I am not sure if my answer below is complete, my math writing is pretty bad so I am not sure if I am covering all the necessary points. Please tell me if you think that the format/style is bad too.
2.F.Suppose f is a one-one function. Show that $f^{-1} \circ f(x) = x$ for all $x$ in $D(f)$.
Let's first remember our definitions:
Definition Inverse Function - Let f be a one-one function with domain D(f) in A and range R(f) in B. If $g = \{(b, a) \in B \times A: (a, b) \in f \}$ then $g$ is a one-one function with domain $D(g) = R(f)$ in $B$ with range $R(g) = D(f)$ in $A$. The function $g$ is called the function inverse to $f$ and we ordinarily denote $g$ by $f^{-1}$\
Definition Composition of Function - Let $f$ be a function with domain $D(f)$ in $A$ and range $R(f)$ in $B$ and let $g$ be a function with domain $D(g)$ in $B$ and range $R(g)$ in $C$. The composition $g \circ f$ is the function from $A$ to $C$ given by\
$g \circ f = \{(a, c) \in A \times C: \text{there exists an element }b \in B \text{ such that }(a,b) \in f \text{ and } (b,c) \in g \}$
(Proof) By the definition 12, we need to prove that if $(x, z) \in f^{-1} \circ f$ then $z = x$. Now let $(x, z)$ be any ordered pair in $f^{-1} \circ f$, by the definition 12 of composition of function, we know that the pair $(x, y) \in f$ and the pair $(y, z) \in f^{-1}$ exist. By the definition of inverse, the function $f^-1$ is the inverse of $f$ if and only if considering the pair $(x, y) \in f$ and $(y, z) \in f^{-1}$, $z = x$, so we proved that for all $(x, z) \in f^{-1} \circ f, z = x.$