$f \circ f^{-1} = i_B$ proof using the fact that $f^{-1} \circ f = i_A$

Question

Suppose f is function from A to B, and suppose that $f^{-1}$ is a function from B to A. Assume $f^{-1} \circ f = i_A$. Then show therefore that $f \circ f^{-1} = i_B$. I tried applying left composition of $f^{-1}$ and right composition of $f$ to both sides of the equation but I still get stuck. $i_B $ and $i_A$ are the identity maps from B to B and A to A, respectively.

Answer 1

Counterexample: take $A = B = \Bbb N = \{1,2,3,\dots\}$

Define $$ f(x) = x+1\\ f^{-1}(x) = \begin{cases} x-1 & x>1\\ 1 & x = 1 \end{cases} $$

Answer 2

Here's a counterexample. For $A:=\{a,b\}$, and $B:=\{a,b,c\}$ consider the maps $$ f:A \to B,\quad g:B\to A $$ defined by $$ f(a)=c,\, f(b)=a,\, g(a)=b,\, g(b)=a, \, g(c)=a. $$ Then $$ g\circ f(a)=g(f(a))=g(c)=a, \, g\circ f(b)=g(f(b))=g(a)=b, $$ i.e. $g\circ f=i_A$. However $f\circ g\ne i_B$ because $$ f\circ g(b)=f(g(b))=f(a)=c \ne b. $$