What does the following summation notation represent?
$\sum\limits_{d_1 \mid a, \; d_2\mid b}f(d_1d_2)=\sum\limits_{d_1\mid a }\sum\limits_{d_2 \mid b}f(d_1)f(d_1)=\sum\limits_{d_1\mid a}f(d_1)\sum\limits_{d_2 \mid b}f(d_2) = F(a)F(b)$.
In particular, the $$\sum\limits_{d_1 \mid a, \; d_2\mid b}f(d_1d_2)$$ part. Is it summing over all possible pairs $(d_1,d_2)$?
Context: This is from the proof of a lemma used to prove that the sum of divisors and number of divisors functions are multiplicative:
Lemma. The map $f(d_1,d_2)=d_1d_2$ between the two sets
$\{d_1\in \mathbb Z^+: d_1\mid a\} \times \{d_2\in\mathbb Z^+ : d_2\mid b\},$ and $\{d\in \mathbb{Z}^+ : d\mid ab\}$ is a bijection if $\gcd(a,b)=1$.
Proof of lemma. Surjectivity: Let $m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_d^{\alpha_d}$, $n=q_1^{\beta_1}q_2^{\alpha_2}\cdots q_3^{\alpha_3}$ (write moduli in their prime factorizations).
Since $\gcd(a,b)=1$, all $p$’s and $q$’s are distinct.
Now consider $d\mid (ab=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_d^{\alpha_d}q_1^{\beta_1}q_2{\beta_2}\cdots q_s^{\beta_s}$). Let $d_1$ contain all powers of $q$’s from $d$ and $d_2$ contain all powers of $p$’s. Then $d_1d_2=d$, $d_1\mid a$, and $d_2\mid b$.
Inj: (any d | ab has a unique preimage)
Suppose $d=d_1d_2=d_1'd_2'$ where $d_1,d_2\mid a$, $d_2,d_2' \mid b$. Now consider $d_1\mid (d_1d_2=d_1'd_2').$ Notice that since $d_1\mid a$ and $d_2'\mid b$, and $\gcd(a,b)=1$, we have $\gcd(d_1,d_2')=1$.
Then by Euclid’s Lemma, $d_1\mid d_1'$,. Analogously, we have $d_1'\mid d_1$ $\Rightarrow d_1=d_1' \Rightarrow d_2=d_2'.$ $\quad\square$
Proof of proposition (*). Based on the lemma above.
Consider $F(n)=F(ab)=\sum\limits_{d\mid ab}f(d)$. Since $\gcd(a,b)=1$, by the lemma above, we can split our divisors $d$ into $d_1$ and $d_2$. Thus, we have $\sum\limits_{d_1 \mid a, \; d_2\mid b}f(d_1d_2)=$ $\sum\limits_{d_1\mid a }\sum\limits_{d_2 \mid b}$$f(d_1)$$f(d_2)$$=\sum\limits_{d_1\mid a}f(d_1)\sum\limits_{d_2 \mid b}f(d_2) = F(a)F(b) \quad\quad\square$
Notation: Summing over $d\mid ab$ = summing over the set $\{d\in\mathbb Z^+: d\mid ab\}$.
Confusion is right. Put the proposition aside for the moment. That hardball is really just a distraction at this point.
You're summing over all possible pairs $(d_1, d_2)$ such that $d_1\mid b_1$ and $d_2\mid b_2$. This is pretty straightforward. Say $b_1=6, b_2=9$, for concreteness. Then we sum over $\{(1,1),(2,1),(3,1),(6,1),(1,3),(2,3),(3,3),(6,3),(1,9),(2,9),(3,9),(6,9)\}$.