Composition of Functions and Invertible Function




In order to understant the composition of functions we have taken three sets in front of us name them as A,B and C.

now if we take a function f which is moving from A to B and then beginning with the stage set B, I take another function g from B to C. If we take a random element , after applying the funtion f it becomes f(x). Now beginning with f(x), this act as the element for the new function g and gives us the final product as g(f(x)), in short the new function which moves from A to C is defined by gof.
So if to define the function we say if 
f: A→B & g: B→C, then
gof: A→C defined by gof(x)= g(f(x)) ∀x∈A
1) f: A→B and g: B→C then,
range of f should be the starting point of the domain for the next function g for gof to exist. Likewise for fog to exist, the range of g should be subset of domain of f.
2) In general fog ≠ gof
Example- Let f: R→R
`"f"(x)= x^2`
g: R→R
g(x)= 2x+1
find fog & gof.
Solution- fog(x)= f(g(x))
                fog(x)= f(2x+1)= (2x+1)^2
`gof(x)= g(f(x))`
`gof(x)= g(x^2)= 2x^2+1` 
Invertible functions- Invertible means the function will be one one or onto both, that means the function will be bijective. 
This means if f: A→B, then there exists `"f"^-1:"B"→"A"`
So, if this is was f(x)= y then `x= "f"^-1(y)`
Example1- If A= {1,2,3}, B= {4,5,6,7} and f={(1,4), (2,5), (3,6)} is a function from A to B. Is it one- one function?
Yes, because every element in set A have different image.
Example2- If f is invertible which is defined as `"f"(x)= (3x-4)/5` then write `"f"^-1 (x).`
Solution- f(x)=y 

`(3x-4)/5 = y`

3x-4 = 5y

`x= (5y+4)/3 = "f"^-1 (y)`
Theorem 1: If f : X → Y, g : Y → Z and h : Z → S are functions, then ho(gof) = (hog)of. 
Proof : We have
 ho(gof) (x) = h(gof(x)) = h(g(f(x))), ∀x in X 
and (hog) of (x) = hog(f (x)) = h(g(f(x))), ∀x in X. 
Hence, ho(gof) = (hog)of.
Theorem 2: Let f : X → Y and g : Y → Z be two invertible functions. Then gof  is also invertible with `(gof)^(–1) = f^(–1)og^(–1).`
 Proof:  To show that gof is invertible with `(gof)^(–1) = f^(–1)og^(–1)`,

it is enough to show that `(f^(–1)og^(–1))o(gof) = "I"_X and (gof)o(f^(–1)og^(–1)) = "I"_Z.`

 Now, `(f^(–1)og^(–1))o(gof) = ((f^(–1)og^(–1)) og) of`,  byTheorem 1 

                                             = `(f^(–1)o(g^(–1)og)) of`, by Theorem 1

                                             = `(f^(–1) oI_Y) of`, by definition of `g^–1` 

                                             = `"I"_X. `

Similarly, it can be shown that `(gof)o(f ^(–1) og ^(–1)) = "I"_Z.` 

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