[proofplan] Fix coprime positive integers $m, n$ and compute $g(mn)$ directly. The key combinatorial fact is that divisors of $mn$ are in bijection with pairs $(d_1, d_2)$ where $d_1 \mid m$ and $d_2 \mid n$, via $(d_1, d_2) \mapsto d_1 d_2$; this bijection requires and uses coprimality. Under this bijection, the sum $\sum_{d \mid mn} f(d)$ becomes a double sum $\sum_{d_1 \mid m} \sum_{d_2 \mid n} f(d_1 d_2)$. Since $d_1 \mid m$ and $d_2 \mid n$ with $(m,n) = 1$ forces $(d_1, d_2) = 1$, the multiplicativity of $f$ gives $f(d_1 d_2) = f(d_1) f(d_2)$, after which the double sum factors as $g(m)g(n)$. [/proofplan]

[step:Set up the divisor-pair bijection for coprime arguments]Fix positive integers $m, n \geq 1$ with $(m, n) = 1$. We claim that the map \begin{align*} \phi: \{(d_1, d_2) \in \mathbb{N} \times \mathbb{N} : d_1 \mid m,\, d_2 \mid n\} &\to \{d \in \mathbb{N} : d \mid mn\} \\ (d_1, d_2) &\mapsto d_1 d_2 \end{align*} is a bijection. First, $\phi$ is well-defined: if $d_1 \mid m$ and $d_2 \mid n$, then $d_1 d_2 \mid mn$ by direct multiplication. For injectivity and surjectivity, let $d \mid mn$. Set $d_1 = (d, m)$ and $d_2 = (d, n)$. Then $d_1 \mid m$, $d_2 \mid n$, and we show $d_1 d_2 = d$. Since $(m, n) = 1$, we have $(d_1, d_2) \mid (m, n) = 1$, so $(d_1, d_2) = 1$. Now $d_1 \mid d$ and $d_2 \mid d$ with $(d_1, d_2) = 1$ imply $d_1 d_2 \mid d$. Conversely, write $d = d_1 d_2'$ where $d_2' = d / d_1$. We show $d_2' = d_2$. Since $d \mid mn$ and $d_1 \mid m$, write $m = d_1 m'$ and $mn = d_1 m' n$; then $d_2' = d / d_1 \mid m' n$. Because $(d_1, n) \mid (m, n) = 1$, we have $(m', n) \mid (m, n) = 1$ (from $m' \mid m$) — more carefully, $(d_2', m') \mid (d_2', m) = (d, m)/d_1 \cdot \text{something}$; we avoid this and instead observe: \begin{align*} d &= d_1 d_2',\quad d_2' \mid mn/d_1 = m'n, \quad (d_2', m) \mid (d_2', d_1 m') = (d_2', d_1) \cdot (d_2'/(d_2',d_1), m') \ldots \end{align*} We take a cleaner route. Observe $d_2' \mid d$ and $d_1 \mid d$ with $(d_1, d_2') \mid (m, n) = 1$, because $d_1 \mid m$ and $d_2' \mid n$ (which we now verify). From $d \mid mn$ and $d_1 = (d, m)$, a standard fact on coprime factorisations gives $d = (d, m)(d, n)$ when $(m, n) = 1$: indeed, $d_1 d_2 \mid d$ (shown above) and $d \mid d_1 d_2$ since $d \mid mn = (d_1 \cdot m/d_1)(d_2 \cdot n/d_2)$ with $(m/d_1, n/d_2) = 1$, combined with $(d, m/d_1) \mid d_2$ and $(d, n/d_2) \mid d_1$ gives $d = d_1 d_2$. Thus $\phi$ is surjective with inverse $d \mapsto ((d, m), (d, n))$. For injectivity: if $d_1 d_2 = d_1' d_2'$ with $d_1, d_1' \mid m$ and $d_2, d_2' \mid n$, then taking $\gcd$ with $m$ gives $d_1 = (d_1 d_2, m) = (d_1' d_2', m) = d_1'$ (using $(d_2, m) = 1$ since $d_2 \mid n$ and $(m, n) = 1$), and likewise $d_2 = d_2'$.[/step]

