Notes On Theorems on Composition of Functions - CBSE Class 12 Maths
Theorem 1 : If f : A â†’ B and g : B â†’ C are one-one, then gof : A â†’ C is also one-one. Proof: A function f : A â†’ 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 âˆˆ A, f(x1) = f (x2) implies x1 = x2. Given that f: A â†’ B and g: B â†’ C are one-one. For any x1, x2 âˆˆ A f(x1)=f(x2) â‡’ x1=x2 â€¦(i) g(x1)=g(x2) â‡’ x1=x2 â€¦(ii) To show: If gof(x1) = gof(x2), then x1 = x2 Let gof(x1) = gof(x2) â‡’ g[f(x1)] = g[f(x2)] â‡’ f(x1) = f(x2) â€¦from (i) â‡’ x1 = x2 â€¦from (ii) Hence, the functions gof: A â†’ C are one-one. Theorem2: If f : A â†’ B and g : B â†’ C are onto, then gof : A â†’ C is also onto. Proof: Let us consider an arbitrary element z âˆˆ C '.' g is onto âˆƒ a pre-image y of z under the function g such that g (y) = z â€¦â€¦â€¦(i) Also, f is onto, and hence, for y ÃŽ B, there exists an element x âˆˆ A such that f (x) = y â€¦â€¦(ii) Therefore, gof (x) = g (f (x)) = g (y) from (ii) = z from (i) Thus, corresponding to any element z âˆˆ C, there exists an element x âˆˆ 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 â†’ Y, g : Y â†’ Z and h : Z â†’ S are functions, then ho(gof ) = (hog) o f. Proof: Let x âˆˆ A LHS: ho(gof ) (x)          = h(gof(x))          = h(g(f(x))), âˆ€ x in X RHS: (hog) of f(x) = hog(f(x)) = h(g(f (x))), âˆ€ x in X. LHS = RHS Hence, ho(gof) = (hog)of. The composition of functions satisfies the associative property.

#### Summary

Theorem 1 : If f : A â†’ B and g : B â†’ C are one-one, then gof : A â†’ C is also one-one. Proof: A function f : A â†’ 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 âˆˆ A, f(x1) = f (x2) implies x1 = x2. Given that f: A â†’ B and g: B â†’ C are one-one. For any x1, x2 âˆˆ A f(x1)=f(x2) â‡’ x1=x2 â€¦(i) g(x1)=g(x2) â‡’ x1=x2 â€¦(ii) To show: If gof(x1) = gof(x2), then x1 = x2 Let gof(x1) = gof(x2) â‡’ g[f(x1)] = g[f(x2)] â‡’ f(x1) = f(x2) â€¦from (i) â‡’ x1 = x2 â€¦from (ii) Hence, the functions gof: A â†’ C are one-one. Theorem2: If f : A â†’ B and g : B â†’ C are onto, then gof : A â†’ C is also onto. Proof: Let us consider an arbitrary element z âˆˆ C '.' g is onto âˆƒ a pre-image y of z under the function g such that g (y) = z â€¦â€¦â€¦(i) Also, f is onto, and hence, for y ÃŽ B, there exists an element x âˆˆ A such that f (x) = y â€¦â€¦(ii) Therefore, gof (x) = g (f (x)) = g (y) from (ii) = z from (i) Thus, corresponding to any element z âˆˆ C, there exists an element x âˆˆ 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 â†’ Y, g : Y â†’ Z and h : Z â†’ S are functions, then ho(gof ) = (hog) o f. Proof: Let x âˆˆ A LHS: ho(gof ) (x)          = h(gof(x))          = h(g(f(x))), âˆ€ x in X RHS: (hog) of f(x) = hog(f(x)) = h(g(f (x))), âˆ€ x in X. LHS = RHS Hence, ho(gof) = (hog)of. The composition of functions satisfies the associative property.

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