Proof of the Dirichlet convolution formula over ordered pairs

Proof

1. Let $f$ and $g$ be arithmetic functions and $n$ a positive integer. Let $D$ be the set of positive divisors of $n$, and $P$ the set of ordered pairs $(a, b)$ of positive integers with $ab = n$.
2. By the definition of the Dirichlet convolution, $(f * g)(n) = \sum_{d \in D} f(d) \, g(n/d)$.
3. For $d \in D$, the number $n/d$ is a positive integer and $d \cdot (n/d) = n$. So $\beta(d) = (d, n/d)$ defines a map $\beta : D \to P$.
4. For $(a, b) \in P$, the equation $n = ab$ shows $a \in D$, and $n/a = b$. So $\beta(a) = (a, b)$, and $\beta$ is surjective.
5. If $\beta(d) = \beta(d')$, then the first coordinates give $d = d'$. So $\beta$ is injective, hence a bijection.
6. Reindexing the finite sum of step 2 along $\beta$ gives $(f * g)(n) = \sum_{(a, b) \in P} f(a) \, g(b)$.