[guided]The central combinatorial identity in this proof is the divisor-pair bijection: divisors of a product of coprime integers factor uniquely into a divisor of each factor. Why is this true and why does coprimality matter? Explicitly, if $d \mid mn$ with $(m, n) = 1$, define \begin{align*} d_1 := (d, m), \qquad d_2 := (d, n). \end{align*} We claim $d = d_1 d_2$. Let us verify both divisibilities. **$d_1 d_2 \mid d$:** Both $d_1 \mid d$ and $d_2 \mid d$, and $(d_1, d_2) \mid (m, n) = 1$, so $(d_1, d_2) = 1$. Two coprime divisors of $d$ multiply to a divisor of $d$: if $d = d_1 u = d_2 v$, then $d_2 \mid d_1 u$, and since $(d_1, d_2) = 1$, $d_2 \mid u$, so $d_1 d_2 \mid d_1 u = d$. **$d \mid d_1 d_2$:** Write $m = d_1 m'$ and $n = d_2 n'$. Then $mn = d_1 d_2 m' n'$, and since $d \mid mn$, we have $d = d_1 d_2 q$ for some rational $q$, specifically $q = d/(d_1 d_2) \in \mathbb{N}$ (from the previous paragraph). So $q \mid m' n'$. But $(q, m) \mid (d/d_2, m)$; since $(d, m) = d_1$ and $(d_2, m) = 1$ (as $d_2 \mid n$ and $(m,n) = 1$), $(d/d_2, m) = (d, m) = d_1$. So $(q, m) \mid d_1$, but also $q = d/(d_1 d_2)$ with $d_1 \mid d$, so $(q, d_1) \mid 1$. Hence $(q, m) = 1$. Symmetrically $(q, n) = 1$, and thus $(q, mn) = 1$. But $q \mid m'n' \mid mn$, forcing $q = 1$. Therefore $d = d_1 d_2$. Conversely, if $d_1 \mid m$ and $d_2 \mid n$, then $d = d_1 d_2 \mid mn$, and $(d, m) = d_1$, $(d, n) = d_2$ (using coprimality of $d_1$ with $n$ and of $d_2$ with $m$, each inherited from $(m,n)=1$). The upshot: the map \begin{align*} \phi: \{(d_1, d_2) : d_1 \mid m,\, d_2 \mid n\} &\longrightarrow \{d : d \mid mn\}, \qquad (d_1, d_2) \mapsto d_1 d_2 \end{align*} is a bijection with inverse $d \mapsto ((d, m), (d, n))$. Coprimality $(m, n) = 1$ is essential — without it, a divisor of $mn$ could split into $d_1, d_2$ in multiple ways, and the bijection breaks down.[/guided]

[step:Convert the divisor sum into a double sum] Using the bijection $\phi$ of the preceding step, the sum defining $g(mn)$ reindexes as a double sum over pairs: \begin{align*} g(mn) = \sum_{d \mid mn} f(d) &= \sum_{\substack{d_1 \mid m \\ d_2 \mid n}} f(d_1 d_2). \end{align*} No divisor of $mn$ is counted twice, and none is missed, because $\phi$ is a bijection. [/step]

[step:Apply multiplicativity of $f$ to factor the integrand]For each pair $(d_1, d_2)$ with $d_1 \mid m$ and $d_2 \mid n$, we have $(d_1, d_2) \mid (m, n) = 1$, hence $(d_1, d_2) = 1$. Since $f$ is multiplicative, by definition this means $f(ab) = f(a) f(b)$ whenever $(a, b) = 1$. Applying this to $a = d_1$, $b = d_2$ gives \begin{align*} f(d_1 d_2) &= f(d_1) f(d_2). \end{align*} Substituting into the double sum: \begin{align*} g(mn) = \sum_{\substack{d_1 \mid m \\ d_2 \mid n}} f(d_1 d_2) &= \sum_{d_1 \mid m} \sum_{d_2 \mid n} f(d_1) f(d_2). \end{align*}[/step]

[guided]A function $f: \mathbb{N} \to \mathbb{C}$ is called multiplicative if $f(ab) = f(a) f(b)$ whenever $(a, b) = 1$. (Distinguish this from totally multiplicative, which drops the coprimality hypothesis.) We must check the coprimality hypothesis before we can apply multiplicativity of $f$. **Verifying $(d_1, d_2) = 1$:** For any common divisor $e$ of $d_1$ and $d_2$, we have $e \mid d_1 \mid m$ and $e \mid d_2 \mid n$, so $e \mid (m, n) = 1$. Hence $e = 1$, and $(d_1, d_2) = 1$. Therefore multiplicativity applies: $f(d_1 d_2) = f(d_1) f(d_2)$. Substituting into the double sum: \begin{align*} g(mn) &= \sum_{d_1 \mid m} \sum_{d_2 \mid n} f(d_1) f(d_2). \end{align*} Note that multiplicativity without the coprimality hypothesis would be much stronger, and it is precisely why pairwise coprimality of $m$ and $n$ is assumed in the definition of multiplicative $g$.[/guided]

[step:Separate the double sum and identify $g(m) g(n)$] Since $f(d_1)$ depends only on $d_1$ and $f(d_2)$ only on $d_2$, the double sum factors: \begin{align*} \sum_{d_1 \mid m} \sum_{d_2 \mid n} f(d_1) f(d_2) &= \left(\sum_{d_1 \mid m} f(d_1)\right) \left(\sum_{d_2 \mid n} f(d_2)\right) = g(m) \cdot g(n). \end{align*} Combining with the previous two steps gives \begin{align*} g(mn) &= g(m) g(n) \quad \text{for all } (m,n) = 1, \end{align*} which is exactly the statement that $g$ is multiplicative. [/step]