Divisor Sum Function is Multiplicative/Proof 2

Proof
Let $a, b$ be coprime integers.

Because $a$ and $b$ have no common divisor, the divisors of $a b$ are integers of the form $a_i b_j$, where $a_i$ is a divisor of $a$ and $b_j$ is a divisor of $b$.

That is, any divisor $d$ of $a b$ is in the form:
 * $d = a_i b_j$

in a unique way, where $a_i \divides a$ and $b_j \divides b$.

We can list the divisors of $a$ and $b$ as
 * $1, a_1, a_2, \ldots, a$

and:
 * $1, b_1, b_2, \ldots, b$

and so their divisor sums are:


 * $\ds \map {\sigma_1} a = \sum_{i \mathop = 1}^r a_i$


 * $\ds \map {\sigma_1} b = \sum_{j \mathop = 1}^s b_j$

Consider all divisors of $a b$ with the same $a_i$.

Their sum is:

Summing for all $a_i$: