(d) $${f_4}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$; $$f_4(1)=d$$, $$f_4(2)=b$$, $$f_4(3)=e$$, $$f_4(4)=a$$, $$f_4(5)=c$$; $$C=\{3\}$$, $$D=\{c\}$$. Let f: X → Y be a function. Therefore the inverse of is given by . Proof: Substitute y o into the function and solve for x. Consider the equation and we are going to express in terms of . If this happens, $$f$$ is not onto. One-To-One Functions | Onto Functions | One-To-One Correspondences | Inverse Functions, if f(a1) = f(a2), then a1 = a2. Since $$\mathbb{R}$$ is closed under subtraction and non-zero division, $$a-\frac{b}{3} \in \mathbb{R}$$ and $$\frac{b}{3} \in \mathbb{R}$$ , thus $$(x,y) \in \mathbb{R} \times \mathbb{R}$$. In other words, Range of f = Co-domain of f. e.g. Using the definition of , we get , which is equivalent to . Then f is one-to-one if and only if f is onto. We also have, for example, $$f\big([\,2,\infty)\big) = [4,\infty)$$. 2. Choose  $$x=\frac{y-11}{5}.$$  Note that if b1 is not equal to b2, that f(a,b1) = f(a,b2), but (a,b1) and (a,b2) are certainly not the same. A function $$f :{A}\to{B}$$ is onto if, for every element $$b\in B$$, there exists an element $$a\in A$$ such that $f(a) = b.$ An onto function is also called a surjection, and we say it is surjective. Find a subset $$B$$ of $$\mathbb{R}$$ that would make the function $$s :{\mathbb{R}}\to{B}$$ defined by $$s(x) = x^2$$ an onto function. (b) $${f_2}:{\{1,2,3,4\}}\to{\{a,b,c,d,e\}}$$; $$f_2(1)=c$$, $$f_2(2)=b$$, $$f_2(3)=a$$, $$f_2(4)=d$$;$$C=\{1,3\}$$, $$D=\{b,d\}$$. ∈ = (), where ∃! (a) Find $$f(3,4)$$, $$f(-2,5)$$, $$f(2,0)$$. f (x 1 ) = x 1. f (x 2 ) = x 2. Since $$u(-2)=u(1)=2$$, the function $$u$$ is not one-to-one. In general, how can we tell if a function $$f :{A}\to{B}$$ is onto? For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. f(a) = b, then f is an on-to function. Consider the example: Proof: Suppose x1 and x2 are real numbers such that f(x1) = f(x2). Maybe it just looks like 2b1 plus 3b2-- I'm just writing a particular case, it won't always be this-- minus b3. For the function $$g :{\mathbb{Z}}\to{\mathbb{Z}}$$ defined by $g(n) = n+3,\nonumber$ we find range of $$g$$ is $$\mathbb{Z}$$, and $$g(\mathbb{N})=\{4,5,6,\ldots\}$$. Theorem: A function is surjective (onto) iff it has a right inverse Proof (⇒): Assume f: A → B is surjective – For every b ∈ B, there is a non-empty set A b ⊆ A such that for every a ∈ A b, f(a) = b (since f is surjective) – Define h : b ↦ an arbitrary element of A b – Again, this is a well-defined function since A b is This is equivalent to the following statement: for every element b in the codomain B, there is exactly one element a in the domain A such that f(a)=b.Another name for bijection is 1-1 correspondence (read "one-to-one correspondence). Simplifying conditions for invertibility. A common proof technique in combinatorics, number theory, and other fields is the use of bijections to show that two expressions are equal. A function ƒ: A → B is onto if and only if ƒ (A) = B; that is, if the range of ƒ is B. x is a real number since sums and quotients (except for division by 0) of real numbers are real numbers. The function is bijective (one-to-one and onto, one-to-one correspondence, or invertible) if each element of the codomain is mapped to by exactly one element of the domain. $\Z_n$ 3. Example: The linear function of a slanted line is onto. In F1, element 5 of set Y is unused and element 4 is unused in function F2. Onto Functions We start with a formal deﬁnition of an onto function. Notice we are asked for the image of a set with two elements. In the first figure, you can see that for each element of B, there is a pre-image or