Let $\mathbb Z^+$ be the set of all positive integers and
let $\mathbb C$ be the set of all complex numbers.
A function $f:\mathbb Z^+ \to \mathbb C$ is called multiplicative if
$f(1) = 1$ and $f(ab) = f(a)f(b)$ for all relatively prime $a$ and $b$.
Let $p_i$ represent the $i^\text{th}$ prime number. Then every positive integer $x$ can be written uniquely as an infinite product
$\displaystyle x = \prod_{i=1}^\infty p_i^{\alpha_i}$
where we require that all but a finite number of the $\alpha_i$ be equal to $0$.
It follows that, if $f$ is a multiplicative function, then
$\displaystyle f(x) = \prod_{i=1}^\infty f(p_i^{\alpha_i})$.
The convolution of two multiplicative functions, say $f$ and $g$,
is defined as $f*g$ where
$$(f*g)(n) = \sum_{ab=n} f(a)g(b)$$
The set $\mathcal F$ of all multiplicative functions is an abelian group with respect to the convolution operator.
Two important multiplicative functions are $\epsilon$ and $\mathbf 1$ defined as
$$\epsilon(n) = \left\{
\begin{array}{ll}
1 & \text{If}\; n = 1\\
0 & \text{If}\; n \ne 1
\end{array} \right.$$
and
$$ \mathbf 1(n) = 1$$
It is easy to prove that $\epsilon$ is the multiplicative identity of the group $[\mathcal F, *]$.
For any multiplicative function $f$, note that
$$\displaystyle (f*\mathbb 1)(n) = \sum_{a|n} f(a)$$
$\mathbf{Theorem. }$ Let $f$ and $g$ be multiplicative functions. Define
$F = f*\mathbf 1$ and $G = g*\mathbf 1$. Define $h = f*g$ and
$H = h*1$. Then $H(n) = F(n)G(n)$ for all positive integers $n$.
In a way, $\; f*1$ behaves very much like a Fourier transform of $\; f$.
It follows that there are times when we know what $F = f*\mathbf 1 $ is and we need to know what $f$ is. This is where $\mu$, the Mobius inversion function, comes to the rescue. $\mu$ is defined as the inverse of $\mathbf 1$. That is
$$\mathbf 1 * \mu = \epsilon$$
We make a few computations. Let $p$ be a prime number and let $\alpha$ be a non negative integer.
\begin{align}
(1*\mu)(1) &= \epsilon(1)\\
\mu(1) &= 1
\end{align}
\begin{align}
(1*\mu)(p) &= \epsilon(p)\\
1 + \mu(p) &= 0\\
\mu(p) &= -1
\end{align}
\begin{align}
(1*\mu)(p^2) &= \epsilon(p^2)\\
1 + \mu(p)+\mu(p^2) &= 0\\
\mu(p^2) &= 0\end{align}
\begin{align}
(1*\mu)(p^3) &= \epsilon(p^3)\\
1 + \mu(p)+\mu(p^2)+\mu(p^3) &= 0\\
\mu(p^3) &= 0\end{align}
We see that
$$\mu(p^\alpha) = \left\{
\begin{array}{rl}
1 & \text{If}\; \alpha = 0\\
-1 & \text{If}\; \alpha = 1\\
0 & \text{If}\; \alpha \ge 2
\end{array} \right.$$
You can infer the usual definition of $\mu$ from this and the fact that $\mu$ is a multiplicative function.