- How do you do a bijective function?
- Injectivity of a function
- Surjectivity of a function
- Function conditioning
- Examples: solved exercises
- Exercise 1
- Exercise 2
- Exercise 3
- Exercise 4
- Proposed exercises
- References
A bijective function is one that meets the double condition of being injective and surjective. That is, all the elements of the domain have a single image in the codomain, and in turn the codomain is equal to the rank of the function (R f).
It is fulfilled by considering a one-to-one relationship between the elements of the domain and codomain. A simple example is the function F: R → R defined by the line F (x) = x
Source: Author
It is observed that for each value of the domain or starting set (both terms apply equally) there is a single image in the codomain or arrival set. In addition, there is no element of the codomain other than image.
In this way F: R → R defined by the line F (x) = x is bijective
How do you do a bijective function?
To answer this, it is necessary to be clear about the concepts relating to Injectivity and Overjectivity of a function, as well as the criteria for conditioning functions in order to adapt them to the requirements.
Injectivity of a function
A function is injective when each of the elements of its domain is related to a single element of the codomain. An element of the codomain can only be the image of a single element of the domain, in this way the values of the dependent variable cannot be repeated.
To consider a function injective, the following must be met:
∀ x 1 ≠ x 2 ⇒ F (x 1) ≠ F (x 2)
Surjectivity of a function
A function is classified as surjective if each element of its codomain is an image of at least one element of the domain.
To consider a function surjective, the following must be fulfilled:
Let F: D f → C f
∀ b ℮ C f E a ℮ D f / F (a) = b
This is the algebraic way to establish that for every “b” that belongs to C f there is an “a” that belongs to D f such that the function evaluated in “a” is equal to “b”.
Function conditioning
Sometimes a function that is not bijective can be subjected to certain conditions. These new conditions can make it a bijective function. All kinds of modifications to the domain and codomain of the function are valid, where the objective is to fulfill the properties of injectivity and surjectivity in the corresponding relationship.
Examples: solved exercises
Exercise 1
Let the function F: R → R be defined by the line F (x) = 5x +1
A:
It is observed that for every value of the domain there is an image in the codomain. This image is unique which makes F an injective function. In the same way, we observe that the codomain of the function is equal to its rank. Thus fulfilling the condition of surjectivity.
Being injective and surjective at the same time we can conclude that
F: R → R defined by the line F (x) = 5x +1 is a bijective function.
This applies to all linear functions (Functions whose highest degree of the variable is one).
Exercise 2
Let the function F: R → R be defined by F (x) = 3x 2 - 2
When drawing a horizontal line, it is observed that the graph is found on more than one occasion. Due to this the function F is not injective and therefore it will not be bijective as long as it is defined in R → R
Similarly, there are codomain values that are not images of any element of the domain. Due to this, the function is not surjective, which also deserves to condition the arrival set.
We proceed to condition the domain and codomain of the function
F: →
Where it is observed that the new domain covers the values from zero to positive infinity. Avoiding the repetition of values that affects injectivity.
Likewise, the codomain has been modified, counting from "-2" to positive infinity, eliminating from the codomain the values that did not correspond to any element of the domain
In this way it can be ensured that F : → defined by F (x) = 3x 2 - 2
It is bijective
Exercise 3
Let the function F: R → R be defined by F (x) = Sen (x)
In the interval the sine function varies its results between zero and one.
Source: Author.
The function F does not correspond to the criteria of injectivity and surjectivity, because the values of the dependent variable are repeated every interval of π. Furthermore, the terms of the codomain outside the interval are not an image of any element of the domain.
When studying the graph of the function F (x) = Sen (x), intervals are observed where the behavior of the curve meets the bijectivity criteria. As for example the interval D f = for the domain. And C f = for the codomain.
Where the function varies results from 1 to -1, without repeating any value in the dependent variable. And at the same time the codomain is equal to the values adopted by the expression Sen (x)
Thus the function F: → defined by F (x) = Sen (x). It is bijective
Exercise 4
State the necessary conditions for D f and C f. So the expression
F (x) = -x 2 be bijective.
Source: Author
The repetition of results is observed when the variable takes opposite values:
F (2) = F (-2) = -4
F (3) = F (-3) = -9
F (4) = F (-4) = -16
The domain is conditioned, limiting it to the right side of the real line.
D f =
In the same way, it is observed that the range of this function is the interval, which when acting as a codomain fulfills the conditions of surjectivity.
In this way we can conclude that
The expression F: → defined by F (x) = -x 2 It is bijective
Proposed exercises
Check if the following functions are bijective:
F: → R defined by F (x) = 5ctg (x)
F: → R defined by F (x) = Cos (x - 3)
F: R → R defined by the line F (x) = -5x + 4
References
- Introduction to Logic and Critical Thinking. Merrilee H. Salmon. University of Pittsburgh
- Problems in Mathematical Analysis. Piotr Biler, Alfred Witkowski. University of Wroclaw. Poland.
- Elements of Abstract Analysis. Mícheál O'Searcoid PhD. Department of mathematics. University college Dublin, Beldfield, Dublind 4
- Introduction to Logic and to the Methodology of the Deductive Sciences. Alfred Tarski, New York Oxford. Oxford University press.
- Principles of mathematical analysis. Enrique Linés Escardó. Editorial Reverté S. A 1991. Barcelona Spain.