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Moosbrugger suggest to talk about the category of $A$-modules. Here is a basic observation which follows from the universal property: If $R$ is a possibly noncommutative $A$-algebra, then there is at most one algebra homomorphism $S^{-1} A \to R$, and it exists iff all elements of $S$ become invertible in $R$. This follows easily from the universal property of $S^{-1} A$ applied to the center of $R$. Now let $M$ be an $A$-module and apply the above to $R=\mathrm{End}_A(M)$. It follows that the category of $S^{-1} A$-modules is equivalent to the category of $A$-modules, on which the elements of $S$ act as isomorphisms.

Therefore a possible formalization of my question might be the following: Let $S \subseteq A$ as above and let $B$ be a commutative $A$-algebra such that scalar restriction $\mathrm{Mod}(B) \to \mathrm{Mod}(A)$ is fully faithful and whose image consists of those $A$-modules on which the elements of $S$ act as isomorphisms. Can we then compute the kernel of $A \to B$ (without refering to the explicit construction of the localization $S^{-1} A$, but only using this statement about module categories, which of course yields $B \cong S^{-1} A$ by Morita).

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Examples. a) Let's start with a related example. If $C$ is a category of algebras of some type, then every continuous functor $C \to \mathrm{Set}$ is representable (special case of SAFT). For example for $C=\mathbb{Ab}$ C=\mathbf{Ab}$the functor$\mathrm{Hom}(A,-) \times \mathrm{Hom}(B,-)$is representable for all abelian groups$A,B$, showing the existence of the coproduct$A + B$. I claim that we can find a description of its elements via the universal property. Namely, choose the coproduct injections$i : A \to A + B$and$j : B \to A + B$. Then$C:=\mathrm{im}(i) + \mathrm{im}(j)$is an abelian group such that$i,j$factor through$C$and still satisfy the universal property - this shows$A = \mathrm{im}(i) + \mathrm{im}(j)$, i.e. every element has the form$i(a) + j(b)$for$a \in A, b \in B$. I claim that$a,b$are unique: By the universal property there is some$f : A + B \to A$with$fi = \mathrm{id}$and$fj=0$. It follows$f(i(a) + j(b))=a$and therefore$a$is unique, similarly$b$. b) Let$M,N$be modules over a commutative ring$k$and define the tensor product$M \otimes_k N$as the classifying object of bilinear$k$-bilinear maps on$M \times N$. If$E$is a generating set of$M$, then the universal property implies that every element has the form$\sum_{e \in E} e \otimes n_e$for some$n_e \in N$(which vanish for almost all$e$). There is a criterion when this element is zero and it can be proved with the universal property of the tensor product, without using its explicit construction (Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs, online). c) If$H$is a subgroup of a group$G_1$as well as of a group$G_2$, then it is clear how to represent elements in the amalgamated sum$G_1 *_H G_2$, which is defined as a pushout. The uniqueness of the representation is shown in Serre's Trees, Section 1.2, by application of the universal property to the symmetric group over all reduced words. There is no need to impose a group operation on the set of reduced words (which would be very tedious) - after this proof you get it for free! Localization. Let's consider a commutative ring$A$and a map$i : S \to |A|$into the underlying set of$A$. Then we can consider the subfunctor of$\mathrm{Hom}(A,-)$which is given by homomorphism homomorphisms$g : A \to B$such that$gi$factors through$B^*$. By general theorems a representing object exists and is usually denoted the localization$S^{-1} A$when$i$is understood. If$S'$denotes the multiplicative closure of the image of$i$, then$S'^{-1} A = S^{-1} A$(they satisfy the same universal property), thus we always may assume that$S$is just a multiplicative closed subset of$A$. Let$\tau : A \to S^{-1} A$be the universal homomorphism which maps$S$to units. Then$S^{-1} A = \{\tau(a) \tau(s)^{-1} : a \in A, s \in S\}$(they satisfy the same universal property). Let es write$\frac{a}{s} = \tau(a) \tau(s)^{-1}$. Thus every element in$S^{-1} A$is some fraction$\frac{a}{s}$. Now when are two such fractions are equal? Of course we know this from the usual construction using equivalence classes of pairs$(a,s)$, but I want to derive avoid this construction and derive it only with the universal property. First observe that$\frac{a}{s} = \frac{a'}{s'}$iff$\frac{sa'-sa'}{1}=0$. Thus it is enough to show$\frac{a}{1} = 0 \Rightarrow sa=0$for some$s \in S$, i.e. that the kernel of$\tau : A \to S^{-1} A$equals$I=\cup_{s \in S} \mathrm{Ann}(s)$. Since$S$is multiplicative,$I$is an ideal, and we may replace$A$by$A/I$, where localization commutes with quotients because of universal properties. Thus we may assume that$I=0$, i.e. that$S$consists of regular elements. Our task is then to show that$A \to S^{-1} A$is injective. But a priori we only know by the universal property that the kernel of$A \to S^{-1} A$equals the intersection of all kernels of homomorphisms$A \to B$which map$S$to units. Of course we have to use this and construct some specific$A \to B$, but I hope that we can either avoid some nasty element construction of$B$or that we can just use a ring which is built up out of$S^{-1} A$, but can be used to show that the kernel is zero - therefore to find a kind of self-referential proof. The examples above suggest that this might be possible after all. Motivation. I hope that a categorical proof makes the usual construction of the localization (via equivalence classes of pairs) redundant. Note also that it is rather nasty to prove all the details (equivalence relation, well-defined addition, well-defined multiplication, universal property) with the usual construction. Also the definition$(a,s) \sim (a',s') \Leftrightarrow \exists t : ts'a=tsa'$is not motivated at all there. A categorical proof should show in particular that this is the right choice, and not some random definition which turns out to be correct only afterwardsafter some computation. But of course I don't claim that a categorical construction of the localization is the easiest or best one. On the other hand, localization could be seen just as a toy example for other, more involved examples, where it is not clear at all how to understand some representing object of some functor, which exists my general nonsense. 6 fixed typo Examples. a) Let's start with a related example. If$C$is a category of algebras of some type, then every continuous functor$C \to \mathrm{Set}$is representable (special case of SAFT). For example for$C=\mathbf{Ab}$C=\mathbb{Ab}$ the functor $\mathrm{Hom}(A,-) \times \mathrm{Hom}(B,-)$ is representable for all abelian groups $A,B$, showing the existence of the coproduct $A + B$. I claim that we can find a description of its elements via the universal property. Namely, choose the coproduct injections $i : A \to A + B$ and $j : B \to A + B$. Then $C:=\mathrm{im}(i) + \mathrm{im}(j)$ is an abelian group such that $i,j$ factor through $C$ and still satisfy the universal property - this shows $A = \mathrm{im}(i) + \mathrm{im}(j)$, i.e. every element has the form $i(a) + j(b)$ for $a \in A, b \in B$. I claim that $a,b$ are unique: By the universal property there is some $f : A + B \to A$ with $fi = \mathrm{id}$ and $fj=0$. It follows $f(i(a) + j(b))=a$ and therefore $a$ is unique, similarly $b$.

