The Möbius function on the divisor poset of $12$ and its agreement with $\mu(1),\mu(2),\mu(3),\mu(4),\mu(6),\mu(12)$

The positive divisors of $12$ are $1,2,3,4,6,12$. In the divisibility order (The divisibility poset of positive integers), the covers are

$1\lessdot2,\quad1\lessdot3,\quad2\lessdot4,\quad2\lessdot6,\quad3\lessdot6,\quad4\lessdot12,\quad6\lessdot12.$

For every comparable $a\mid b$, The number-theoretic Möbius function is the poset Möbius function of divisibility: $\mu(n)=\mu_{\mid}(1,n)$ gives $\mu_{\mid}(a,b)=\mu(b/a)$. Hence the full table is obtained from

$\mu(1)=1,\quad \mu(2)=-1,\quad \mu(3)=-1,\quad \mu(4)=0,\quad \mu(6)=1,\quad \mu(12)=0.$

In particular, the row from $1$ is $(1,-1,-1,0,1,0)$ in the divisor order listed above. The recurrence checks the less immediate values: $1-1-1+\mu(1,6)=0$ gives $\mu(1,6)=1$, and $1-1-1+0+1+\mu(1,12)=0$ gives $\mu(1,12)=0$ (The Möbius recurrence: $\mu_P(x,x)=1$ and both interval sums of $\mu_P$ vanish when $x<y$).