If the Function f : A-> B is both one to one and onto i.e bijective ,then we can find a function g: B-> A
such that
g(y)=x when y=f(x) i.e $g \circ f=I_A$
and
f(g(x)) =y i.e $f \circ g=I_B$
The function g is called the inverse of f and is denoted as f-1. The function f(x) is called invertible function
Another definition of Invertible function
A Function f : A-> B is invertible if we can find a function g: B- > A such that
$f \circ g=I_B$ and $g \circ f=I_A$
and if f is invertible, then f must be one-one and onto
How to find the Inverse function
Method I
let f(x) be the function f(x)=4x -2
Consider an arbitrary element y then by definition y=4x-2
Find out the value of x in terms of y
$y=4x-2$
$y+2=4x$
$x= \frac {y+2}{4}$
Now the invertible function can be defined as
$g(x) = \frac {x+2}{4}$
Method II
let f(x) be the function f(x)=4x -2
Consider an arbitrary element y then by definition y=4x-2
Swap the position of x and y
$y=4x-2$
$x=4y -2$
$y= \frac {x+2}{4}$
Now the invertible function can be defined as
$g(x) = \frac {x+2}{4}$
How to Proof for function to be invertable
(1) We need to proof that the function is one-one and onto One-one statement
Let f: A -> B, a function from a set A to a set B, f is called a one-to-one function or injection, if, and only if, for all elements a1 and a2 in A,
if f(a1) = f(a2), then a1 = a2 onto statment
A function f: A-> B is said to be onto(surjective) if every element of B is the image of some element of A under f, i,e
for every y ∈ B, there exists a element x in A where f(x)=y
(2)We can find the inverse g for a function f : A -> B and if
$g \circ f=I_A$ and $g \circ f=I_A$, then it is invertible
Inverse of Composite function
Theorem of Inverse of Composite function states that
if f : X -> Y and g : Y -> Z be two invertible functions. Then $g \circ f$ is also invertible with $(g \circ f)^{–1} = f^{–1} \circ g^{-1}$ Proof
for $g \circ f$ to be invertible with $(g \circ f)^{–1} = f^{–1} \circ g^{-1}$
we just need to prove that
$(f^{–1} \circ g^{-1}) \circ (g \circ f) = I_X$
and $(g \circ f) \circ (f^{–1} \circ g^{-1})=I_Z$
We will use theorem of composition of three functions
$h \circ (g \circ f) =(h \circ g ) \circ f$
$(f^{–1} \circ g^{-1}) \circ (g \circ f) = ((f^{–1} \circ g^{-1}) \circ g) \circ f$
$=(f^{–1} \circ (g^{-1}) \circ g)) \circ f$
$=(f^{–1} \circ I_Y) \circ f$
$=I_X$
Similarly other can be proved
Solved Examples
Example
A set A is defined as A={a,b,c}
Let f: A-> A be the function defined as are
(1) f={(a,a),(b,b),(c,c)}
(2) f={(a,b),(b,a),(c,c)}
(3) f={(a,c),(b,c),(c,a)}
Find if all these function defined are invertible Solutions
(1) The neccesary condition for invertibleness is one on one and onto
This function is clearly one on one and onto,so it is invertible
(2) This function is clearly one on one and onto,so it is invertible
(3) This function is not one on one and neither onto,so it is not invertible
