$\gcd(m,n) = 1$ and $\gcd (mn,a)=1$ implies $a \cong 1 \pmod{ mn}$ I have $m$ and $n$ which are relatively prime to one another and $a$ is relatively prime to $mn$
and after alot of tinkering with my problem i came to this equality:
$a \cong 1 \pmod m \cong 1 \pmod n$
why is it safe to say that $a \cong 1 \pmod {mn}$?..

Hint $\rm\,\ \ m,n\,|\,a\!-\!1\!\iff\! lcm(m,n)\,|\,a\!-\!1\ \ $ [proof below] 
And, further, recall that $\rm\,\ lcm(m,n) = 
\dfrac{mn}{gcd(m,n)}\ [= mn\ \ if\ \ gcd(m,n)=1]$
Here's a proof of $\rm\ m,n\,|\,b\:\Rightarrow\:mn\,|\,bd,\ \ d = gcd(m,n) [= mx+ny\ $ by Bezout] 
$\rm\begin{eqnarray} 
\rm\quad m,n\,|\,b\:\Rightarrow\:mn\,|\,bm,bn &\Rightarrow&\,\rm mn\ |\ \,bmx\, +\, bny \ =\ b(mx\!+\!ny)\, = bd\quad [Bezout\ form] \\
\rm\quad\ \ m,n\,|\,b\:\Rightarrow\:mn\,|\,bm,bn &\Rightarrow&\,\rm mn\ |\ \gcd(bm, bn)\, =\: b \gcd(m,n) = bd\quad [GCD\ form] \\
\end{eqnarray}$
