The mapping f: A > B is called an onto function if the set B is entirely used up, i.e if every element of B is the image of atleast one element of A.
For every b € B there exists atleast one element a € Asuch that f(a) = b
If function f: A > B is not onto, that is some of the elements of B remain unmated, then f is called an into function.
Remarks:
 An onto function is also called surjective function or a surjection.
 A oneone function may be both onto and into.
 If A and B are finite sets and f: A>B is surjective, then n(B) <= n(A).
 If n(A) = m, n(B) = m, then the possible number of possible surjective mappings from A to B is m!.
: Let A = {2,2, 3,3}, B = {4,9} and f: A >B be a function defined by f(x) = x^{2}, then f is onto because f(2) = 4, f(3) = 9, f(3) = 9 i.e f(A) = {4,9} = B.
If A and B are finite sets and f: A >B is surjective, then n(B) <= n(A).
If n(A) = m and n(B)  m, then the possible number of possible surjective mappings frm A to B is m!.
Example: Let A = {2, 2, 3, 3}, B = {4,9} and f: A > B be a function defined by f(x) = x^{2}, then f is onto, because f(2) and f(2) = 4 and f(3) and f(3) = 9,i.e f(A) = {4,9} = B
A function f: N >N defined by f(x) = 7x is not an onto function, because f(N) = {7,14,21,..} is not equal to N(codomain).
Method to find surjectivity of s function
f: R > R defined by f(x) = x^{3} + 5 for all x € R.
 Choose any arbitrary element y in B.
 Put f(x) = y
 Solve the equation f(x) = y for x and obtain x in terms of y. Let x = g(y).
 If for all values of y € B, the values of x obtained from x = g(y) are in A, then f is onto. If there are some y € B for which x, given by x = g(y), is not in A, then f is not onto.
Solution: Let there be an arbitrary element in R. Then f(x) = y = x^{3} + 5
x = (y 5)^{1/3}.
Now for all y € R, (y5)^{1/3} is a real number. So for all y € R (codomain), there exists x = (y  5)^{1/3} in R (domain) such that f(x) = x^{3} + 5
Here f: R > R is an onto function.
Directions: Answer the following.
