What one-one asks of distinct inputs, and what many-one allows

What to assume they know

• Functions from Class XI: domain, co-domain, range, and the image of an element
• The identity, constant, polynomial, rational, modulus and signum functions and their graphs, from Class XI
• Adding two functions defined on the same domain
• Parity of a natural number, squares, cubes and fourth powers
• The greatest integer of a real number
• Reading a graph, and what it means for a point to lie on one

What they should be able to do

• State Definition 5 in both its forms and explain why the second form is the one a proof uses
• Prove a stated function one-one by starting from an equality of two images and deriving equality of the two inputs
• Show a stated function many-one by exhibiting one pair of distinct inputs with a common image
• Read an arrow diagram and decide injectivity from the arrowheads alone
• Explain why the same rule can be one-one on one domain and many-one on another, and give the chapter's own instance
• Handle a function defined by cases, showing that no collision can occur across the cases as well as within them
• Decide injectivity for the standard modulus, signum and greatest-integer functions, and name the collision in each
• Count the one-one functions from a small finite set to itself

Where it usually goes wrong

• "I checked several pairs and none collided, so the function is one-one." Checking pairs can only ever refute, never establish, unless the domain is finite and you check all of them. Miscellaneous Exercise Q5 is the case where exhaustive checking is legitimate — four inputs — and every other example in this topic is not.
• "Many-one is a separate property I have to prove." It is the negation of one-one, so proving it means producing one colliding pair and stopping. Students who try to argue that "many inputs go to each output" are proving something stronger than the definition asks.
• "One-one is a property of the formula." It is a property of the formula together with the domain. Exercise 1.2 Q2 puts squaring on N and on Z side by side, and the verdict flips. The formula is the same character for character.
• "If a function is defined in two cases, checking each case separately is enough." Example 12 is the counter-lesson: the interesting collision would be between an odd input and an even one, across the case boundary, and the chapter spends most of its argument ruling that out.
• "Squaring is one-one because every number has one square." Every function sends each input to one output; that is what makes it a function. One-one is the reverse question — whether each output comes from one input — and students routinely answer the first question when asked the second.
• "A graph that keeps rising and then falls can still be one-one somewhere, so it is one-one." Injectivity is a claim about the whole declared domain. The parabola of Fig 1.4 is one-one on the non-negative reals and the chapter still calls it many-one, because R is what was declared.
• "Adding two one-one functions gives a one-one function." Miscellaneous Example 26 kills this in one line with sine and cosine on a quarter turn.
• "The signum function must be one-one because it looks like it has three separate branches." Three branches, but only a handful of output values, so collisions are unavoidable the moment the domain is infinite.

Questions to check understanding

• Prove that a given function is one-one, starting from an assumed equality of images
• Show that a given function is many-one by exhibiting a colliding pair
• Check the injectivity of a family of functions differing only in domain — the form of Exercise 1.2 Q2
• Classify a stated function as one-one, as onto, or as both, and justify each half separately
• Choose the correct classification of a function from four options
• Decide whether two functions given by different formulas on a small domain are the same function
• Count the one-one functions between two small finite sets

Examples worth working on the board

Values marked verified are worked out here on the chapter's own data; no answer key was consulted, and the chapter prints no answers to its exercises.

