THEOREM: if $F$ and $G$ are one to one then $G \circ F$ is also one to one and $(G \circ F)^\neg$ = $F^\neg \circ G^\neg$
PROOF:
if $F: A\rightarrow B$, $G: B \rightarrow C$ and
$\forall a, a' \in A \ \ F(a)=F(a') \Rightarrow a=a'$ then F is one to one and if
$\forall b, b' \in B \ \ G(b)=G(b') \Rightarrow b=b'$ then G is one to one by the definition of one to one.
If $(G \circ F)(a)=(G \circ F)(a') \Rightarrow G(F(a))= G(F(a'))$ then $F(a)= F(a')$ since $G$ is one to one.
If $F(a)=F(a')$ then $a=a'$ since $F$ is one to one
Because $G \circ F$ is one to one it is also invertible so $(G \circ F)^\neg$ exist now if we compute $\begin{align}((G\circ F)\circ (F^\neg\circ G^\neg))(a)&=\\ ((G\circ (F\circ F^\neg)\circ G^\neg))(a)&=\\ ((G\circ G^\neg))(a)&=a\end{align} $
So, $(F^\neg\circ G^\neg)$ is the inverse of $G \circ F$ therefore $(G \circ F)^\neg=(F^\neg\circ G^\neg)$
QED
I feel like the first part is ok but the 2nd part is all messed up ... what did I do wrong?