a matching element in Set A. But the definition of "onto" is that every point in Rm is mapped to from one or more points in Rn. The French word sur means over or above, and relates to the fact that the image of the domain of a surjective function completely covers the function's codomain. We do not want any two of them sharing a common image. Prove that h is not one-to-one by giving a counter example. (a) Find $$f(C)$$. Mathematically, if the rule of assignment is in the form of a computation, then we need to solve the equation $$y=f(x)$$ for $$x$$. A function is surjective or onto if the range is equal to the codomain. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. So the discussions below are informal. [5.1] Informally, a function from A to B is a rule which assigns to each element a of A a unique element f(a) of B. Oﬃcially, we have Deﬁnition. When $$f$$ is a surjection, we also say that $$f$$ is an onto function or that $$f$$ maps $$A$$ onto $$B$$. 6. Define the $$r :{\mathbb{Z}\times\mathbb{Z}}\to{\mathbb{Q}}$$ according to $$r(m,n) = 3^m 5^n$$. The horizontal line y = b crosses the graph of y = f(x) at precisely the points where f(x) = b. Determine $$f(\{(0,2), (1,3)\})$$, where the function $$f :\{0,1,2\} \times\{0,1,2,3\} \to \mathbb{Z}$$ is defined according to. In an onto function, codomain, and range are the same. Clearly, f : A ⟶ B is a one-one function. That is, combining the definitions of injective and surjective, ∀ ∈, ∃! In words : ^ Z element in the co -domain of f has a pre -]uP _ Mathematical Description : f:Xo Y is onto y x, f(x) = y Onto Functions onto (all elements in Y have a We will de ne a function f 1: B !A as follows. Otherwise, many-one. Deﬁnition 2.1. The GCD and the LCM; 7. If f : A -> B is an onto function then, the range of f = B . is also onto. Relating invertibility to being onto and one-to-one. A function $$f : A \to B$$ is said to be bijective (or one-to-one and onto) if it is both injective and surjective. Let b 2B. exercise $$\PageIndex{2}\label{ex:ontofcn-02}$$, exercise $$\PageIndex{3}\label{ex:ontofcn-03}$$. Given a function $$f :{A}\to{B}$$, and $$C\subseteq A$$, the image of  $$C$$ under  $$f$$ is defined as $f(C) = \{ f(x) \mid x\in C \}.$ In words, $$f(C)$$ is the set of all the images of the elements of $$C$$. $$r:{\mathbb{Z}_{36}}\to{\mathbb{Z}_{36}}$$; $$r(n)\equiv 5n$$ (mod 36). Then, we have. Onto Function. 1.1. . If $$x\in f^{-1}(D)$$, then $$x\in A$$, and $$f(x)\in D$$. Let f: X → Y be a function. It is clearly onto, because, given any $$y\in[2,5]$$, we can find at least one $$x\in[1,3]$$ such that $$h(x)=y$$. A function f is said to be one-to-one (or injective) if f(x 1) = f(x 2) implies x 1 = x 2. All of the vectors in the null space are solutions to T (x)= 0. We also say that $$f$$ is a one-to-one correspondence . $$s :{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$s(n)\equiv n+5$$ (mod 10). Surjective (onto) and injective (one-to-one) functions. I leave as an exercise the proof that fis onto. If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. For example, if C (A) = Rk and Rm is a subspace of Rk, then the condition for "onto" would still be satisfied since every point in Rm is still mapped to by C (A). Finding an inverse function for a function given by a formula: Example: Define f: R R by the rule f(x) = 5x - 2 for all x -1. f -1(y) = x    such that    f(x) = y. 2.1. . In this case, the function f sets up a pairing between elements of A and elements of B that pairs each element of A with exactly one element of B and each element of B with exactly one element of A. The Phi Function—Continued; 10. Therefore $$f$$ is onto, by definition of onto. In