• Fig 1.2, panels (i) to (iv) (Part I p. 8). Four arrow diagrams sharing the same source set. Read off the page image, the arrows are: (i) the source holds 1, 2, 3, 4 and the target holds a, b, c, d, e, f; the arrows run 1 to a, 2 to b, 3 to d and 4 to c, so the last two cross; (ii) the same two sets; the arrows run 1 to b, 2 to b, 3 to c and 4 to d; (iii) the source holds 1, 2, 3, 4 and the target holds a, b, c; the arrows run 1 to a, 2 to a, 3 to b and 4 to c; (iv) the source holds 1, 2, 3, 4 and the target holds a, b, c, d; the arrows run 1 to b, 2 to c, 3 to a and 4 to d, with three of them crossing. Verified against Definition 5: no target takes two arrows in panels (i) or (iv), so those two are one-one; b takes two in panel (ii) and a takes two in panel (iii), so each of those is many-one. The chapter reaches the same four verdicts on p. 7.
• Definition 5 (§1.3, Part I p. 7). The chapter gives the condition in words — distinct inputs get distinct images — and immediately restates it as an implication running the other way, from an equality of two images to an equality of the two inputs. It then names the negation many-one. Section 3 of the explanation exists to show these are the same condition and that only the second is workable.
• Example 7 (Part I p. 8). A is a class of fifty students, and f sends each student to their roll number, landing in N. Verified: no two students of one class carry one roll number, so an equality of roll numbers forces the students to be the same — the function is one-one. Note that the argument is a fact about school administration, and that is exactly the point: the proof shape is the same whatever supplies the reason.
• Example 8 (Part I p. 8). f from N to N sends x to 2x. Verified: if 2x₁ = 2x₂ then x₁ = x₂, so f is one-one. Two lines, one cancellation, and the whole argument is the contrapositive form in action.
• Example 11 and Fig 1.4 (Part I p. 9). f from R to R sends x to x squared; it fails both tests, injectivity and surjectivity alike. The collision the chapter names is at –1 and 1, both of which square to 1. The figure is a parabola opening upward with the two points marked and a caption stating that 1 and –1 share an image; the values x = –1 and x = 1 are labelled on the horizontal axis.
• Example 12 (Part I pp. 9–10). f from N to N sends an odd x to x + 1 and an even x to x – 1. The chapter's injectivity argument has three parts: if one input were odd and the other even, the supposed equality would force their difference to be 2, which contradicts one being one more and the other one less; and within the odd case and within the even case the cancellation is immediate. Verified: the function swaps 1 with 2, 3 with 4, 5 with 6, and so on, so it is one-one and the cross-case collision genuinely cannot happen.
• Exercise 1.2 Q2 (Part I p. 10), five functions differing only in domain, co-domain and exponent: (i) from N to N, squaring; (ii) from Z to Z, squaring; (iii) from R to R, squaring; (iv) from N to N, cubing; (v) from Z to Z, cubing. Verified, injectivity column only: (i) is one-one, because two natural numbers with equal squares are equal; (ii) is not, since –1 and 1 both square to 1; (iii) is not, for the same collision; (iv) is one-one; (v) is one-one, because cubing preserves sign and a negative cube can never equal a positive one. Items (i) against (ii) are the sharpest thing on the page: the rule is identical and only the domain changed.
• Exercise 1.2 Q3 (Part I p. 10). The greatest integer function from R to R, sending x to the greatest integer at most x. Verified: 1.2 and 1.5 both go to 1, so it is many-one.
• Exercise 1.2 Q4 (Part I p. 11). The modulus function from R to R. Verified: –1 and 1 both go to 1, so it is many-one.
• Exercise 1.2 Q5 (Part I p. 11). The signum function from R to R, printed as three cases. Verified: whichever way the third case is read, two distinct inputs share an output — see the note below about how that case is printed — so the function is many-one. Use 1 and 2 as the colliding pair, since both are positive and both go to 1; that pair works under either reading and avoids the printed slip entirely.
• Exercise 1.2 Q6 (Part I p. 11). A = {1, 2, 3}, B = {4, 5, 6, 7}, and f is the three-pair function sending 1 to 4, 2 to 5 and 3 to 6. Verified: the three images are distinct, so f is one-one; and 7 receives nothing, which the question does not ask about. Use it to make the point that the two questions of this module are independent.
• Exercise 1.2 Q7 and Q11 and Q12 (Part I p. 11). Q7(i): from R to R, x goes to 3 – 4x. Q7(ii): from R to R, x goes to 1 + x squared. Q11: x goes to x to the fourth. Q12: x goes to 3x. Verified, injectivity only: 3 – 4x is one-one, since equal outputs cancel to equal inputs; 1 + x squared is not, since 1 and –1 both give 2; the fourth power is not, since 1 and –1 both give 1; 3x is one-one.
• Miscellaneous Exercise Q2 (Part I p. 15). The cubing function from R to R is injective. Verified: the cube is strictly increasing on the reals, so distinct inputs cannot share a cube.
• Miscellaneous Exercise Q5 (Part I p. 15). A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2}. One function sends x to x squared minus x; the other sends x to twice the absolute value of x minus one half, then subtracts 1. The question is whether the two are equal, and the hint defines equal functions as those agreeing at every point of the common domain. Verified: the first gives 2, 0, 0, 2 at –1, 0, 1, 2. The second gives, at those same four inputs, 3 – 1 = 2, 1 – 1 = 0, 1 – 1 = 0 and 3 – 1 = 2. The four values agree, so the two functions are equal. Note in passing that both are many-one, since –1 and 2 share an image and so do 0 and 1. This is the explanation's best illustration of section 2 — a four-element domain is small enough to check every input, and that is precisely why the general case needs the other reading of the definition.
• Miscellaneous Example 26 (Part I p. 15). On the closed interval from 0 to half of pi, sine is one-one and so is cosine, and yet their sum is not. Verified: the chapter's own witness is the pair of endpoints — at 0 the sum is 1 + 0, and at half of pi it is 0 + 1, so both give 1. Being one-one is not preserved by adding functions, which students assume without noticing.
• Miscellaneous Example 22 (Part I p. 14). The number of one-one functions from {1, 2, 3} to itself is 6, because such a function is a rearrangement of three symbols and there are 3 factorial of those.

Figures to have open

• Redraws of Fig 1.2 panels (i) to (iv) (Part I p. 8) with the arrows exactly as listed above. These are the chapter's own diagrams and the whole of §1.3 refers back to them; the explanation needs all four, and needs the arrowheads accurate, because panels (i) and (ii) differ only in where two arrows land.
• A redraw of Fig 1.4 (Part I p. 9): a parabola through the origin opening upward, with the points at x = –1 and x = 1 marked and joined by a horizontal cut at height 1. From the textbook's figure, redrawn; the horizontal cut is the explanation's addition.
• Graphs of the modulus, signum and greatest-integer functions for section 10. Standard schematics; this chapter prints none of them, and Class XI does.
• A list of the six rearrangements of three symbols for section 12. Not in the book.

Where this sits in the book

• NCERT Class 12 Mathematics, Part I, Chapter 1 "Relations and Functions", §1.3 Types of Functions, opening paragraphs and Definition 5, p. 7
• Fig 1.2, panels (i) to (iv), p. 8, with Examples 7 and 8
• Examples 11 and 12 with Fig 1.4, pp. 9–10
• Exercise 1.2, questions 2 to 7, 11 and 12, pp. 10–11
• Miscellaneous Example 22, p. 14; Miscellaneous Example 26 and Miscellaneous Exercise questions 2 and 5, p. 15