Onto as the demand that nothing in the codomain be left unhit

What to assume they know

• What one-one asks of distinct inputs, and what many-one allows — one-one and many-one, and the shape of an injectivity proof
• Domain, co-domain and range from Class XI, and the difference between the last two
• Solving a linear equation for the unknown, and dividing by a non-zero real
• Parity of a natural number; that a real square is never negative
• The absolute value of a real number
• The identity function

What they should be able to do

• State Definition 6 and describe the two-step shape of a surjectivity proof: take an arbitrary element of the co-domain, produce an input that reaches it
• Use the range-equals-co-domain remark to convert a surjectivity question into a question about the range
• Show a function is not onto by naming one element of the co-domain and proving nothing maps to it
• Explain why the same formula is onto with one co-domain and not with another, citing the chapter's own instance
• Explain why shrinking the domain can destroy surjectivity, and give the chapter's exercise where it does
• Produce a function that is onto and not one-one, and say why that is possible on an infinite set
• Restrict a co-domain so that a given function becomes onto by construction

Where it usually goes wrong

• "Onto is a property of the formula, like being a polynomial." It is a property of the formula, the domain and the co-domain together. Doubling on the naturals is not onto; doubling on the reals is. The characters of the formula are identical in Examples 8 and 9, printed one page apart, which is why the chapter puts them where it does.
• "Range and co-domain are the same thing." The co-domain is declared; the range is discovered. The remark on Part I p. 8 says the function is onto exactly when the two coincide, and that sentence would be empty if they always did.
• "To prove a function is onto I should show its outputs stay inside the co-domain." That is what makes it a function at all, not what makes it onto. The demand runs the other way: every element of the co-domain must be produced. Miscellaneous Exercise Q1 needs both directions, and students almost always do only the easy one.
• "To disprove onto I need to show most elements are missed." One unreached element is enough, and the chapter always produces exactly one — 1 in Example 8, 51 in Example 7, –2 in Example 11, 3 in Miscellaneous Example 25.
• "Only the co-domain can affect surjectivity, so changing the domain is safe." Exercise 1.2 Q1 refutes this directly: leave the co-domain alone, shrink the domain from the non-zero reals to the naturals, and the function stops being onto.
• "A function that is onto must be one-one, because it uses everything up." Example 10 folds 1 and 2 together and still reaches every natural number. On an infinite set there is room to waste an input and still cover.
• "Squaring is onto the reals because every real has a square." Again the direction is reversed. The question is whether every real is a square, and the negatives are not.
• "If I add two onto functions I get an onto function." Miscellaneous Example 25 is the counterexample, and it uses the simplest function in the chapter.

Questions to check understanding

• Prove a function onto by taking an arbitrary element of the co-domain and exhibiting an input that reaches it
• Show a function is not onto by naming one unreached element and proving it is unreached
• Check the surjectivity of a family of functions differing only in domain and co-domain — the form of Exercise 1.2 Q2
• Decide whether a function stays onto when its domain is replaced — the form of Exercise 1.2 Q1
• Justify a stated restriction on a domain or a co-domain, saying what would go wrong without it
• Show a function between two intervals is onto, where the co-domain has been chosen to match the range

