= Solution
An <arithmetic function> $f$ is <multiplicative arithmetic function>[multiplicative] when $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are <coprime integers>[coprime]. If $m$ and $n$ are coprime, every divisor $d$ of $mn$ has a unique factorization $d=d_1d_2$ with $d_1\mid m$ and $d_2\mid n$. Hence the <Dirichlet convolution> of multiplicative functions satisfies
$$
\begin{aligned}
(f\star g)(mn)
&=\sum_{d_1\mid m}\sum_{d_2\mid n}
f(d_1d_2)g\!\left(\frac m{d_1}\frac n{d_2}\right)\\
&=(f\star g)(m)(f\star g)(n),
\end{aligned}
$$
and $(f\star g)(1)=1$, so it is multiplicative.
Let $\varepsilon$ be the <identity for Dirichlet convolution>, with $\varepsilon(1)=1$ and $\varepsilon(n)=0$ for $n>1$. The <Möbius function> satisfies $\mu\star1=\varepsilon$. Therefore, using commutativity and associativity,
$$
f=\mu\star g
\quad\Longrightarrow\quad
f\star1=g\star(\mu\star1)=g\star\varepsilon=g.
$$
This is the <Möbius inversion formula>.
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