From Mac Lane's Categories for the Working Mathematician:
Let $S: D \rightarrow C$.
A universal arrow arrow from $S$ to $c$ is a pair $\langle r,v \rangle$ consisting of an object $r \in D$ and an arrow $v : Sr \rightarrow c$ with codomain $c$ such that to every pair $\langle d,f \rangle$ with $f : Sd \rightarrow c$ there is a unique $f' : d \rightarrow r$ with $f = v \circ S f'$.
He then says:
For the projections $p : a \times b \rightarrow a$ and $q : a \times b \rightarrow b$, for any other pair of arrows $f : c \rightarrow a$ and $g : c \rightarrow a$ to $a$ and $b$ there is a unique $h: c \rightarrow a \times b$ with $ph = f, qh = g$. This can be made into a universal arrow by introducing a diagonal functor $\Delta : C \rightarrow C \times C$ with $\Delta c = \langle c, c \rangle$. Then the pair $\langle f,g \rangle : \Delta c \rightarrow \langle a, b \rangle$ in $C \times C$, and $\langle p, q \rangle$ is a universal arrow from $\Delta$ to the object $\langle a, b \rangle$.
How is $\langle p,q \rangle$ a universal arrow from $\Delta$ to $\langle a, b \rangle$?
By the definition Mac Lane defines if it's a universal arrow from $\Delta$ to $\langle a,b \rangle$ then it's an object $r \in C$ and an arrow $v : \Delta r \rightarrow \langle a, b \rangle$ such that for every pair $\langle d, f \rangle$ with $f : \Delta d \rightarrow \langle a, b \rangle$ there's a unique $f' : d \rightarrow r$ with $f = v \circ \Delta f'$.
But this doesn't seem to match what he wrote in the second paragraph included above about projections. What is the object $r$ and what is the arrow $v$ and an object $d$ for the universal arrow "$\langle p,q \rangle$"?
Best Answer
The $r$ is $a\times b$, and $\langle p,q\rangle:\Delta(a\times b)\to \langle a,b\rangle$ is the universal arrow in question. For any other $f=\langle f_1,f_2\rangle:\Delta d\to \langle a, b\rangle$ in $C\times C$, the unique $v:d\to a\times b$ in $C$ is precisely the map usually labeled $\langle f_1,f_2\rangle:d\to a\times b$ (the double duty on the angle brackets is a little unfortunate here).