$\operatorname{Frac}R/p\cong R_p/pR_p$ 
Let ${\mathfrak p}$ be a prime ideal of a ring $R$. Show that there is a canonical isomorphism $\operatorname{Frac}R/\mathfrak p\to R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p}$.

Since this exercise is from Bosch's book, that encourages the use of universal arrows, I wanted to know if this exercise can be done without working with the elements. I tried as follows.
$K:=\operatorname{Frac}R/\mathfrak p$ and $L:=R_{\mathfrak p}/{\mathfrak p}R_{\mathfrak p}$. Let $i:R\to R_p$, $u:R_{\mathfrak p}\to L$ be respectively the localization map and the projection, and set $g:=u\circ i$; proposition 1.2.5 (iii) says $\operatorname{ker}g={\mathfrak p}$, so we have that $g =j\circ v$, where $v:R\to R/{\mathfrak p}$ is the projection and $j:R/{\mathfrak p}\to L$ is the unique injection coming from the universal property of kernels. But $L$ is a field so, if $l:R/p\to K$ is the localization map, there is a unique map $f:K\to L$ (injective, as $K$ is a field) such that $f\circ l=j$. Now it should be left to prove  that $f$ is surjective. The map $l\circ v:R\to K$ is equal to $h\circ i$, for a unique $h:R_{\mathfrak p}\to K$; plus it's easy to see that $g=f\circ h\circ i$. The map $w$ such that $g=w\circ i$ is unique, and so $u=f\circ h$; being $u$ surjective, also $f$ is.
Did I make any mistake? I don't know how to  draw  such complex diagrams in latex, so I uploaded a picture; I know it is not a great choice but I tried to be as clear as possible so that the picture shouldn't be necessary to understand the post.

 A: Your proof is correct. It can, however, be improved by constructing explicitly the inverse to $f$, and also dropping the requirements that you are taking the quotient by a prime ideal or localizing at its complement.
Explcitily, let $i\colon R\twoheadrightarrow S^{-1}R$ be the localization of $R$ at a subset $S\subseteq R$, and let $u\colon S^{-1}R\twoheadrightarrow S^{-1}R/i(I)=L$ be the quotient of $S^{-1}R$ by the ideal in $S^{-1}R$ generated by the set $i(I)$. Then by definition (no need to quote a proposition) $g=u\circ i$ satisfies $g(r)=u(i(r))=0$ , i.e $I\subseteq\ker g$. Consequently, $g=j\circ v$ for a unique ring homomorphism $j\colon R/I\to L$, where $v\colon R\twoheadrightarrow R/I$ is the quotient map.
Similarly, $l\colon R/I\twoheadrightarrow (S/I)^{-1}(R/I)=K$ is the localization of $R/I$ at the image $v(S)$ of $S$ under $v\colon R\twoheadrightarrow R/I$, then $l\circ v(s)$ is invertible for each $s\in S$, whence there $l\circ v=h\circ i$ for a unique ring homomorphism $h\colon S^{-1}R\to K$.
Observe that we now have two ways of completing the morphisms $R/I\leftarrow R\to S^{-1}R$ to commutative squares: via $l\colon R/I\to K\leftarrow S^{-1}R:h$ and via $j\colon R/I\to L\leftarrow S^{-1}R:u$.
It turns out that both commutative squares satisfy the univeral property of being pushout squares. That is, for any completion of $R/I\leftarrow R\to S^{-1}R$ to a commutative square by $R/I\to T\leftarrow S^{-1}R$, there exist unique ring homomorphism $K\to T$ and $L\to T$ such that $R/I\to T$ factors as $R/I\to K\to T$ and $R/I\to L\to T$ and $S^{-1}R\to T$ factors as $S^{-1}R\to K\to T$ and $S^{-1}R\to L\to T$.
The construction of $K\to T$ is basically the same argument as the one with which you construct $f\colon K\to L$, while the construction of $L\to T$ is similar but swapping the order of the universal properties used (quotient first, then localization for localization first, then quotient).
Once you prove these universal properties, you obtain $f\colon K\to L$ and $f^{-1}\colon L\to K$ whose composites have to be the identity morphisms, and hence are isomorphisms.
