Simulation · Maths · Class 12

One-one, onto, or both?

From the lesson Onto and bijective functions in Relations and Functions. Change the values and watch what happens.

The idea behind it

NCERT §1.3

f : X → Y is onto (surjective) if nothing in Y is left without a preimage: for every y ∈ Y there is an x ∈ X with f(x) = y. Equivalently, the range of f is the whole codomain Y.
To test onto, solve f(x) = y for x and check that the x found lies in the domain. For f : R → R, f(x) = 2x, x = y/2 is real for every real y, so f is onto; for f : N → N, f(x) = 2x, y = 1 would need x = 1/2 ∉ N, so f is not onto.
f is one-one and onto (bijective) when it is both. f(x) = 2x is bijective from R to R but only one-one from N to N.
One-one and onto are independent. The roll-number function into N is one-one but not onto; the map N → N sending both 1 and 2 to 1 and every larger x to x − 1 is onto but not one-one; x² on R is neither, since −2 is never an output.
f : N → N, f(x) = x + 1 for odd x and x − 1 for even x, swaps 1 ↔ 2, 3 ↔ 4, 5 ↔ 6 and so on. It is one-one and onto: an odd y is the image of y + 1, and an even y is the image of y − 1.
The codomain decides onto. x² is not onto as a map R → R, but it is onto as a map R → [0, ∞), because every y ≥ 0 equals (√y)².
Adding functions can break both properties. The identity I_N is onto N, but I_N + I_N, which sends x to 2x, misses 3. sin x and cos x are each one-one on [0, π/2], yet sin x + cos x equals 1 at both x = 0 and x = π/2.

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