This question is motivated by lhf's comment here .

"It'd be nice to relate this formula with the natural mapping $U_{mn} \to U_m \times U_n$ by proving that the kernel has size $d$ and the image has index $\varphi(d)$."

Here, $U_k$ denotes the group of units of the ring $\mathbb{Z} / k \mathbb{Z}$ whenever $k$ is a positive integer.

I'm trying to prove the formula $$ \varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)} $$ by considering the natural map $\eta\colon U_{mn} \to U_m \times U_n$ (i.e. the map sending $\overline{x} \mapsto (\overline{x},\overline{x})$, where the bar denotes reduction mod $mn$, $m$, or $n$, respectively).

I've been able to show that the kernel has the right size as follows:

The kernel of $\eta$ consists of the elements $\overline{x} \in U_{mn}$ with $x \equiv 1 \bmod m$ and $x \equiv 1 \bmod n$. The integers $x$ which satisfy these conditions are those of the form $x = \frac{mn}{d}k + 1$ for $k \in \mathbb Z$. On the other hand, any such integer $x$ is relatively prime to $mn$, and hence gives and element $\overline{x} \in U_{mn}$. Therefore, $\ker \eta$ consists of the $d$ distinct elements $\overline{x}$, where $x = \frac{mn}{d}k + 1$ and $k \in \{1,\ldots,d\}$.

Once it has been shown that the image has index $\varphi(d)$, the first isomorphism theorem gives $$ \frac{U_{mn}}{\ker \eta} \cong Im(\eta), $$ and so $$ \frac{\varphi(mn)}{d} = \frac{|U_{mn}|}{|\ker \eta|} = |Im(\eta)| = \frac{|U_m \times U_n|}{|U_m \times U_n:Im(\eta)|} = \frac{\varphi(m)\varphi(n)}{\varphi(d)}, $$ or $$ \varphi(mn) = \varphi(m)\varphi(n) \frac{d}{\varphi(d)}. $$

I'm having trouble showing the image has the right index.

I've noticed that $\eta(\overline{x}) = \eta(\overline{x + \frac{mn}{d}})$, so the image consists the images of the elements $\overline{x}$ with $1 \leq x < \frac{mn}{d}$. I'm not sure if this is going anywhere, though. Any suggestions?