Onto Function

A function is something, which relates elements/values of one set to the elements/values of another set, in such a way that elements of the second set is identically determined by the elements of the first set. A function has many types which define the relationship between two sets in a different pattern. They are various types of functions like one to one function, onto function, many to one function, etc.

Onto Function Definition (Surjective Function)

Onto function could be explained by considering two sets, Set A and Set B which consist of elements. If for every element of B there is at least one or more than one element matching with A, then the function is said to be onto function or surjective function. The term for the surjective function was introduced by Nicolas Bourbaki.

In the first figure, you can see that for each element of B there is a pre-image or a matching element in Set A, therefore, its an onto function. But if you see in the second figure, one element in Set B is not mapped with any element of set A, so it’s not an onto or surjective function.

Properties of a Surjective Function (Onto)

We can define onto function as if any function states surjection by limit its codomain to its range.
The domain is basically what can go into the function, codomain states possible outcomes and range denotes the actual outcome of the function.
Every onto function has a right inverse
Every function with a right inverse is a surjective function
If we compose onto functions, it will result in onto function only.

Onto Function Example Questions

Example: Let A={1,5,8,9) and B{2,4} And f={(1,2),(5,4),(8,2),(9,4)}. Then prove f is a onto function.

Solution: From the question itself we get,

A={1,5,8,9)

B{2,4}

& f={(1,2),(5,4),(8,2),(9,4)}

So, all the element on B has a domain element on A or we can say element 1 and 8 & 5 and 9 has same range 2 & 4 respectively.

Therefore, f: A \(\rightarrow\) B is an surjective fucntion.

Hence, the onto function proof is explained.

Number of Onto Functions (Surjective functions) Formula

If we have to find the number of onto function from a set A with n number of elements to set B with m number of elements, then;

\((^{m}_{0})m^{n}\,-\,(^{m}_{1})(m-1)^{n}\,+\,(^{m}_{2})(m-2)^{n}\,-\,(^{m}_{3})(m-3)^{n}\,+\,.\,.\,.\,.\,.\,.\)

When n<m, the number of onto function = 0

And when n=m, number of onto function = m!

We can also write the number of surjective functions for a given domain and range as;

mⁿ– m + m ((m – 1)ⁿ– (m – 1))