for localizations, in order to see that $a/1=0 \in S^{-1} A$ if and only if a is annihilated by some $s \in S$. I've already found out many basic results about localizations just using the universal property (e.g. "coherence isomorphisms", behavior under colimits), and using that I can reduce all to the fact that $S^{-1} A$ is a flat $A$-module, but this also seems to be hard without elements. a major step would be the case of an integral domain.

EDIT: A new improved version of this question can be found here.