Questions: Function Types and Bijections

Question 1

Consider the function f: ℝ → ℝ defined by f(x) = x². Which statement is correct?

• A. f is injective but not surjective, because distinct positive inputs give distinct outputs
• B. f is surjective but not injective, because every non-negative real number is an output
• C. f is neither injective nor surjective
• D. f is bijective, because every input has exactly one output

Answer: (C)

f fails injectivity: f(2) = f(−2) = 4, so two different inputs give the same output. f fails surjectivity: no real number maps to −1 (no x satisfies x² = −1), so negative numbers are not in the image. Option A is wrong because f(2) = f(−2). Option B confuses the image (non-negative reals) with the codomain (all reals). Option D confuses 'each input has one output' (the definition of being a function at all) with bijectivity.

Question 2

Which modification would turn f(x) = x² into a bijective function?

• A. Restrict the domain to all integers
• B. Restrict the domain to [0, ∞) and the codomain to [0, ∞)
• C. Restrict only the codomain to [0, ∞), keeping the domain as ℝ
• D. Restrict only the domain to [0, ∞), keeping the codomain as ℝ

Answer: (B)

Restricting domain to [0, ∞) makes f injective (non-negative inputs give distinct outputs). Restricting codomain to [0, ∞) makes f surjective (every non-negative number is now hit). Both restrictions together produce a bijection. Option C fails: restricting only the codomain to [0, ∞) fixes surjectivity but f(2)=f(−2) still breaks injectivity. Option D fails: restricting only the domain to [0, ∞) makes it injective but negative numbers remain in the codomain with no preimage, so it is not surjective. This illustrates why domain and codomain are both part of the function's definition.

Question 3

A function can be injective without being surjective, and surjective without being injective — these are independent properties.

• T. True
• F. False

Answer: True

Injectivity and surjectivity are genuinely independent. The function f(x) = eˣ from ℝ to ℝ is injective (distinct inputs give distinct outputs) but not surjective (negative numbers have no preimage). A constant function f(x) = 0 from ℝ to ℝ is surjective only if the codomain is {0}, but fails injectivity for any domain with more than one element. A bijection requires both properties simultaneously, and many functions have neither.

Question 4

If a function f: A → B is injective, then most element of B should be the output of some element of A.

• T. True
• F. False

Answer: False

This is the definition of surjectivity, not injectivity. Injectivity says: different inputs give different outputs (if f(x) = f(y) then x = y). It says nothing about whether every element of B is reached. A function can be injective while leaving many elements of B with no preimage — for instance, f: {1,2} → {1,2,3} defined by f(1)=1, f(2)=2 is injective but 3 ∈ B has no preimage. Confusing these two properties is one of the most common errors in beginning set theory.

Question 5

Why do mathematicians define the cardinality of infinite sets using bijections rather than by direct counting?

Model answer: Counting works for finite sets because we can match elements to natural numbers 1, 2, 3, ... and read off the size. But for infinite sets, there is no 'last number' to read off — direct counting never terminates. A bijection provides a size-comparison without requiring a count: if a bijection f: A → B exists, then A and B are matched element-for-element with nothing left over on either side, so they have the same cardinality. Crucially, this definition reveals that not all infinite sets are the same size — there is no bijection between ℕ and ℝ, so these infinite sets have different cardinalities.

The bijection-based definition of cardinality is one of Cantor's great insights. The natural numbers and the even numbers are both countably infinite (there is a bijection n ↦ 2n), but the natural numbers and the real numbers are NOT in bijection — there are 'more' reals than naturals (Cantor's diagonal argument). This would be invisible to any approach based on naive counting.