other words, if each b ∈ B there exists at least one a ∈ A such that. Lemma 2. So what is the inverse of ? We want to know if it contains elements not associated with any element in the domain. List all the onto functions from $$\{1,2,3,4\}$$ to $$\{a,b\}$$? Since f is injective, this a is unique, so f 1 is well-de ned. Determine which of the following functions are onto. (c) $${f_3}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$; $$f_3(1)=b$$, $$f_3(2)=b$$, $$f_3(3)=b$$, $$f_3(4)=a$$, $$f_3(5)=d$$; $$C=\{1,3,5\}$$, $$D=\{c\}$$. Example: Define f : R R by the rule f(x) = 5x - 2 for all xR. It is like saying f(x) = 2 or 4 . That's the $$x$$ we want to choose so that $$g(x)=y$$. The quadratic function $f:\R\to [1,\infty)$ given by $f(x)=x^2+1$ is onto. Find $$u([\,3,5))$$ and $$v(\{3,4,5\})$$. But g : X ⟶ Y is not one-one function because two distinct elements x1 and x3have the same image under function g. (i) Method to check the injectivity of a functi… Put y = f (x) Find x in terms of y. In the example of functions from X = {a, b, c} to Y = {4, 5}, F1 and F2 given in Table 1 are not onto. One-to-one functions focus on the elements in the domain. Find $$r^{-1}(D)$$, where $$D=\{3,9,27,81,\ldots\,\}$$. Let f : A !B. (a) $${f_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d\}}$$; $$f_1(1)=b$$, $$f_1(2)=c$$, $$f_1(3)=a$$, $$f_1(4)=a$$, $$f_1(5)=c$$; $$C=\{1,3\}$$, $$D=\{a,c\}$$. In addition to finding images & preimages of elements, we also find images & preimages of sets. A function g : B !A is the inverse of f if f g = 1 B and g f = 1 A. Theorem 1. Proof: Let y R. (We need to show that x in R such that f(x) = y.) exercise $$\PageIndex{6}\label{ex:ontofcn-6}$$. Proof: Let y R. (We need to show that x in R such that f(x) = y.). A bijective function is also called a bijection. such that $$f(x)=y$$. Is it possible for a function from $$\{1,2\}$$ to $$\{a,b,c,d\}$$ to be onto? In exploring whether or not the function is an injection, it might be a good idea to uses cases based on whether the inputs are even or odd. f : A B can be both one-to-one and onto at the same time. A function is one to one if f(x)=f(y) implies that x=y, onto if for all y in the domain there is an x such that f(x) = y, and it's bijective if it is both one to one and onto. Prove that g is not onto by giving a counter example. In other words, if every element in the codomain is assigned to at least one value in the domain. Putting f (x 1 ) = f (x 2 ) x 1 = x 2. https://goo.gl/JQ8Nys The Composition of Surjective(Onto) Functions is Surjective Proof. The quadratic function $f:\R\to\R$ given by $f(x)=x^2+1$ is not. Explain. The proof of g is an onto function from Y 2 to X 2 is quite similar Please work from MH 3100 at Nanyang Technological University Explain. How would you go about proving that the function f:(0,1) -> R, defined as f(x) = (x-1/2)/[x(x-1)] is onto? $$t :{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$t(n)\equiv 3n+5$$ (mod 10). If the function satisfies this condition, then it is known as one-to-one correspondence. • Yes. Diode in opposite direction? Range is the number of elements in Set B which have their relative elements in set A. A function is surjective or onto if each element of the codomain is mapped to by at least one element of the domain. … Hands-on exercise $$\PageIndex{1}\label{he:ontofcn-01}$$. Any function induces a surjection by restricting its co exercise $$\PageIndex{10}\label{ex:ontofcn-10}$$, Give an example of a function $$f :\mathbb{N}\to \mathbb{N}$$ that is. So, given an arbitrary element of the codomain, we have shown a preimage in the domain. Let f : A !B be bijective. We already know that f(A) Bif fis a