Question about universal property of the localization

Question

I have this doubt, that arises quite often when I work with localizations. Let $S\subset A$ be a multiplicatively closed set of a ring, and let $i:A\to S^{-1}A$ be the canonical homomorphism. Suppose that the ring homomorphism $f:A\to B$ induces a homomorphism $f':S^{-1}A\to B$ by the universal property of the localization. In general, to uniquely determine a homomorphism, it suffices to know how it is defined on the underlying set of its domain; however here the underlying set of $S^{-1}A$ is a set modulo an equivalence relation, say $X/\sim$. The fact that $f=f'\circ i$ tells me how $f'$ is necessarily defined on $X$ (uniqueness); since an homomorphism $S^{-1}A\to B$ must exist, can I conclude that $f'$ (that until now was a map $X\to B$) induces actually a well defined homomorphism $S^{-1}A\to B$, without making any check? Or a similar situation is that I define from nowhere a map $g:X\to B$, and it is such that $g\circ i=f$; can I conclude that $g$ is the unique homomorphism induced? (This seems the case of the second link below).

For example, I had this doubt in this question (https://math.stackexchange.com/questions/4346953/isomorphism-between-localizations), where I already made it explicit; also in the answer to this question (Induced map between localizations), it was proven that $\varphi$ is well defined, but not a homomorphism. Can you give me a clarification about this? Thanks in advance

Answer 1

Your interpretation of $S^{-1}A$ as equivalence classes is an artifact of your construction. The universal property guarantees that the $f'$ is well-defined on the equivalence classes.

In more detail, if $f\colon A\to B$ is such that for each $s\in A$ there exists $b_s\in B$ such that $f(s)b_s=1$, then we have a unique ring homomorphism $A\{x_s:s\in S\}\to B$ generated by $x_s\mapsto b_s$ and $a\mapsto f(a)$, where $A\{x_s:s\in S\}$ is the free ring expressions in variables $x_s$ with coefficients in $A$. Note that $f\colon A\to B$ factors through this ring homomorphism as $A\to A\{x_s:s\in S\}\to B$, where $A\to A\{x_s:s\in S\}$ sends $a\mapsto a1$.

The kernel of the ring homomorphism $A\{x_s:s\in S\}\to B$ includes the ideal generated by $\langle sx_s-1\rangle$. In fact, it is precisely that ideal, as the image of each $s\in A$ under the resulting quotient satisfies $sx_s=1$. But the ring homomorphisms $A\{x_s:s\in S\}\to B$ factors uniquely through the quotient by its kernel, whence its quotient is $S^{-1}A$ with $i\colon A\to A\{x_s:s\in S\}\to S^{-1}A$ and $f$ factoring uniquely as $f'i$ for $f'\colon S^{-1}A\to B$.

Notice that $f'$ is already defined (and a ring homomorphism) on the quotient, so on the set of equivalence classes that you're considering. In particular, you don't need to check it's well-defined again.

Answer 2

The universal property of localization is the following: Given some multiplicatively closed set $S\subset A$, then for any ring homomorphism $f:A\to B$, s.t. $f(S)\subset B^\times$ there is a unique ring homomorphism $f':S^{-1}A\to B$ s.t. $f' \circ i=f$. That this holds true needs a proof. So once you check that this property holds, i.e. check that $f'$ is a ring homomorphism, check that $f' \circ i=f$ and so on.

The uniqueness bit is quite easy to see. Suppose $g:S^{-1}A\to B$ is some ring homomorphism s.t. $g\circ i=f$, then

$$ g(\frac{a}{s})=g(\frac{a}{1}\cdot\frac{1}{s})=g(i(a))\cdot g(\frac{s}{1})^{-1}=g(i(a))\cdot g(i(s))^{-1}=f(a)\cdot f(s)^{-1},\quad \forall a\in A,s\in S,$$ so $g$ is unqiue.