Theorem   An arithmetic function $f$ has a convolution inverse if and only if $f(1) \neq 0$

Proof. If $f$ has a convolution inverse $g$ then $f*g=\varepsilon$ where $\varepsilon$ denotes the convolution identity function. Thus, $1=\varepsilon(1)=(f*g)(1)=f(1)g(1)$ and it follows that $f(1) \neq 0$

Conversely, if $f(1) \neq 0$ then an arithmetic function $g$ must be constructed such that $(f*g)(n)=\varepsilon(n)$ for all $n \in \mathbb{N}$ This will be done by induction on $n$

Since $f(1) \neq 0$ we have that $\displaystyle \frac{1}{f(1)} \in \mathbb{C}$ Define $\displaystyle g(1)=\frac{1}{f(1)}$

Now let $k \in \mathbb{N}$ with $k>1$ and $g(1), \dots, g(k-1)$ be such that $(f*g)(n)=\varepsilon(n)$ for all $n \in \mathbb{N}$ with $n<k.$ Define

$$g(k)=-\frac{1}{f(1)}\sum_{d|k \text{ and } d<k} f \left( \frac{k}{d} \right) g(d).$$

Then