Divisor Sum Function is Multiplicative

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

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

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

The result follows from Sum Over Divisors of Multiplicative Function.