b) Let $M,N$ be modules over a commutative ring $k$ and define the tensor product $M \otimes_k N$ as the classifying object of $k$-bilinear bilinear mapson $M \times N$. If $E$ is a generating set of $M$, then the universal property implies that every element has the form $\sum_{e \in E} e \otimes n_e$ for some $n_e \in N$ (which vanish for almost all $e$). There is a criterion when this element is zero and it can be proved with the universal property of the tensor product, without using its explicit construction (Pierre Mazet, Caracterisation des Epimorphismes par relations et generateurs, online).

c)

If $H$ is a subgroup of a group $G_1$ as well as of a group $G_2$, then it is clear how to represent elements in the amalgamated sum $G_1 *_H G_2$, which is defined as a pushout. The uniqueness of the representation is shown in Serre's Trees, Section 1.2, by application of the universal property to the symmetric group over all reduced words. There is no need to impose a group operation on the set of reduced words (which would be very tedious) - after this proof you get it for free!

Localization. Let's consider a commutative ring $A$ and a map $i : S \to |A|$ into the underlying set of $A$. Then we can consider the subfunctor of $\mathrm{Hom}(A,-)$ which is given by homomorphisms homomorphism $g : A \to B$ such that $gi$ factors through $B^*$. By general theorems a representing object exists and is usually denoted the localization $S^{-1} A$ when $i$ is understood. If $S'$ denotes the multiplicative closure of the image of $i$, then $S'^{-1} A = S^{-1} A$ (they satisfy the same universal property), thus we always may assume that $S$ is just a multiplicative closed subset of $A$. Let $\tau : A \to S^{-1} A$ be the universal homomorphism which maps $S$ to units. Then $S^{-1} A = \{\tau(a) \tau(s)^{-1} : a \in A, s \in S\}$ (they satisfy the same universal property). Let es write $\frac{a}{s} = \tau(a) \tau(s)^{-1}$. Thus every element in $S^{-1} A$ is some fraction $\frac{a}{s}$. Now when are two such fractions are equal? Of course we know this from the usual construction using equivalence classes of pairs $(a,s)$, but I want to avoid this construction and derive it this only with the universal property.

First observe that $\frac{a}{s} = \frac{a'}{s'}$ iff $\frac{sa'-sa'}{1}=0$. Thus it is enough to show $\frac{a}{1} = 0 \Rightarrow sa=0$ for some $s \in S$, i.e. that the kernel of $\tau : A \to S^{-1} A$ equals $I=\cup_{s \in S} \mathrm{Ann}(s)$. Since $S$ is multiplicative, $I$ is an ideal, and we way may replace $A$ by $A/I$, where localization commutes with quotients because of universal properties. Thus we may assume that $I=0$, i.e. that $S$ consists of regular elements. Our task is then to show that $A \to S^{-1} A$ is injective. But a priori we only know by the universal property that the kernel of $A \to S^{-1} A$ equals the intersection of all kernels of homomorphisms $A \to B$ which map $S$ to units. Of course we have to use this and construct some specific $A \to B$, but I hope that we can either avoid some nasty element construction of $B$ or that we can just use a ring which is built up out of $S^{-1} A$, but can be used to show that the kernel is zero- therefore to find a kind of self-referential proof. The examples above suggest that this might be possible after all.

Motivation. I hope that a categorical proof makes the usual construction of the localization (via equivalence classes of pairs) redundant. Note also that it is rather nasty to prove all the details (equivalence relation, well-defined addition, well-defined multiplication, universal property) with the usual construction. Also the definition $(a,s) \sim (a',s') \Leftrightarrow \exists t : ts'a=tsa'$ is not motivated at all there. A categorical proof should show in particular that this is the right choice, and not some random definition which turns out to be correct only after some computationafterwards. But of course I don't claim that a categorical construction of the localization is the easiest or best one.

On the other hand, localization could be seen just as a toy example for other, more involved examples, where it is not clear at all how to understand some representing object of some functor, which exists my general nonsense.

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