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## 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) = x

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.

Let's say that f(x) = x

^{3}-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.