Hmm, I can't follow what you did there (without making a great effort).

I'll just show my proof (using the definition of f from my previous post):$\displaystyle x \mod m \equiv m^{\phi(n)} + n^{\phi(m)} \mod m \equiv 1 \mod m$

$\displaystyle x \mod n \equiv m^{\phi(n)} + n^{\phi(m)} \mod n \equiv 1 \mod n$

Therefore:$\displaystyle f: (x \mod {mn}) \mapsto (1 \mod m, 1 \mod n)$

Since we also have from the definition of f:$\displaystyle f: (1 \mod {mn}) \mapsto (1 \mod m, 1 \mod n)$

and since f is bijective (it's an isomorphism), it follows that$\displaystyle x \mod {mn} \equiv 1 \mod {mn}$

$\displaystyle \square$