MAT148H1 Chap.3 Functions, One-to-One Maps and Inverses
Functions, One-to-One Maps and Inverses
What one-to-one really says
A function f with domain D is one-to-one, also called injective, when different inputs always produce different outputs. Written with quantifiers, the definition reads: for all x₁, x₂ in D, if x₁ ≠ x₂ then f(x₁) ≠ f(x₂). Every piece of that sentence carries weight. Without the universal quantifier the symbols x₁ and x₂ are undefined, and the sentence means nothing.
Replacing "for all" with "there exist" only asks for two points with different images, something a function on the set {1, 2, 3} sending 1 and 3 to the same value still satisfies. Reversing the implication gives a statement that every function satisfies, because two different outputs can never come from the same input.
Squaring is the classic failure: f(−2) = 4 = f(2), so x² is two-to-one on the real line, and sine is infinitely many-to-one because it repeats every 2π.
Proving injectivity with the contrapositive
The definition is rarely proved in its original direction. Instead you use the logically equivalent contrapositive: for all x₁, x₂ in D, if f(x₁) = f(x₂) then x₁ = x₂. The proof template is short.
Fix two arbitrary inputs u and v in the domain, assume their outputs are equal, and manipulate the equation until it forces u = v. For a linear map cx with c ≠ 0, the equation cu = cv becomes c(u − v) = 0, and since c is not zero, u − v must be zero. Rational functions usually yield to cross-multiplication, and polynomials to factoring out (u − v).
When the leftover factor can vanish, the domain matters: x² − 4x fails on all of ℝ but becomes injective once the domain is cut down to [2, ∞).
A geometric shortcut, the horizontal line test, gives intuition: a graph is one-to-one exactly when no horizontal line meets it more than once.
Building the inverse and its domain
If f is injective with range R, the inverse f⁻¹ is the function from R back to D defined by f⁻¹(y) = x exactly when f(x) = y. Consequently f⁻¹(f(x)) = x for every x in D and f(f⁻¹(y)) = y for every y in R.
To compute a formula, write y = f(x) and solve for x. The step where solving becomes illegal, typically a division by an expression that can vanish, reveals the values missing from the range, and those values are excluded from the domain of the inverse.
Geometrically the inverse is the reflection of the graph in the line y = x; that reflection turns the horizontal line test into the vertical line test, which explains why only injective functions possess inverse functions.
Where these ideas return later
Injectivity also behaves well under composition: when two one-to-one maps are chained and the composite takes one value at inputs u and v, the outer map forces the inner outputs at u and v to coincide, and the inner map then forces u = v, so the composite is again one-to-one.
Later in the course the Inverse Function Theorem states that when f is continuously differentiable at c with f′(c) ≠ 0, then near c the function is injective with a continuously differentiable inverse whose derivative equals 1 divided by f′(f⁻¹(x)).
That theorem is the engine behind derivatives of arcsine, arctangent and the logarithm, so careful domain and range bookkeeping now pays off directly in the unit on transcendental functions.
What this chapter covers
- 01
Quantified definition of one-to-one
- 02
Contrapositive form used in proofs
- 03
Counterexamples that break injectivity
- 04
Restricting a domain to recover injectivity
- 05
Inverse function, domain and range
- 06
Composition of injective functions
Worked example · free
Injectivity and inverse of a rational function
- 1Let u, v be in the domain, so u ≠ 1 and v ≠ 1, and suppose the outputs at u and v coincide. Both denominators are nonzero, so cross-multiply: (2u + 3)(v − 1) = (2v + 3)(u − 1).
- 2Expand: 2uv − 2u + 3v − 3 = 2uv − 2v + 3u − 3. Cancel 2uv − 3 to get 5v = 5u, hence u = v. So f is injective.
- 2Solve y = (2x + 3)/(x − 1): yx − y = 2x + 3, so x(y − 2) = y + 3 and x = (y + 3)/(y − 2) whenever y ≠ 2.
- 1Check y = 2 is never attained: it would force 2x + 3 = 2x − 2, that is 3 = −2. So the range of f is ℝ without 2, and f⁻¹ has domain ℝ without 2 and range ℝ without 1. Test: f(2) = 7 and f⁻¹(7) = 10/5 = 2.
Key terms
- Injective Function
- An injective function is a map that sends distinct inputs to distinct outputs, so equal outputs can only come from equal inputs. It is the same notion as one-to-one and is required for an inverse function.
- Contrapositive
- A contrapositive is the logically equivalent statement formed from P implies Q by writing not Q implies not P. For injectivity it turns the definition into the usable form: equal outputs imply equal inputs.
- Inverse Function
- An inverse function is the map f⁻¹ from the range of an injective f back to its domain, defined by f⁻¹(y) = x exactly when f(x) = y, so composing in either order returns the input.
- Horizontal Line Test
- The horizontal line test is a graphical check stating that a function is one-to-one when every horizontal line meets its graph at most once. It offers intuition but does not replace an algebraic proof.
Functions, One-to-One Maps and Inverses FAQ
How do you prove a function is one-to-one?
Take two arbitrary points u and v from the domain and assume their images agree. Then use algebra such as cross-multiplication or factoring out (u − v) to force u = v. Because a and b were arbitrary, this covers every pair simultaneously and completes the proof.
What is the easiest way to show a function is not injective?
Exhibit one explicit pair of different inputs from the domain with identical outputs, for example g(0) = g(4) = 0 when g(x) = x² − 4x. One concrete pair is enough because injectivity is a universal claim, and a single counterexample falsifies it.
Why must a function be one-to-one to have an inverse?
Reversing a many-to-one map would send a single output back to several inputs, so the reversed rule fails to be a function. Reflecting the graph in y = x shows this: a horizontal line meeting the graph twice becomes a vertical line meeting the reflection twice.
Exam move
Memorise the contrapositive form word for word, then practise on three families: a rational function (cross-multiply), a polynomial (factor out a − b), and a function that fails (find one explicit pair). For every inverse, write the domain and range in a separate line and verify one numerical value both ways.
Finally, rewrite the definition with each quantifier slip from the lecture list and explain aloud why each version is wrong.
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