Prove Localization in CRing is Epimorphism Give a commutative ring $R$ and a multiplicative subset $S$ of $A$, we have the normal localization map $\lambda_S: R \rightarrow S^{-1}R$. How does one prove that this is an epimorphism? 
So given some $f, g: S^{-1}R \rightarrow C$ such that $f \circ \lambda_S = g \circ \lambda_S$, I can see that $f$ and $g$ agree on elements with denominator 1. But I need to show that they agree on all elements in $S^{-1}R$. 
 A: If $s\in S$, then you already have that $f(\frac{s}{1}) = g(\frac{s}{1})$. Now note that this is a unit, hence its image is a unit in $C$. Units have unique inverses, and the image of the inverse is the inverse of the image, so
$$f\left(\frac{1}{s}\right) = \left(f\left(\frac{s}{1}\right)\right)^{-1} = \left( g\left(\frac{s}{1}\right)\right)^{-1} = g\left(\frac{1}{s}\right).$$
Now, given an arbitrary element $\frac{r}{s}\in S^{-1}R$, we have $f(\frac{r}{s}) = f(\frac{r}{1}\frac{1}{s}) = f(\frac{r}{1})f(\frac{1}{s}) = g(\frac{r}{1})g(\frac{1}{s}) = g(\frac{r}{s})$.
But in fact you do not need to assume that $S$ includes $1$, or that $C$ has a $1$, or that the maps must be unital. The conclusion follows using a Zigzag: given $t\in S$, the canonical map $\lambda_S$ given by $\lambda_S(r) = \frac{rt}{t}$ is a well-defined ring homomorphism that does not depend on the choice of $t$, and we have 
$$\begin{align*}
f\left(\frac{r}{s}\right) &= f\left(\frac{1}{s}\frac{rt}{t}\right)\\
&= f\left(\frac{1}{s}\right)f\left(\lambda_S(r)\right)\\
&= f\left(\frac{1}{s}\right)g\left(\lambda_S(r)\right)\\
&= f\left(\frac{1}{s}\right)g\left(\frac{rt}{t}\right)\\
&= f\left(\frac{1}{s}\right)g\left(\frac{srt}{st}\right)\\
&= f\left(\frac{1}{s}\right)g\left(\frac{st}{t}\right)g\left(\frac{r}{s}\right)\\
&= f\left(\frac{1}{s}\right)g\left(\lambda_S(s)\right)g\left(\frac{r}{s}\right)\\
&= f\left(\frac{1}{s}\right)f\left(\lambda_S(s)\right)g\left(\frac{r}{s}\right) \\
&= f\left(\frac{1}{s}\right)f\left(\frac{st}{t}\right)g\left(\frac{r}{s}\right)\\
&= f\left(\frac{st}{st}\right)g\left(\frac{r}{s}\right)\\
&= f\left(\frac{t}{t}\right)g\left(\frac{r}{s}\right)\\
&= f\left(\lambda_S(1)\right)g\left(\frac{r}{s}\right)\\
&= g\left(\lambda_S(1)\right)g\left(\frac{r}{s}\right)\\
&= g\left(\frac{t}{t}\right)g\left(\frac{r}{s}\right)\\
&= g\left(\frac{tr}{ts}\right) = g\left(\frac{r}{s}\right).
\end{align*}$$
It's the same as the argument to show that two (not necessarily unital) ring homomorphisms from $\mathbb{Q}$ that agree on $\mathbb{Z}$ must in fact be equal. See this answer
(You can even avoid assuming that $R$ itself has a $1$ by replacing $\frac{1}{s}$ with $\frac{t}{st}$ throughout.)
