$\gcd(m,n) = 1$ 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 $a \cong 1 \pmod {mn}$?..
 A: It looks as if you are asking the following. Suppose that $m$ and $n$ are relatively prime. Show that if $a\equiv 1\pmod{m}$ and $a\equiv 1\pmod{n}$, then $a\equiv 1\pmod{mn}$.
So we know that $m$ divides $a-1$, and that $n$ divides $a-1$. We want to show that $mn$ divides $a-1$.
Let $a-1=mk$. Since $n$ divides $a-1$, it follows that $n$ divides $mk$. But $m$ and $n$ are relatively prime, and therefore $n$ divides $k$. so $k=nl$ for some $l$, and therefore $a-1=mnl$. 
Remark: $1.$ There are many ways to show that if $m$ and $n$ are relatively prime, and $n$ divides $mk$, then $n$ divides $k$. 
One of them is to use Bezout's Theorem: If $m$ and $n$ are relatively prime, there exist integers $x$ and $y$ such that $mx+ny=1$. 
Multiply through by $k$. We get $mkx+nky=k$. By assumption, $n$ divides $mk$, so $n$ divides $mkx$. Clearly, $n$ divides $nky$. So $n$ divides $mkx+nky$, that is, $n$ divides $k$. 
$2.$ Note that there was nothing special about $1$. Let $m$ and $n$ be relatively prime. If $a\equiv c\pmod{m}$ and $a\equiv c\pmod{n}$, then $a\equiv c\pmod{mn}$. 
A: $\gcd(m,n)=1$ implies there is an $x$ and $y$ so that
$$
mx+ny=1\tag{1}
$$
$a\equiv1\pmod{m}$ and $a\equiv1\pmod{n}$ imply there is a $j$ and $k$ so that
$$
a-1=jm=kn\tag{2}
$$
multiplying $(1)$ by $j$ and using $(2)$
$$
nkx+jny=j\tag{3}
$$
Plugging $(3)$ back into $(2)$ yields
$$
a-1=jm=n(kx+jy)m\tag{4}
$$
which implies that $a\equiv1\pmod{mn}$.

Idea of the Preceding Argument
The idea of $(3)$ is to show that since $\gcd(m,n)=1$, by $(1)$, and $jm$ is a multiple of $n$, by $(2)$, we have that $j$ is a multiple of $n$. Thus, $a-1=jm$ is a multiple of $mn$, by $(4)$.
A: 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] 
$\rm\begin{eqnarray} 
\rm\quad m,n\,|\,b\:\Rightarrow\:mn\,|\,bm,bn &\Rightarrow&\,\rm mn\ |\ \,bmx\, +\, bny \ =\ b(mx\!+\!ny)\, = bd\quad [Bezout\ form] \\
\rm\quad\ \ m,n\,|\,b\:\Rightarrow\:mn\,|\,bm,bn &\Rightarrow&\,\rm mn\ |\ \gcd(bm, bn)\, =\: b \gcd(m,n) = bd\quad [GCD\ form] \\
\end{eqnarray}$
