Inverse of completely multiplicative function(s) If $f$ is completely multiplicative, prove that
$$(f \cdot g)^{-1}=f \cdot g^{-1},$$
for every arithmetical function $g$ with $g(1) \neq 0$. 
My main problem with this is that I can't even see why it is true! As far as I understand it, we have something like
$$(f \cdot g)^{-1}=f^{-1} \cdot g^{-1}= \mu f \cdot g^{-1},$$
since $f$ is completely multiplicative. But this can't be right, I don't see how we could get $\mu f$ to be $f$. I'm almost certain that I have confused something with something else, but luckily I have this wonderful community to set me straight! So my question is, what am I doing wrong?
EDIT: Copied the question verbatim from the book (Apostol, Introduction to Analytic Number Theory, p.49)
 A: What is unclear to me is how you go from $(fg)^{-1}$ to $f^{-1}g^{-1}$. Given that we are dealing with Dirichlet inverses, this is not obvious to me. In fact, it's not true.  For example, if $f=g=u$, then $(fg)^{-1} = u^{-1} = \mu$, but $f^{-1}g^{-1} = \mu\mu\neq \mu$. 
So that is where the error in your reasoning comes, I think.
Remember that since ${}^{-1}$ represents the Dirichlet inverse, which is the inverse of convolution, you know that
$$(f*g)^{-1} = g^{-1}*f^{-1} = f^{-1}*g^{-1},$$
but you are trying to apply this to the pointwise product, which is incorrect.

As to proving the desired identity, if $h$ is an arithmetic function with $h(1)\neq 0$, then $h^{-1}$ is given by:
$$ h^{-1}(1) = \frac{1}{h(1)},\qquad h^{-1}(n) = \frac{-1}{h(1)}\sum_{{d|n},\,{d\lt n}}h\left(\frac{n}{d}\right)h^{-1}(d),\quad n\gt 1.$$
If $h$ is completely multiplicative then you have $f^{-1}(n)=\mu(n)f(n)$ for all $n\geq 1$.
So we verify that the identity holds for $f$ completely multiplicative and $g$ an arithmetic function with $g(1)\neq 0$.
At $n=1$, we have
$$(f\cdot g)^{-1}(1) = \frac{1}{f(1)g(1)} = \frac{1}{g(1)}$$
(since $f(1)=1$ must hold); and
$$(f\cdot g^{-1})(1) = f(1)g^{-1}(1) = \frac{1}{g(1)}.$$
If $n\gt 1$ and the result hold for all integers smaller than $n$, then we have:
$$\begin{align*}
(f\cdot g)^{-1}(n) &= \frac{-1}{(f\cdot g)(1)} \sum_{d|n, d\lt n}(f\cdot g)\left(\frac{n}{d}\right)(f\cdot g)^{-1}(d)\\
&= \frac{-1}{g(1)} \sum_{d|n\,d\lt n} f\left(\frac{n}{d}\right)g\left(\frac{n}{d}\right) f(d)g^{-1}(d)\\
&= \frac{-1}{g(1)}\sum_{d|n\,d\lt n}\frac{f(n)}{f(d)}g\left(\frac{n}{d}\right)f(d)g^{-1}(d)\\
&= \frac{-f(n)}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac {n}{d}\right)g^{-1}(d)\\
&= f(n)\left(-\frac{1}{g(1)}\sum_{d|n\,d\lt n}g\left(\frac{n}{d}\right)g^{-1}(d)\right)\\
&= f(n)g^{-1}(n)\\
&= (f\cdot g^{-1})(n).
\end{align*}$$
We used the fact that $f$ is completely multiplicative to get that $f\left(\frac{n}{d}\right) = \frac{f(n)}{f(d)}$. Equivalently, you can note that $f\left(\frac{n}{d}\right)f(d) = f(n)$. 
This proves the first part of the exercise. 
The second part of the exercise asks you to show that if $f$ is multiplicative and
$$(f\cdot \mu^{-1})^{-1} = f\cdot \mu$$
holds, then $f$ is completely multiplicative.
Since $f$ is completely multiplicative if and only if $f^{-1} = \mu\cdot f$, you are being asked to show that $f\cdot \mu^{-1} = f$. That is, you need to show that $\mu^{-1}(n) = 1 = u(n)$.
But we know this, since $\sum_{d|n}\mu(d) = I(n)$ (the identity of the convolution), so $u$ and $\mu$ are inverses of each other. 
A: Möbius function $\mu(n)$ is defined as follows:


*

*$\mu(n) = 1$ if $n$ is a square-free positive integer with an even number of prime factors.

*$\mu(n) = −1$ if $n$ is a square-free positive integer with an odd number of prime factors.

*$\mu(n) = 0$ if $n$ is not square-free
This means that your statement is true only if $n$ is a square-free positive integer with an even number of prime factors.