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) and (iii) (Part I p. 8). Read off the page image: in panel (i) the target holds a, b, c, d, e, f and the four arrows land on a, b, d and c, leaving e and f with nothing pointing at them. In panel (iii) the target holds only a, b, c and the four arrows land on a, a, b and c, so every element of the target is reached. The chapter reaches the same two verdicts on p. 7. Panel (iii) is the one to dwell on: it is onto and it has a collision, which is the whole content of section 8.
• Definition 6 and the Remark (Part I pp. 7–8). The definition asks that every element of the target be the image of something in the source, and states it as an existence claim — for each y there is to be an x with the right image. The remark on p. 8 then says the same thing in one line by equating the range with the co-domain. Section 3 exists to make the student see these as one statement, not two facts.
• Example 8 (Part I p. 8). f from N to N sends x to 2x. Verified: it is not onto, and the chapter's witness is 1, because no natural number doubles to 1. Every odd number is an equally good witness.
• Example 9 and Fig 1.3 (Part I p. 9). f from R to R sends x to 2x. Verified: it is onto, and the proof is the two-step shape — take any real y, offer half of y as the input, and check that doubling it returns y. Compare directly with Example 8: the formula is identical and only the domain and co-domain moved. The figure is a straight line through the origin of positive slope, drawn on labelled axes and annotated with the equation.
• Example 7 (Part I p. 8). Fifty students mapped to their roll numbers, with N as the declared co-domain. Verified: the chapter's witness is 51, which is nobody's roll number; the range is the fifty numbers from 1 to 50 and the co-domain is infinite. If the co-domain had been declared as the numbers from 1 to 50, the same function would be onto without a single change to the rule. That observation is added here, and it is the cleanest possible statement of the thesis.
• Example 10 (Part I p. 9). f from N to N sends 1 and 2 both to 1, and sends every x above 2 to x – 1. Verified: it is onto — for any natural y other than 1, the input y + 1 exceeds 2 and maps to y; and 1 is reached from 1. It is not one-one, because 1 and 2 collide. Put this next to Example 8 and you have one function with each half of the pair, both on the same infinite set.
• Example 11 (Part I p. 9). f from R to R sends x to x squared. The chapter says –2 is not the image of anything and leaves the reason in brackets as a question. Verified answer to the bracket: the square of a real is never negative, so no real input can produce –2; the range is the non-negative reals, and the whole lower half of the co-domain goes unreached.
• Exercise 1.2 Q1 (Part I p. 10). f defined on the non-zero reals, sending x to its reciprocal, with the non-zero reals as co-domain; the question then asks whether the same verdict holds when the domain is replaced by N and the co-domain is left alone. Verified: on the non-zero reals it is onto, since any non-zero y is reached from its own reciprocal, and one-one, since equal reciprocals force equal inputs. With domain N it stays one-one and stops being onto — 2 is in the co-domain and would need an input of one half, which is not a natural number. This is the only place in the chapter where surjectivity dies from a change to the domain, and it deserves its own section, because students carry away that onto is about the co-domain alone.
• Exercise 1.2 Q2 (Part I p. 10), five functions: (i) N to N, squaring; (ii) Z to Z, squaring; (iii) R to R, squaring; (iv) N to N, cubing; (v) Z to Z, cubing. Verified, surjectivity column only: not one of the five is onto. For (i) and (ii) and (iii), 2 is not a square in the relevant set — and in (ii) and (iii) the negatives are missed as well. For (iv) and (v), 2 is not a cube. So the injectivity verdicts vary across the five and the surjectivity verdicts do not vary at all, which is a useful thing to show as a table.
• Exercise 1.2 Q7 (Part I p. 11). (i) from R to R, x goes to 3 – 4x; (ii) from R to R, x goes to 1 + x squared. Verified: (i) is onto, since a target y is reached from the input formed by subtracting y from 3 and dividing by 4; (ii) is not, since every output is at least 1 and so 0 is unreached.
• Exercise 1.2 Q10 (Part I p. 11). The domain is the reals with 3 removed and the co-domain the reals with 1 removed; f sends x to the quotient of x – 2 by x – 3. The question asks for both properties. Verified: it is one-one — cross-multiplying an assumed equality of two images and cancelling the product term leaves the two inputs equal. It is onto — solving the defining equation for x gives the quotient of 3y – 2 by y – 1, which is defined exactly because 1 was removed from the co-domain, and which can never equal 3, since that would force –2 and –3 to be equal. Both printed exclusions are load-bearing and each does a different job: 3 leaves the domain because the formula has no value there, and 1 leaves the co-domain because nothing maps to it. Treat them as content, not fine print.
• Miscellaneous Example 25 (Part I pp. 14–15). The identity function on N is onto. Adding it to itself gives the doubling function, which is not: the chapter's witness is 3. Verified: 3 is odd, so no natural doubles to it. Surjectivity is not preserved by adding functions, and this is the mirror of the sine-and-cosine example that shows injectivity is not preserved either.
• Miscellaneous Exercise Q1 (Part I p. 15). f is declared from R to the set of reals strictly between –1 and 1, sending x to the quotient of x by one plus the absolute value of x; the exercise asks for one-one and onto. Verified: every output does lie strictly between –1 and 1, because the numerator is smaller in size than the denominator. Given a target y in that open interval, the input formed as the quotient of y by one minus the absolute value of y reaches it — substituting gives back y after the denominators cancel. So the function is onto, and it is onto because the co-domain was declared to be exactly the range and nothing more. This is the closing example for the whole topic.
• Exercise 1.2 Q12 (Part I p. 11) and Q11. Tripling on the reals is onto; the fourth power on the reals is not, since no real fourth power is negative. Use them as quick checks after section 5.

Figures to have open

• Redraws of Fig 1.2 panels (i) and (iii) (Part I p. 8), with the arrows exactly as listed in the Worked examples. The unreached elements must be visibly distinct from the reached ones — that contrast is the topic.
• A redraw of Fig 1.3 (Part I p. 9): a straight line of positive slope through the origin on labelled axes, annotated with its equation. From the textbook's figure, redrawn.
• A co-domain-versus-range diagram — two nested regions that can be shown moving into coincidence. An added device, carrying sections 3 and 12.
• A number-line strip of the naturals with alternate elements lit, for section 4. Standard schematic; the chapter draws nothing here.

Where this sits in the book

• NCERT Class 12 Mathematics, Part I, Chapter 1 "Relations and Functions", §1.3 Types of Functions, Definition 6, p. 7, and the Remark on p. 8
• Fig 1.2, panels (i) and (iii), p. 8; Examples 7 and 8, p. 8
• Examples 9, 10 and 11 with Fig 1.3, p. 9
• Exercise 1.2, questions 1, 2, 7, 10, 11 and 12, pp. 10–11
• Miscellaneous Example 25, pp. 14–15, and Miscellaneous Exercise question 1, p. 15