Summary

Videos

References

**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 x_{1}, x_{2} âˆˆ A, f(x_{1}) = f (x_{2}) implies x_{1} = x_{2}.

Given that f: A â†’ B and g: B â†’ C are one-one.

For any x_{1}, x_{2} âˆˆ A

f(x_{1})=f(x_{2}) â‡’ x_{1}=x_{2 } â€¦(i)

g(x_{1})=g(x_{2}) â‡’ x_{1}=x_{2 } â€¦(ii)

To show: If gof(x_{1}) = gof(x_{2}), then x_{1} = x_{2}

Let gof(x_{1}) = gof(x_{2})

â‡’ g[f(x_{1})] = g[f(x_{2})]

â‡’ f(x_{1}) = f(x_{2}) â€¦from (i)

â‡’ x_{1} = x_{2} 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.

**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 x_{1}, x_{2} âˆˆ A, f(x_{1}) = f (x_{2}) implies x_{1} = x_{2}.

Given that f: A â†’ B and g: B â†’ C are one-one.

For any x_{1}, x_{2} âˆˆ A

f(x_{1})=f(x_{2}) â‡’ x_{1}=x_{2 } â€¦(i)

g(x_{1})=g(x_{2}) â‡’ x_{1}=x_{2 } â€¦(ii)

To show: If gof(x_{1}) = gof(x_{2}), then x_{1} = x_{2}

Let gof(x_{1}) = gof(x_{2})

â‡’ g[f(x_{1})] = g[f(x_{2})]

â‡’ f(x_{1}) = f(x_{2}) â€¦from (i)

â‡’ x_{1} = x_{2} 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.