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!

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!