Originally Posted by
Sudharaka Dear 450081592,
1) Suppose a,b $\displaystyle \in{Dom (f_og)}$ such that,
$\displaystyle f_o{g(a)}=p$ and $\displaystyle f_o{g(b)}=p$
Then, f[g(a)]=f[g(b)]
g(a)=g(b) since f is an one to one function.
a=b since g is an one to one function.
Therefore $\displaystyle f_o{g(a)}=f_o{g(b)}\Rightarrow{a=b}$
Hence $\displaystyle f_o{g}$ is an one to one function.
2) Supose g(a) = c and f(c) = b,
Therefore, $\displaystyle f_og(a)=b$
Since $\displaystyle f_og$ is one to one it is invertible,
$\displaystyle (f_og)^{-1}(b)=a$-----------A
Since f and g are one to one they are invertible.
$\displaystyle g^{-1}(c)=a$ and $\displaystyle f^{-1}(b)=c$
Therefore $\displaystyle g^{-1}[f^{-1}(b)]=a$-------B
From A and B,
$\displaystyle g^{-1}_of^{-1}(b)=(f_og)^{-1}(b)$ $\displaystyle \forall$ $\displaystyle b\in Dom{(f_og)}$