$\gcd(m,n) = 1$ 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 $a \cong 1 \pmod {mn}$?..

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 $a\equiv c\pmod{mn}$. 
