One to one functions 
Suppose $f$ is a one-one function. Show that $f^{-1} \circ f(x) = x$
  for all $x\in \mathbb{D}(f)$ and $f \circ f^{-1}(y) =y$ for all $y\in
 \mathbb{R}(f)$.

What I understand from this is that since $x\in \mathbb{D}(f)$ and $f$ is one-one, then there exist elements $x_1$,$x_2$ where $f(x_1) = f(x_2)$.
 A: Let $ (X,Y,f) $ be a function (defined as an ordered triple). Define $ f^{-1} \subseteq Y \times X $ by
$$
f^{-1} \stackrel{\text{def}}{=} \{ (y,x) \in Y \times X ~|~ (x,y) \in f \}.
$$
Suppose that $ (X,Y,f) $ is one-to-one, i.e.,
$$
\forall x_{1},x_{2} \in X, ~ \forall y \in Y: \quad (x_{1},y),(x_{2},y) \in f ~ \Longrightarrow ~ x_{1} = x_{2}.
$$
Then $ (\text{Range}(f),X,f^{-1}) $ is a function, which allows us to solve the problem as follows.

Let $ x \in \text{Dom}(f) $. Then there exists a $ y \in Y $ such that
  
  
*
  
*$ (x,y) \in f $, equivalently, $ (y,x) \in f^{-1} $; hence,
  
*$ y = f(x) $ and $ x = {f^{-1}}(y) $, as both $ f $ and $ f^{-1} $ are functions.
Therefore, $ x = {f^{-1}}(y) = {f^{-1}}(f(x)) $.



Let $ y \in \text{Range}(f) $. Then there exists an $ x \in X $ such that
  
  
*
  
*$ (x,y) \in f $, equivalently, $ (y,x) \in f^{-1} $; hence,
  
*$ y = f(x) $ and $ x = {f^{-1}}(y) $, as both $ f $ and $ f^{-1} $ are functions.
Therefore, $ y = f(x) = f({f^{-1}}(y)) $.

A: Restrict the target to the range of $f$. Then $f$ is a bijection and the result follows.
P.S. I agree with Gerry Myerson's comment above.

Chapter 1. of Algebra of Marty Isaacs has more detailed explanation for this stuff.
