# Question about injection, surjection, bijection, and mapping

## Main Question or Discussion Point

f(x) is a bijection if and only if f(x) is both a surjection and a bijection. Now a surjection is when every element of B has at least one mapping, and an injection is when all of the elements have a unique mapping from A, and therefore a bijection is a one-to-one mapping.

Let's say that f(x) = x3-x+1.

It's easy to see that it is not injective by showing that f(1)=f(-1)=1
Since the function is defined for all x, it is surjective (-inf, +inf)

Then the example finds an interval such that the function is a bijection on a mapping of S to S such that S is a subset of R.

The example problem says that the function is a bijective on the interval [1,+inf). As I see it, it would also be surjective on the interval (-inf,-1].

Is that correct and the book just doesn't mention it since it just asks for "a single" interval? or is there a reason that this interval does not work?

It would seem that the union of [-inf, -1) (1, +inf] would be the interval on which the function is a bijection.

Thanks a lot for your time.

Related Set Theory, Logic, Probability, Statistics News on Phys.org
CRGreathouse
Homework Helper
$(-\infty,-1]\cup[1,+\infty)$ is not an interval, so it doesn't work. But other answers are possible, for example S = {1}.

$(-\infty,-1]\cup[1,+\infty)$ is not an interval, so it doesn't work. But other answers are possible, for example S = {1}.

Okay cool. The question does state an interval which is more than a single point.

So $(-\infty,-1]$ would be an acceptable interval then is the conclusion?

HallsofIvy
Homework Helper
Yes, it would. In fact, since the "turning points" are at $x= \pm\sqrt{3}/3$, "maximal" intervals on which f(x)= x3- x+ 1 is an injection are
$$\left[-\infty, -\frac{\sqrt{3}}{3}\right]$$
$$\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right]$$
and
$$\left[\frac{\sqrt{3}}{3}, \infty\right]$$

CRGreathouse
Homework Helper
Okay cool. The question does state an interval which is more than a single point.
{1} = [1, 1] is an interval.

HallsofIvy
Homework Helper
Yes, of course. The point was that the question asked to find "an interval which was more than a single point". It wouldn't be necessary to add "which was more than a single point" if there were no intervals containing only a single point.

CRGreathouse