≠ fog . Solution We have gof ( x ) = g ( f ( x )) = g (cos x ) = (cos x ) = cos x . Similarly, fog ( x ) = f ( g ( x )) = f ( x ) = cos ( x ). Note that 3cos x ≠ cos x , for x = .
Hence, gof ≠ fog . Definition A function f : X → Y is defined to be invertible , if there exists a function g : Y → X such that gof = I X and fog = I Y . The function g is called the inverse of f and is denoted by f – . Thus, if f is invertible, then f must be one-one and onto and conversely, if f is one-one and onto, then f must be invertible.
This fact significantly helps for proving a function f to be invertible by showing that f is one-one and onto, specially when the actual inverse of f is not to be determined. Example Let f : N → Y be a function defined as f ( x ) = x + , where, Y = { y ∈ N : y = x + for some x ∈ N }. Show that f is invertible. Find the inverse.
Solution Consider an arbitrary element y of Y. By the definition of Y, y = x + , for some x in the domain N . This shows that ( ) y . Define g : Y → N by ( ) ( ) y g y .
Now, gof ( x ) = g ( f ( x )) = g ( x + ) = ( ) + and fog ( y ) = f ( g ( y )) = f ( ) ( ) y y = + = y – + = y . This