well-de ned function. Hence there is no integer n for g(n) = 0 and so g is not onto. So surely Rm just needs to be a subspace of C (A)? It fails the "Vertical Line Test" and so is not a function. In your case, A = {1, 2, 3, 4, 5}, and B = N is the set of natural numbers (? CS 441 Discrete mathematics for CS M. Hauskrecht Bijective functions Theorem: Let f be a function f: A A from a set A to itself, where A is finite. This means a formal proof of surjectivity is rarely direct. Congruence; 2. Monday: Functions as relations, one to one and onto functions What is a function? exercise $$\PageIndex{1}\label{ex:ontofcn-01}$$. Example $$\PageIndex{2}\label{eg:ontofcn-02}$$, Consider the function $$g :\mathbb{R} \times \mathbb{R} \to{\mathbb{R}}$$ defined by $$g(x,y)=\frac{x+y}{2}.$$. By the theorem, there is a nontrivial solution of Ax = 0. By definition, to determine if a function is ONTO, you need to know information about both set A and B. The Fundamental Theorem of Arithmetic; 6. When depicted by arrow diagrams, it is illustrated as below: A function which is a one-to-one correspondence. Because $f(0,2)=0+2=2, \qquad\mbox{and}\qquad f(1,3)=1+3=4,$ we determine that $$f(\{(0,2),(1,3)\}) = \{2,4\}$$.a Set, Given a function $$f :{A}\to{B}$$, and $$D\subseteq B$$, the preimage  $$D$$ of under  $$f$$ is defined as $f^{-1}(D) = \{ x\in A \mid f(x) \in D \}.$ Hence, $$f^{-1}(D)$$ is the set of elements in the domain whose images are in $$C$$. (a) $$f(C)=\{0,2,4,9\}$$. Please Subscribe here, thank you!!! For each of the following functions, find the image of $$C$$, and the preimage of $$D$$. Suppose that T (x)= Ax is a matrix transformation that is not one-to-one. Therefore, if f-1 (y) ∈ A, ∀ y ∈ B then function is onto. ( a, b) ∈ R × R since 2 x ∈ R because the real numbers are closed under multiplication and 0 ∈ R. g ( a, b) = g ( 2 x, 0) = 2 x + 0 2 = x . We will de ne a function f 1: B !A as follows. Is it onto? Determining whether a transformation is onto. This key observation is often what we need to start a proof with. y = 2x + 1. Construct a one-to-one and onto function $$f$$ from $$[1,3]$$ to $$[2,5]$$. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. (b) $$f_2(C)=\{a,c\}$$ ; $$f_2^{-1}(D)=\{2,4\}$$ Prove that it is onto. Since x 1 = x 2 , f is one-one. This means that the null space of A is not the zero space. Proof: Let y R. (We need to show that x in R such that f(x) = y. Example: Define g: Z Z by the rule g(n) = 2n - 1 for all n Z. The preimage of $$D\subseteq B$$ is defined as $$f^{-1}(D) = \{x\in A \mid f(x)\in D\}$$. Since f is surjective, there exists a 2A such that f(a) = b. (b)  $$u^{-1}((2,7\,])=(-3,-\frac{4}{3}]$$ and $$v^{-1}((2,7\,]=\{-2\})$$. If $$y\in f(C)$$, then $$y\in B$$, and there exists an $$x\in C$$ In mathematics, a bijective function or bijection is a function f : A → B that is both an injection and a surjection. An onto function is also called surjective function. Onto Functions We start with a formal deﬁnition of an onto function. (It is also an injection and thus a bijection.) Here, y is a real number. Its graph is displayed on the right of Figure 6.5. (We need to show x1 = x2 .). Hands-on exercise $$\PageIndex{2}\label{he:ontofcn-02}$$. Watch the recordings here on Youtube! What is the difference between "Do you interest" and "...interested in" something? (a) $$f_1(C)=\{a,b\}$$ ; $$f_1^{-1}(D)=\{2,3,4,5\}$$ Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. Sal says T is Onto iff C (A) = Rm. The symbol $$f^{-1}(D)$$ is also pronounced as “$$f$$ inverse of $$D$$.”