Divisor Sum Function is Multiplicative

Theorem
The sigma function:
 * $\displaystyle \sigma: \Z^*_+ \to \Z^*_+: \sigma \left({n}\right) = \sum_{d \backslash n} d$

is multiplicative.

Proof
Let $I_{\Z^*_+}: \Z^*_+ \to \Z^*_+$ be the identity function:
 * $\forall n \in \Z^*_+: I_{\Z^*_+} \left({n}\right) = n$.

Thus we have:
 * $\displaystyle \sigma \left({n}\right) = \sum_{d \backslash n} d = \sum_{d \backslash n} I_{\Z^*_+} \left({d}\right)$.

But from Identity Function is Completely Multiplicative, $I_{\Z^*_+}$ is multiplicative.

The result follows from Sum Over Divisors of Multiplicative Function.