Functions — one-one, into & onto

A function can be thought of as a machine that takes an input and produces a unique output. But how does this machine map its inputs to its outputs? Does every possible output get produced? Does any output get produced by more than one input? Answering these questions allows us to classify functions into distinct categories, which is crucial for understanding more advanced concepts like inverse functions and calculus.

{KEY: type=concept | title=The Four Fundamental Mappings | text=Every function f: A → B can be classified into one of four types based on how elements from the domain A are mapped to the codomain B. This classification depends on two independent properties: whether the mapping is one-to-one or many-to-one, and whether it is onto or into.}

Understanding these classifications is not just an academic exercise. In computer science, a bijective function is like a perfect encryption key—each plaintext has a unique ciphertext, and every possible ciphertext can be decrypted back to a unique plaintext. In contrast, a many-to-one function is like a hash function (e.g., MD5, SHA-256) where multiple different inputs (like files or passwords) can produce the same output hash, but you can't go backward.

One-to-One (Injective) vs. Many-to-One Functions

This classification focuses on the uniqueness of inputs for each output.

One-one Function (Injection)

A function f: A → B is called one-to-one or injective if different elements in the domain A have different images in the codomain B.

In simple terms, no two inputs map to the same output.

• Analytical Definition: f(x₁) = f(x₂) ⇒ x₁ = x₂ for all x₁, x₂ in A. Alternatively, x₁ ≠ x₂ ⇒ f(x₁) ≠ f(x₂)
• Calculus Method: If the function f(x) is differentiable, it is one-one on an interval if it is strictly monotonic on that interval. This means f'(x) > 0 (strictly increasing) or f'(x) < 0 (strictly decreasing) for all x in the interval.
• Graphical Method: Horizontal Line Test. A function is one-one if and only if no horizontal line intersects its graph more than once.

Many-one Function

A function f: A → B is called many-to-one if there exist at least two distinct elements in the domain A which have the same image in B.

It's the logical opposite of a one-one function. If a function is not one-one, it must be many-one.

• Analytical Definition: There exist x₁, x₂ in A such that x₁ ≠ x₂ but f(x₁) = f(x₂).
• Examples:
  • f(x) = x²: Here, f(-2) = 4 and f(2) = 4. Different inputs (-2 and 2) lead to the same output (4).
  • f(x) = |x|: f(-3) = 3 and f(3) = 3.
  • f(x) = sin(x): sin(0) = 0, sin(π) = 0, sin(2π) = 0.

{{VISUAL: diagram: Two arrow diagrams side-by-side. The left diagram shows a one-one function from set A={1,2,3} to B={a,b,c,d}, with 1→a, 2→c, 3→b. The right diagram shows a many-one function from A={1,2,3} to B={a,b}, with 1→a, 2→a, 3→b.}}

{CALLOUT: type=exam | text=For JEE, the calculus method is a powerful shortcut. If you can quickly find the derivative and check its sign, you can determine injectivity much faster than using the algebraic f(x₁) = f(x₂) method, especially for complex polynomial, trigonometric, or exponential functions.}

Onto (Surjective) vs. Into Functions

This classification focuses on the coverage of the codomain.

Onto Function (Surjection)

A function f: A → B is called onto or surjective if every element in the codomain B is the image of at least one element in the domain A.

In simpler terms, the function "covers" the entire codomain. No element in the codomain is left unused.

• Analytical Definition: The Range of the function is equal to its Codomain.
  • Range(f) = Codomain(f)
• Method to Check:
  1. Let y be an arbitrary element in the codomain B.
  2. Put f(x) = y.
  3. Solve the equation for x in terms of y.
  4. If for every value of y in the codomain B, you can find a corresponding value of x in the domain A, the function is onto.

Into Function

A function f: A → B is called into if there exists at least one element in the codomain B which is not the image of any element in the domain A.

It's the logical opposite of an onto function.

• Analytical Definition: The Range of the function is a proper subset of its Codomain.
  • Range(f) ⊂ Codomain(f) and Range(f) ≠ Codomain(f).

{{VISUAL: diagram: Two arrow diagrams side-by-side. The left diagram shows an onto function from set A={1,2,3,4} to B={a,b}, with 1→a, 2→a, 3→b, 4→b. The right diagram shows an into function from A={1,2,3} to B={a,b,c,d}, with 1→a, 2→c, 3→a. Elements 'b' and 'd' in B are not images.}}

Bijective Function (One-one and Onto)

A function that is both injective (one-one) and surjective (onto) is called a bijective function or a one-to-one correspondence. These functions create a perfect pairing between the elements of the domain and the codomain.

• Importance: A function f: A → B has an inverse function f⁻¹: B → A if and only if f is bijective. This is a critical concept for many areas of mathematics.

Solved Examples

Let's apply these concepts to some problems.

Example 1: Linear Function

Given: A function f: R → R defined by f(x) = 3x + 5.

To Find: Determine if f(x) is one-one, many-one, onto, or into.

