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Relations and Functions - One-to-one and onto functions

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of a Function: A function f:ArightarrowBf: A \\rightarrow B is a specific type of relation where every element in set AA (domain) is mapped to exactly one unique element in set BB (codomain). Visually, this means every point in the domain set has one, and only one, outgoing arrow pointing to an element in the codomain.

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One-to-one (Injective) Function: A function is injective if distinct elements in the domain have distinct images in the codomain. Mathematically, f(x1)=f(x2)f(x_1) = f(x_2) must imply x1=x2x_1 = x_2. Visually, no two arrows from the domain point to the same element in the codomain. On a graph, this is confirmed by the Horizontal Line Test: any horizontal line should intersect the graph at most once.

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Many-to-one Function: If a function is not one-to-one, it is many-to-one. This occurs when two or more distinct elements in the domain map to the same element in the codomain (e.g., f(1)=5f(1) = 5 and f(2)=5f(2) = 5). Visually, you will see multiple arrows converging onto a single point in the codomain, and a horizontal line would intersect the graph at two or more points.

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Onto (Surjective) Function: A function f:ArightarrowBf: A \\rightarrow B is surjective if every element in the codomain BB is the image of at least one element in the domain AA. This means the Range of the function is exactly equal to its Codomain (Range=BRange = B). Visually, every element in the codomain set has at least one arrow pointing to it, and the graph covers the entire vertical range defined by the codomain.

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Into Function: A function is called an 'into' function if it is not onto. This means there is at least one element in the codomain BB that does not have a pre-image in the domain AA. In this case, the Range is a proper subset of the Codomain (RangesubsetBRange \\subset B). Visually, there will be 'leftover' elements in the codomain set with no incoming arrows.

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Bijective Function: A function that is both one-to-one (injective) and onto (surjective) is called a bijective function or a 'one-to-one correspondence'. These functions are always invertible. Visually, this creates a perfect pairing where every element of AA is matched with exactly one unique element of BB, and no elements in BB are left unmatched.

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Composition and Invertibility: For a function f:ArightarrowBf: A \\rightarrow B to have an inverse f−1:BrightarrowAf^{-1}: B \\rightarrow A, it must be bijective. If the function is not one-to-one, the reverse mapping would not be a function (one input would go to multiple outputs). If it is not onto, the reverse mapping would have inputs with no outputs.

📐Formulae

Condition for Injectivity: f(x1)=f(x2)impliesx1=x2forallx1,x2inDomainf(x_1) = f(x_2) \\implies x_1 = x_2 \\forall x_1, x_2 \\in Domain

Condition for Surjectivity: f(A)=Bf(A) = B or Range=CodomainRange = Codomain

Number of injective functions from AA to BB (where n(A)=m,n(B)=nn(A)=m, n(B)=n): nPm=fracn!(n−m)!^nP_m = \\frac{n!}{(n-m)!} if ngeqmn \\geq m, else 00

Number of bijective functions from AA to BB (where n(A)=n(B)=nn(A) = n(B) = n): n!n!

Number of onto functions from AA to BB (where n(A)=m,n(B)=nn(A)=m, n(B)=n): sumr=0n(−1)r,nCr,(n−r)m\\sum_{r=0}^{n} (-1)^r \\, ^nC_r \\, (n-r)^m

💡Examples

Problem 1:

Show that the function f:mathbbRrightarrowmathbbRf: \\mathbb{R} \\rightarrow \\mathbb{R} defined by f(x)=2x+3f(x) = 2x + 3 is a bijective function.

Solution:

Step 1: Check for Injectivity (One-to-one). Let f(x1)=f(x2)f(x_1) = f(x_2) for some x1,x2inmathbbRx_1, x_2 \\in \\mathbb{R}.\n2x1+3=2x2+32x_1 + 3 = 2x_2 + 3\n2x1=2x22x_1 = 2x_2\nx1=x2x_1 = x_2.\nSince f(x1)=f(x2)impliesx1=x2f(x_1) = f(x_2) \\implies x_1 = x_2, the function is injective.\nStep 2: Check for Surjectivity (Onto). Let yy be an arbitrary element in the codomain mathbbR\\mathbb{R}.\nSet y=f(x)=2x+3y = f(x) = 2x + 3.\nSolve for xx: x=fracy−32x = \\frac{y-3}{2}.\nSince yy is a real number, x=fracy−32x = \\frac{y-3}{2} is also a real number (belongs to the domain).\nf(x)=f(fracy−32)=2(fracy−32)+3=y−3+3=yf(x) = f(\\frac{y-3}{2}) = 2(\\frac{y-3}{2}) + 3 = y - 3 + 3 = y.\nThus, for every yy in the codomain, there exists an xx in the domain such that f(x)=yf(x) = y. The function is surjective.\nConclusion: Since ff is both injective and surjective, it is bijective.

Explanation:

To prove bijectivity, we must algebraically demonstrate that the function satisfies both the one-to-one condition and the onto condition.

Problem 2:

Check the injectivity and surjectivity of the function f:mathbbZrightarrowmathbbZf: \\mathbb{Z} \\rightarrow \\mathbb{Z} given by f(x)=x2f(x) = x^2.

Solution:

Step 1: Check Injectivity. Let x1=1x_1 = 1 and x2=−1x_2 = -1.\nf(1)=(1)2=1f(1) = (1)^2 = 1 and f(−1)=(−1)2=1f(-1) = (-1)^2 = 1.\nHere f(1)=f(−1)f(1) = f(-1) but 1neq−11 \\neq -1.\nTherefore, ff is not injective (it is many-to-one).\nStep 2: Check Surjectivity. The codomain is mathbbZ\\mathbb{Z} (all integers).\nConsider y=−2y = -2 in the codomain. Is there an xinmathbbZx \\in \\mathbb{Z} such that x2=−2x^2 = -2?\nNo integer squared results in a negative number.\nSince there is no xinmathbbZx \\in \\mathbb{Z} such that f(x)=−2f(x) = -2, the element −2-2 has no pre-image.\nTherefore, ff is not surjective (it is an 'into' function).

Explanation:

Using counter-examples is the most efficient way to show that a function fails to be injective or surjective. In this case, the square function on integers fails both.