Lemma 23.4.13.
Assume as above that \(f^{-1}\) is multiplicative for inputs less than \(mn\text{,}\) with \(\gcd(m,n)=1\text{.}\) Then
\begin{equation*}
f^{-1}(mn)=-\sum_{\substack{(ac)(bd)=(m)(n)\\ab < mn,\; a\mid m, \; b\mid n}}f^{-1}(a)f^{-1}(b)f(c)f(d)\text{.}
\end{equation*}
Assume that \(m,n>1\) and coprime. By the definition of inverse, we have
\begin{equation*}
0=(f^{-1}\star f)(mn)=\left[\sum_{x<mn,\; xy=mn}\left(f^{-1}(x)f(y)\right)\right]+f^{-1}(mn)f(1)\text{.}
\end{equation*}
By assumption, every function in this expression (both \(f\) and \(f^{-1}\)) is multiplicative on the values in question, with the possible exception of \(f^{-1}(mn)\text{.}\)
We can use this effectively because each summand is for a divisor \(x\mid mn\text{,}\) which we can write as \(xy=mn\text{.}\) Since \(m\) and \(n\) are coprime, both \(x\) and \(y\) are themselves products of coprime divisors dividing \(m\) and \(n\) respectively.
So let \(x=ab\) and \(y=cd\text{,}\) where \(a,c\mid m\) and \(b,d\mid n\text{.}\) Then, as everything is multiplicative, \(f^{-1}(x)f(y)=f^{-1}(a)f^{-1}(b)f(c)f(d)\text{.}\)
Since by the previous lemma \(f(1)=1\text{,}\) we can subtract the summation from both sides of the equation whose left-hand side is zero at the beginning of this lemma’s proof, yielding
\begin{equation*}
f^{-1}(mn)=-\sum_{\substack{(ac)(bd)=(m)(n)\\ab < mn,\; a\mid m, \; b\mid n}}f^{-1}(a)f^{-1}(b)f(c)f(d)\text{.}
\end{equation*}
We now write all this in terms of things we already can evaluate.
If the sum in question were summed over every \(ab\leq mn\) instead of \(ab<mn\text{,}\) it would easily simplify as a product:
\begin{equation*}
\sum_{\substack{(ac)(bd)=(m)(n)\\a\mid m, \; b\mid n}}f^{-1}(a)f^{-1}(b)f(c)f(d)=\sum_{ac=m}f^{-1}(a)f(c)\sum_{bd=n}f^{-1}(b)f(d)
\end{equation*}
The sum in
Lemma 23.4.13 only lacks the term with
\(a=m,b=n\text{,}\) in fact. So
\begin{equation*}
\sum_{\substack{(ac)(bd)=(m)(n)\\ab < mn,\; a\mid m, \; b\mid n}}f^{-1}(a)f^{-1}(b)f(c)f(d)=
\end{equation*}
\begin{equation*}
\left[\sum_{ac=m}f^{-1}(a)f(c)\sum_{bd=n}f^{-1}(b)f(d)\right]-\left(f^{-1}(m)f^{-1}(n)f(1)f(1)\right)
\end{equation*}
Now we can plug this back into the previous characterization of \(f^{-1}(mn)\text{:}\)
\begin{equation*}
f^{-1}(mn)=- \left[\sum_{ac=m}f^{-1}(a)f(c)\sum_{bd=n}f^{-1}(b)f(d)-f^{-1}(m)f^{-1}(n)f(1)f(1)\right]
\end{equation*}
Since \(m,n>1\text{,}\) the individual sums may be rewritten as
\begin{equation*}
(f^{-1}\star f)(m)=I(m)=0=I(n)=(f^{-1}\star f)(n)
\end{equation*}
That means we achieve the desired result
\begin{equation*}
f^{-1}(mn)=f^{-1}(m)f^{-1}(n)f(1)f(1)=f^{-1}(m)f^{-1}(n)
\end{equation*}
This follows since \(u\) is multiplicative (trivially) and \(\mu=u^{-1}\text{.}\)