. Example 7 . To prove that a function is not surjective, simply argue that some element of cannot possibly be the output of the function . (c) $$f_3(C)=\{b,d\}$$ ; $$f_3^{-1}(D)=\emptyset$$ • If no horizontal line intersects the graph of the function more than once, then the function is one-to-one. The preimage of $$D$$ is a subset of the domain $$A$$. Why has "pence" been used in this sentence, not "pences"? Take any real number, x ∈ R. Choose ( a, b) = ( 2 x, 0) . Demonstrate $$x$$ is indeed an element of the domain, $$A.$$. We say f is onto, or surjective, if and only if for any y ∈ Y, there exists some x ∈ X such that y = f(x). That is, the function is both injective and surjective. If f is one-to-one and onto, then its inverse function g is defined implicitly by the relation g(f(x)) = x. A function is said to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. Into Function : Function f from set A to set B is Into function if at least set B has a element which is not connected with any of the element of set A. Therefore, it is an onto function. And it will essentially be some function of all of the b's. So, total numbers of onto functions from X to Y are 6 (F3 to F8). Let’s take some examples. For functions from R to R, we can use the “horizontal line test” to see if a function is one-to-one and/or onto. Since f is injective, this a is unique, so f 1 is well-de ned. If X has m elements and Y has 2 elements, the number of onto functions will be 2 m-2. (d) $$f_4(C)=\{e\}$$ ; $$f_4^{-1}(D)=\{5\}$$. We need to find an $$x$$ that maps to $$y.$$ Suppose  $$y=5x+11$$; now we solve for $$x$$ in terms of $$y$$. The key question is: given an element $$y$$ in the codomain, is it the image of some element $$x$$ in the domain? The function $$u :{\mathbb{R}}\to{\mathbb{R}}$$ is defined as $$u(x)=3x+11$$, and the function $$v :{\mathbb{Z}}\to{\mathbb{R}}$$ is defined as $$v(x)=3x+11$$. I'm writing a particular case in here, maybe I shouldn't have written a particular case. Many-one Function : If any two or more elements of set A are connected with a single element of set B, then we call this function as Many one function. In arrow diagram representations, a function is onto if each element of the co-domain has an arrow pointing to it from some element of the domain. Have questions or comments? However, g(n) 0 for any integer n. 2n  = 1       by adding 1 on both sides, n  = 1/2      by dividing 2 on both sides. Proving or Disproving That Functions Are Onto. So, every element in the codomain has a preimage in the domain and thus $$f$$ is onto. In other words, ƒ is onto if and only if there for every b ∈ B exists a ∈ A such that ƒ (a) = b. Thus, we have found an $$x \in \mathbb{R}$$ such that $$g(x)=y.$$ We now review these important ideas. Fix any . f : N → N (There are infinite number of natural numbers) f : R → R (There are infinite number of real numbers ) f : Z → Z (There are infinite number of integers) Steps : How to check onto? Also, if the range of $$f$$ is equal to $$B$$, then $$f$$ is onto. Example: Define f : R R by the rule f(x) = 5x - 2 for all x R. Prove that f is onto. That is, y=ax+b where a≠0 is a surjection. The following arrow-diagram shows onto function. Therefore, do not merely say “the image.” Be specific: the image of an element, or the image of a subset. If we can always express $$x$$ in terms of $$y$$, and if the resulting $$x$$-value is in the domain, the function is onto. Prove:’ 1.’The’composition’of’two’surjective’functions’is’surjective.’ 2.’The’composition’of’two’injectivefunctionsisinjective.’ Determine which of the following are onto functions. Let f 1(b) = a. Now, since the real numbers are closed under subtraction and non-zero division, $$x \in \mathbb{R}.