I'm currently going through the proof of the GPY theorem that the EH conjecture implies that prime gaps are bounded, following the proof outlined here but an confused on page 24 where it says that $\displaystyle\frac{r}{\phi(r)}\sum_{n: r \mid n} \frac{y(n)}{\phi_\omega(n/r)} \sum_{d: d/r \mid n/r}{\mu(d/r)\frac{\omega^*(d/r)d/r}{\phi(d/r)} \omega(n/d)} = r \sum_{n: r\mid n}^\prime{\frac{y(n)}{\phi(n)}}$
We know that $\phi_\omega, \omega, \omega^*$ are all multiplicative, that $\phi$ is Euler totient function and $\mu$ is the mobius function, and that $\sum^\prime$ is the sum restricted to square-free integers. $\phi_\omega(p) = p -\omega(p)$ for primes p, extended multiplicatively onto composite numbers. $\omega(n)$ is defined to be to be the size of the set $\Omega(n)$ of congruence classes m (mod n) for which n divides some fixed polynomial $P(x)$ evaluated at m, i.e $ n \mid P(m)$ and $\omega^*(p) = \omega(p) -1$ on primes also extended multiplicatively. This feels like it should be a simple cancellation as the author does not elaborate any further. The two ideas were either that this looks a lot like a convolution, and the second was that the totient function has the following property
The following derivation would work if $y(n)$ is zero when $n$ is non-squarefree (which is often the case in sieve problems):
$$ \begin{aligned} &\sum_{d:d/r|n/r}\mu(d/r){\omega^*(d/r)d/r\over\phi(d/r)}\omega(n/d)=\sum_{t|n/r}\mu(t){\omega^*(t)t\over\phi(t)}\omega\left(n/r\over t\right) \\ &=\omega\left(\frac nr\right)\sum_{t|n/r}\mu(t){\omega^*(t)t\over\phi(t)\omega(t)} =\omega\left(\frac nr\right)\prod_{p|(n/r)}\left(1-{\omega^*(p)p\over(p-1)\omega(p)}\right) \\ &={\omega(n/r)\over\phi(n/r)}\prod_{p|(n/r)}\left(p-1-{p\omega(p)-p\over\omega(p)}\right)={\omega(n/r)\over\phi(n/r)}\prod_{p|(n/r)}{p-\omega(p)\over\omega(p)}={\phi_\omega(n/r)\over\phi(n/r)}. \end{aligned} $$