Onto and Into Functions; Bijections

Transcript

0:00Introduction

Hello everyone! Okay, one quick question before we start. Is f of x equals x squared an onto function? Some of you are saying yes. Some of you are saying no. And look, the funny thing is, both groups can be right! How can one formula give two different answers? Today we will find out. And this is not a small topic. In JEE, questions on types of functions test exactly this idea, again and again. So, let's begin.

0:31Onto, into and bijection

In the last video, you saw one-one functions. Today we look at the other side, the codomain, okay? So, f from A to B is called onto, or surjective, if every element y of B has at least one pre-image. Pre-image means an element of A whose arrow lands on that y. The curly E you see on screen means belongs to. Now look at the four pictures on screen. In each one, the left oval is A and the right oval is B, okay? In which pictures does every dot on the right get at least one arrow? Yes, the second and the third! In the first and the last, one dot on the right is left alone. No arrow comes to it. So onto simply means this. The range, the set of images that are actually used, equals the codomain, the whole of set B, okay? Nothing in B is left out. Clear? And if even one element of B has no pre-image, then the range is strictly smaller than the codomain, and f is called into. Like the first and the last pictures, where one dot is left alone. And a function that is both one-one and onto is called a bijection, or a one-one correspondence. That is the third picture: a perfect pairing. Every element of A gets its own partner in B, and nobody in B is left out. Absolutely perfect!

1:54How to test onto

So how do we test onto in a question? Just two steps. Step one: find the range of f, using the methods of chapter three. Step two: compare that range with the codomain given in the question. Not the codomain you imagine, the one actually written in the question, okay? Onto, if and only if the two are equal. The double arrow you see on screen means if and only if. Range equal to the codomain, onto. Range smaller, into. Simple!

2:23Careful: the codomain decides

Alright, now back to our puzzle. f of x equals x squared, as a map from R to R, okay? The double struck R means all real numbers, and the small arrow just says from one set to the other. Is it onto? x squared takes so many values, so yes, right? No, no, careful! See, a square is never negative, is it? So the range is zero to infinity, with zero included; that is what the square bracket means, and the sideways eight is infinity. Every negative number in the codomain has no pre-image. The range is not equal to the codomain; that is the equals sign with a slash through it. So as a map from R to R, x squared is into. But now change only the codomain. Same x squared, as a map from R to zero, infinity. Now the range equals the codomain, so it is onto! Puzzle solved: one group was thinking of the codomain R, the other group of zero to infinity. Both were right. So onto is not a property of the formula alone. Always, always read the codomain in the question. This is the JEE trap here: students check only the formula and lose easy marks. Got it?

3:40Topper's trick: for finite sets, just count

Now a trick that toppers use when A and B are finite sets. Before any work, just count the elements. For one-one, every element of A needs its own different image in B, right? So which set must be bigger? Yes, set B! n of A must be less than or equal to n of B, where n of A means the number of elements in A, okay? For onto, every element of B needs at least one arrow, and each element of A sends only one arrow. So now A must be the bigger one: n of A greater than or equal to n of B, okay? Both together, for a bijection? n of A equals n of B, exactly. So if the counts do not match, there is no bijection at all. No calculation needed! And one more beautiful result. When A and B are finite and have the same number of elements, one-one and onto become the same thing. Check one, and you get the other free! But remember, only for finite sets of equal size, okay?

4:44Example 1: classify 2x+3, x^2, x^3 and e^x from R to R

Now, example one. Classify each of these functions from R to R, okay? Two x plus three, x squared, x cubed, and e to the power x. For each one: one-one or many-one, onto or into? Pause the video, and try it first. Done? Okay, let's check. Part one, two x plus three. It is one-one. For onto, put y equals two x plus three, and solve for x. The double arrow pointing right means implies: x equals y minus three, over two. For every real y, this x exists, so every y has a pre-image. Onto! One-one and onto, so a bijection. Part two, x squared, our friend from the start. It is many-one, and the range is zero to infinity, but the codomain here is all real numbers. So, into. Exactly what we saw. Part three, x cubed. Tell me, does every real number have a real cube root? Yes, absolutely, negative numbers too! So every real number has a pre-image under x cubed: onto. And it is one-one. So x cubed is a bijection. Part four, e to the x. One-one, yes. But its range is zero to infinity, with round brackets, so zero is not included. Zero and the negative numbers have no pre-image, so into. Two bijections and two into functions. Clear?

6:18Example 2: JEE Main 2019, classify 2x/(x-1)

Now a harder one, from JEE Main two thousand nineteen. Let A be the set of real numbers that are not positive integers, and let f from A to R be two x, over x minus one. Is it injective? Is it surjective? Pause the video, and try it. Done? Okay, let's check. Look, first rewrite it: two x over x minus one equals two, plus two over x minus one. Now x appears only once, and everything becomes easy. Injective, that is one-one, first. Suppose f of x one equals f of x two. Subtract two from both sides, divide by two, and one over x one minus one equals one over x two minus one. So x one equals x two. Injective! Now surjective, that is onto. The codomain is all real numbers. Many students say: it is one-one, the formula looks harmless, so it must be onto as well, a bijection, right? Hmm, wait. Look at two, plus two over x minus one. Which value can it never take? Yes, two! For f of x to be two, two over x minus one must be zero. But a fraction with numerator two can never be zero. Impossible! So the number two in the codomain has no pre-image at all. The range is not the whole codomain, so f is not surjective. Final answer: injective but not surjective, that is, one-one into. See? One small rewrite did all the work. Got it?

7:54Quick revision: range vs codomain, counting for bijection

So, quick revision. Onto means the range equals the codomain; otherwise the function is into. Always read the codomain, not just the formula. And for finite sets, a bijection needs n of A equal to n of B, okay? Now go and try the practice problems on this page. You will find them easy, I promise. All the best!