So suppose we are given some other integer \\(c\\), and arrows \\(\tau_1: c \rightarrow m\\) , \\(\tau_2: c \rightarrow n\\).
That means that \\(c\\) divides \\(m\\) and \\(c\\) divides \\(n\\), which is to say that \\(c\\) is a common divisor of \\(m\\) and \\(n\\).
Moreover, gcd(m,n) is the _greatest_ common divisor of \\(m\\) and \\(n\\), which means that c divides gcd(m,n).
So there is an arrow \\(h: c \rightarrow gcd(m,n)\\).
Recall that the universal property required the existence of precisely such an arrow \\(h\\). Check!