Let $R$ be a commutative unital ring and let $D$ be a multiplicative subset of $R$ containing $1$. Define $D^{-1}R$ in the usual way (as the quotient $(R\times D)/\sim$ with $(r,d) \sim (s, e) \iff x(er - ds) = 0$ for some $x \in D$) and call these equivalence classes $r/d$.
The construction comes from Dummit and Foote; there is a commutative ring $D^{-1}R$ and a ring homomorphism $\pi : R \to D^{-1}R$ satisfying the universal property: for any homomorphism $\psi:R \to S$ of commutative rings sending $1$ to $1$ and such that $\psi(d)$ is a unit in $S$ for every $d \in D$ there is a $\textbf{unique}$ homomorphism $\Psi:D^{-1}R \to S$ such that $\Psi \circ \pi = \psi.$
The existence is fairly easy to understand but the uniqueness seems oddly handwavy. In the text, it is remarked that $\Psi$ is unique because every element of $D^{-1}R$ can be written as a product $(r/1)(d/1)^{-1}$. The text then goes on to explain that the value of $\Psi$ on an element of the form $x/1$ is uniquely determined by $\psi$, since $\Psi(x/1) = \Psi(\pi(x)) = \psi(x)$. Finally, the text explains that the value of $\Psi$ on any element $u^{-1}$ is uniquely determined by $\Psi(u)$ for any unit $u$.
I don't see any explanation in the conclusions that these values are uniquely determined; the text simply SAYS that they are uniquely determined without any actual proof (it seems).
Am I missing something that is obvious? I understand that homomorphisms send units to units so this seems reasonable, but simply saying that the images of these elements under $\Psi$ are uniquely determined seems uniquely handwavy.
In order to prove uniqueness, let $\Psi,\Psi':D^{-1}R\to S$ be two ring homomorphisms such that $\Psi'\circ\pi=\psi$ and $\Psi\circ\pi=\psi$. Let $x\in D^{-1}R$. Then there exists $r\in R$ and $d\in D$ such that $x=r/d$ hence $(d/1)x=(r/1)$ that's $\pi(d)x=\pi(r)$. Then $\Psi\circ\pi(d)\Psi(d)=\Psi\circ\pi(r)$ from which $\psi(d)\Psi(x)=\psi(r)$ in $S$. Similarly, $\psi(d)\Psi'(x)=\psi(r)$, thus $\psi(d)\Psi'(x)=\psi(d)\Psi(x)$. Since $\psi(d)$ is invertible in $S$ (recall $d\in D$), we get $\Psi(x)=\Psi'(x)$. By the arbitrariety of $x$, we get $\Psi=\Psi'$.