Set theory: proof that g is the inverse of f using identity functions

Leave $ f: X rightarrow Y $ Y $ g: Y rightarrow X $. Yes $ g circ f = i_x $ Y $ f circ g = i_y $, so $ g $ is the inverse of $ f $. Note that $ i_x = x $ for all $ x in X $ Y $ i_y = y $ for all $ y in Y $.

So far, I have shown that whenever $ (x, y) in f $ so $ (y, x) in g $. However, my teacher says there is more to the test than this. I checked the definition, which says "We leave $ f: X rightarrow Y $ Be a function with the property that $ {(y, x) in Y times X | (x, y) in f } $ it is also a function. "I imagine that something is missing but I do not know how to focus it.


| $