Question about injection, surjection, bijection, and mapping

In summary, a function f(x) is a bijection if and only if it is both a surjection and an injection. An example of a function that is not injective is f(x) = x3-x+1 because f(1) = f(-1) = 1. However, it is surjective over the interval (-∞, ∞). To find an interval on which the function is a bijection, an example problem suggests the interval [1, ∞). However, other intervals such as (-∞, -1] can also work. It is important to note that (-∞, -1] is not an interval. The correct intervals for this function to be an injection are [-
  • #1
Nok1
18
0
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.
 
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  • #2
[itex](-\infty,-1]\cup[1,+\infty)[/itex] is not an interval, so it doesn't work. But other answers are possible, for example S = {1}.
 
  • #3
CRGreathouse said:
[itex](-\infty,-1]\cup[1,+\infty)[/itex] 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 [itex](-\infty,-1][/itex] would be an acceptable interval then is the conclusion?
 
  • #4
Yes, it would. In fact, since the "turning points" are at [itex]x= \pm\sqrt{3}/3[/itex], "maximal" intervals on which f(x)= x3- x+ 1 is an injection are
[tex]\left[-\infty, -\frac{\sqrt{3}}{3}\right][/tex]
[tex]\left[-\frac{\sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right][/tex]
and
[tex]\left[\frac{\sqrt{3}}{3}, \infty\right][/tex]
 
  • #5
Nok1 said:
Okay cool. The question does state an interval which is more than a single point.

{1} = [1, 1] is an interval.
 
  • #6
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.
 
  • #7
HallsofIvy said:
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.

Yes, but that was posted only after my post.
 
  • #8
Sorry about being unclear in the beginning. Thanks for the replies.
 

Related to Question about injection, surjection, bijection, and mapping

1. What is the difference between injection, surjection, and bijection?

Injection, surjection, and bijection are all types of functions in mathematics. An injection is a function where each input has a unique output, meaning no two inputs can map to the same output. A surjection is a function where every element in the output has at least one corresponding input. A bijection is a combination of both an injection and surjection, meaning each input has a unique output and every element in the output has at least one corresponding input.

2. How do I know if a function is injective, surjective, or bijective?

To determine if a function is injective, you can use the horizontal line test. Draw a horizontal line anywhere on the graph of the function. If the line intersects the graph in more than one point, the function is not injective. To determine if a function is surjective, you can check if every element in the output has at least one corresponding input. Finally, to determine if a function is bijective, it must pass both the horizontal line test and have every element in the output with at least one corresponding input.

3. What is a mapping in mathematics?

In mathematics, a mapping is a relationship between two sets of elements, where each element in the first set is paired with a unique element in the second set. This relationship is often represented as a function, where the first set is the domain and the second set is the range. Mappings are used to describe the relationships between different mathematical objects, such as numbers or geometric shapes.

4. How can I use injections, surjections, and bijections in real-life situations?

Injections, surjections, and bijections are used in various fields such as computer science, economics, and biology. In computer science, they are used to optimize algorithms and data structures. In economics, they are used to model relationships between supply and demand. In biology, they are used to study the interactions between different species in an ecosystem. They can also be used in everyday life, such as in creating unique identification numbers or mapping routes for navigation.

5. What are some properties of bijections?

Bijections have several important properties, including being both injective and surjective, meaning each element in the domain has a unique corresponding element in the range. Additionally, bijections are invertible, meaning they can be reversed or undone. This property is useful in solving equations and finding the inverse of a function. Bijections also preserve equality, meaning if two elements in the domain are equal, their corresponding elements in the range will also be equal.

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