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Theorems on Composition of Functions

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Theorems on Composition of Functions - Lesson Summary

Theorem 1 : If f : A → B and g : B → C are one-one, then gof : A → C is also one-one.

Proof:
A function f : A → B is defined to be one-one, if the images of distinct elements of A under f are distinct, i.e. for every x1, x2 ∈ A, f(x1) = f (x2) implies x1 = x2.

Given that f: A → B and g: B → C are one-one.

For any x1, x2 ∈ A

f(x1)=f(x2) ⇒ x1=x2 …(i)

g(x1)=g(x2) ⇒ x1=x2 …(ii)

To show: If gof(x1) = gof(x2), then x1 = x2

Let gof(x1) = gof(x2)

⇒ g[f(x1)] = g[f(x2)]

⇒ f(x1) = f(x2) …from (i)

⇒ x1 = x2 …from (ii)

Hence, the functions gof: A → C are one-one.

Theorem2: If f : A → B and g : B → C are onto, then gof : A → C is also onto.

Proof:

Let us consider an arbitrary element z ∈ C

'.' g is onto ∃ a pre-image y of z under the function g such that g (y) = z ………(i)

Also, f is onto, and hence, for y Î B, there exists an element x ∈ A such that f (x) = y ……(ii)

Therefore, gof (x) = g (f (x)) = g (y) from (ii)

= z from (i)

Thus, corresponding to any element z ∈ C, there exists an element x ∈ A such that gof (x) = z.

Hence, gof is onto.

Note: In general, if gof is one-one, then f is one-one. Similarly, if gof is onto, then g is onto.

The composition of functions can be considered for n number of functions.

Theorem 3: If f : X → Y, g : Y → Z and h : Z → S are functions, then ho(gof ) = (hog) o f.

Proof: Let x ∈ A

LHS: ho(gof ) (x)

         = h(gof(x))

         = h(g(f(x))), ∀ x in X

RHS: (hog) of f(x)

= hog(f(x))

= h(g(f (x))), ∀ x in X.

LHS = RHS

Hence, ho(gof) = (hog)of.

The composition of functions satisfies the associative property.

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