Exercise 4.2.10 (If $f$ is totally multiplicative, $F$ needs not also be totally multiplicative)

Proof. Take $f:{\mathbb{N}}^{\text{∗}}\to {\mathbb{N}}^{\text{∗}}$ defined by $f(n)=1$. Then $f$ is totally multiplicative. But then $F(n)=d(n)$ for all $n\in {\mathbb{N}}^{\text{∗}}$, and $d$ is not totally multiplicative: for instance, $3=d(9)\ne d{(3)}^2=4$.

If $f$ is totally multiplicative, $F$ needs not also be totally multiplicative, where $F$ is defined by $F(n)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}f(d)$ for all $n\in {\mathbb{N}}^{\text{∗}}$. □