Question

Prove that the inverse of one-one onto mapping is unique.

Answer 1

Suppose $g_1$ and $g_2$ are both inverses to $f$. Then $$ g_1=g_1\circ i_B=g_1\circ (f\circ g_2)=(g_1\circ f)\circ g_2=i_A\circ g_2= g_2, $$ proving the theorem.$\square$Because of theorem  we can talk about "the'' inverse of $f$, assuming it has one; we write $f^{-1}$ for the inverse of $f$. Note well that this extends the meaning of "$f^{-1}$'', in a potentially confusing way.