In mathematics, functions help us understand how one quantity depends on another. But not all functions behave the same way when connecting inputs to outputs. Some functions match each input to a unique output, some make sure every output is used, and some do both!
To study these differences, we classify functions into three types: injective (one-one), surjective (onto), and bijective (both one-one and onto).
These types help us understand how functions work and are especially important in higher-level math like algebra, calculus, and computer science.
1. Injective Functions (One-to-One)
A function is called one-to-one if for all elements a and b in A, if f(a) = f(b), then it must be the case that a = b. It never maps distinct elements of its domain to the same element of its co-domain. 
Properties of Injective Functions:
- An injective function does not map two different elements in the domain to the same element in the codomain.
- If a function is injective, it has an inverse function on its image.
2. Surjective Functions (Onto)
If every element b in B has a corresponding element a in A such that f(a) = b. It is not required that a is unique; The function f may map one or more elements of A to the same element of B.
Properties of Surjective Functions
- A surjective function covers the entire codomain.
- If a function is surjective, it can have an inverse function if restricted to its range.
3. Bijective Functions (One-to-One Correspondence)
A function is a Bijective function if it is both one to one and onto function.
Properties of Bijective Functions
- A bijective function has both an inverse function and a unique mapping for each element in the domain and codomain.
- If a function is bijective, it establishes a one-to-one correspondence between the sets A and B.
Composition of Functions
Let g be a function from B to C and f be a function from A to B, the composition of f and g, which is denoted as fog(a) = f(g(a)).
Properties of Function Composition
- fog ≠gof
- f-1 of = f-1 (f(a)) = f-1(b) = a.
- fof-1 = f(f-1 (b)) = f(a) = b.
- If f and g both are one to one function, then fog is also one to one.
- If f and g both are onto function, then fog is also onto.
- If f and fog both are one to one function, then g is also one to one.
- If f and fog are onto, then it is not necessary that g is also onto.
- (fog)-1 = g-1 o f-1
Applications in Engineering
Understanding injective, surjective, and bijective functions is crucial in various engineering disciplines:
- Data Encryption: In computer engineering, bijective functions are used in encryption algorithms to ensure that each plaintext maps to a unique ciphertext and vice versa, enabling secure communication.
- Signal Processing: In signal processing, injective functions help in ensuring that distinct signals remain distinct after transformation, which is essential for accurate signal reconstruction.
- Network Design: In network design and analysis, surjective functions are used to ensure that resources are optimally utilized, and every demand is met by at least one resource.
- Control Systems: In control systems engineering, bijective functions are important for designing controllers that map system states to control actions uniquely and efficiently.
Solved Examples - Classes of Functions
Problem 1: Determine if the function f(x) = 2x+1 is injective.
Solution:
To check if f is injective, assume f(x1) = f(x2)): 2x1+1 = 2x2+1
Subtract 1 from both sides: 2x1 = 2x2
Divide by 2: x1 = x2
Sincex1 = x2​ follows from f(x1) = f(x2), f is injective.
Problem 2: Determine if the function f(x) = x2 for x∈R is surjective when the codomain is R.
Solution:
For f(x) = x2 to be surjective, every y∈R must have a corresponding x∈R such that x2 = y. However, x2 ≥ 0 for all real x, so f(x) = x2 cannot produce negative values. Therefore, f is not surjective when the codomain is R.
If the codomain were [0,∞), then f would be surjective.
Problem 3: Determine if the function f(x) = 3x − 2 is bijective if the domain and codomain are both R.
Solution:
- Injective Check: As shown in Example Problem 1, f(x) = 3x−2 is injective.
- Surjective Check: For any y∈R, solve y = 3x − 2 for x: x = y+2/3. ​Since x is always a real number for any real y, f is surjective.
Since f is both injective and surjective, it is bijective.