Since $\varphi$ (Euler's totient function) is the Dirichlet product of $\mu$ and $N$, we can use properties of inverses to show that
$$\varphi^{-1} (n) = \sum_{d|n} d \mu(d).$$
Furthermore, for multiplicative functions $f$, we can find that
$$\sum_{d|n} \mu(d) f(d) = \prod_{p|n} (1 - f(p)).$$
As a result, we can find that
$$\varphi^{-1} (n) = \prod_{p|n} (1 - p).$$
I understand how this formula for $\varphi^{-1}$ was derived, but I have no idea what it actually means.
When $\varphi(6) = 6 \cdot \frac{1}{2} \cdot \frac{2}{3} = 2$, but why isn't $\varphi^{-1} (2) = 6$? Instead, $\varphi^{-1} (2) = (1 - 2) = -1$. I also understand that the $\varphi$ function is not injective, so how do I relate $-1$ to all $n$ such that $\varphi(n) = 2$?