Solution:

1. Check for One-one / Many-one: Let x₁, x₂ be two elements in the domain R such that f(x₁) = f(x₂).

   3x₁ + 5 = 3x₂ + 5
   

2. Subtracting 5 from both sides gives:

   3x₁ = 3x₂
   

3. Dividing by 3 gives:

   x₁ = x₂
   

   Since f(x₁) = f(x₂) implies x₁ = x₂, the function is one-one (injective).

4. Check for Onto / Into: Let y be an arbitrary element in the codomain R. We set f(x) = y.

   y = 3x + 5
   

5. Now, we solve for x in terms of y.

   x = (y - 5) / 3
   

6. For any real number y in the codomain R, the value (y - 5) / 3 is also a real number, which is in the domain R. This means every element in the codomain has a pre-image. Therefore, Range = Codomain (R). The function is onto (surjective).

Final Answer: The function f(x) = 3x + 5 is one-one and onto, so it is a bijective function.

Example 2: Quadratic Function

Given: A function f: R → R defined by f(x) = x² + 2.

To Find: Classify the function.

Solution:

1. Check for One-one / Many-one: Let's test two distinct inputs. Consider x₁ = -1 and x₂ = 1.

   f(-1) = (-1)² + 2 = 1 + 2 = 3
   

   f(1) = (1)² + 2 = 1 + 2 = 3
   

2. We found two distinct inputs, -1 and 1, that map to the same output, 3. Since -1 ≠ 1 but f(-1) = f(1), the function is many-one.

3. Check for Onto / Into: The codomain is given as R (all real numbers). Let's find the range of f(x). We know that for any real number x, x² ≥ 0.

   x² + 2 ≥ 0 + 2
   

   f(x) ≥ 2
   

4. The range of the function is [2, ∞). The codomain is R. Since [2, ∞) is a proper subset of R, the Range ≠ Codomain. For example, y = 0 is in the codomain but there is no real x such that x² + 2 = 0. Therefore, the function is into.

Final Answer: The function f(x) = x² + 2 from R to R is many-one and into.

Example 3: Rational Function

Problem: Check if the function f: R - {-2} → R - {1} defined by f(x) = (x - 3) / (x + 2) is bijective.

Let's use the whiteboard to break this down step-by-step. This type of problem, with restricted domains and codomains, is very common in JEE. We need to be careful with our algebraic manipulation.

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Example 4: Trigonometric Function

Given: A function f: [0, π] → [-1, 1] defined by f(x) = cos(x).

To Find: Classify the function.

Solution:

1. Check for One-one / Many-one: We can use the calculus method.

   f'(x) = -sin(x)
   

2. In the interval (0, π), sin(x) is always positive. Therefore, f'(x) = -sin(x) is always negative (< 0) in (0, π).

3. Since f'(x) < 0, the function is strictly decreasing on the interval [0, π]. A strictly monotonic function is always one-one (injective).

4. Check for Onto / Into: The codomain is given as [-1, 1]. We need to find the range of cos(x) for x in [0, π].

5. As x varies from 0 to π, the value of cos(x) starts at cos(0) = 1, decreases, and ends at cos(π) = -1. The function covers all values between -1 and 1, inclusive.

   Range(f) = [-1, 1]
   

6. Since Range = Codomain ([-1, 1]), the function is onto (surjective).

Final Answer: The function f(x) = cos(x) on the given domain and codomain is bijective.

Tips & Tricks for Classification

Common Mistakes to Avoid

Tricky Practice Problems

Problem 1

Let f: R → R be defined by f(x) = (eˣ - e⁻ˣ) / (eˣ + e⁻ˣ). The function f(x) is: (A) one-one and into (B) one-one and onto (C) many-one and into (D) many-one and onto

💡 Answer: (A) The function is f(x) = tanh(x). Its derivative f'(x) = sech²(x) is always positive, so it is one-one. The range of tanh(x) is (-1, 1), which is a proper subset of the codomain R. Therefore, it is into.

Problem 2

Let A = {1, 2, 3} and B = {a, b}. The total number of surjective (onto) functions from A to B is: (A) 8 (B) 6 (C) 2 (D) 0

💡 Answer: (B) Total functions from A to B = |B|^|A| = 2³ = 8. Functions that are not onto (i.e., into) are those where all elements map to 'a' only (1 case) or all elements map to 'b' only (1 case). So, 2 functions are not onto. Number of onto functions = Total functions - Into functions = 8 - 2 = 6.

Problem 3

The function f: [2, ∞) → [5, ∞) is defined by f(x) = x² - 4x + 9. This function is: (A) one-one and into (B) bijective (C) many-one and onto (D) many-one and into

💡 Answer: (B) Complete the square: f(x) = (x - 2)² + 5. The vertex of the parabola is at (2, 5). For the domain x ≥ 2, the function is strictly increasing, so it is one-one. The minimum value is f(2) = 5. The maximum value is ∞. So the range is [5, ∞), which is equal to the codomain. Thus, it is onto. Since it's one-one and onto, it is bijective.

Mini Cheatsheet

{FLASHCARD: q="What are the three main ways to check if a function is one-one (injective)?" | a="1. Algebraic: Show f(x₁) = f(x₂) ⇒ x₁ = x₂. 2. Graphical: Use the Horizontal Line Test. 3. Calculus: Show the derivative f'(x) is either always positive or always negative in the domain."}