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!