$$ Hands-on exercise $$\PageIndex{3}\label{he:ontofcn-03}$$. \end{aligned}\] Since preimages are sets, we need to write the answers in set notation. Public Key Cryptography; 12. If such a real number x exists, then 5x -2 = y and x = (y + 2)/5. We need to show that b 1 = b 2. That is, y=ax+b where a≠0 is a surjection. In other words, nothing is left out. Thus, for any real number, we have shown a preimage $$\mathbb{R} \times \mathbb{R}$$ that maps to this real number. Given a function $$f :{A}\to{B}$$, the image of $$C\subseteq A$$ is defined as $$f(C) = \{f(x) \mid x\in C\}$$. What are One-To-One Functions? Put f (x 1 ) = f (x 2 ), If x 1 = x 2 , then it is one-one. The term surjective and the related terms injective and bijective were introduced by Nicolas Bourbaki, a group of mainly French 20th-century mathematicians who, under this pseudonym, wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. Write something like this: “consider .” (this being the expression in terms of you find in the scrap work) Show that . But 1/2 is not an integer. Also given any IMG SRC="images/I>b in B, there is an element a in A such that f(a) = b as f is onto and there is only one such b as f is one-to-one. Example $$\PageIndex{4}\label{eg:ontofcn-04}$$, Is the function $${u}:{\mathbb{Z}}\to{\mathbb{Z}}$$ defined by, $u(n) = \cases{ 2n & if n\geq0 \cr -n & if n < 0 \cr} \nonumber$. Consider the function . f: X → YFunction f is onto if every element of set Y has a pre-image in set Xi.e.For every y ∈ Y,there is x ∈ Xsuch that f(x) = yHow to check if function is onto - Method 1In this method, we check for each and every element manually if it has unique imageCheckwhether the following areonto?Since all Here is a brief overview of surjective, injective and bijective functions: Surjective: If f: P → Q is a surjective function, for every element in … Onto Function A function f: A -> B is called an onto function if the range of f is B. Prove that f is onto. The co-domain of g is Z by the definition of g and 0 Z. The function $$g$$ is both one-to-one and onto. For example sine, cosine, etc are like that. Let f : A ⟶ B and g : X ⟶ Y be two functions represented by the following diagrams. f is onto y   B, x  A such that f(x) = y. Conversely, a function f: A B is not onto y in B such that x  A,  f(x) y. To decide if this function is onto, we need to determine if every element in the codomain has a preimage in the domain. Conversely, a function f: A B is not a one-to-one function elements a1 and a2 in A such that f(a1) = f(a2) and a1 a2. So, total numbers of onto functions from X to Y are 6 (F3 to F8). (It is also an injection and thus a bijection.) Given a function $$f :{A}\to{B}$$, and $$C\subset A$$, since $$f(C)$$ is a subset of $$B$$, the preimage of this subset is indicated by the notation $$f^{-1}(f(C))$$. If f and fog are onto, then it is not necessary that g is also onto. So let me write it this way. The image of an ordered pair is the average of the two coordinates of the ordered pair. Thus, f : A ⟶ B is one-one. $$f :{\mathbb{Z}_{10}}\to{\mathbb{Z}_{10}}$$; $$h(n)\equiv 3n$$ (mod 10). A function is not onto if some element of the co-domain has no arrow pointing to it. Now we much check that f 1 is the inverse of f. All elements in B are used. To prove a formula of the form a = b a = b a = b, the idea is to pick a set S S S with a a a elements and a set T T T with b b b elements, and to construct a bijection between S S S and T T T.. 2. is onto (surjective)if every element of is mapped to by some element of . Let x ∈ A, y ∈ B and x, y ∈ R. Then, x is pre-image and y is image. In particular, the preimage of $$B$$ is always $$A$$. Deﬁnition 2.1. So let f 1(b 1) = f 1(b 2) = a for some b 1;b 2 2Band a2A. If there is a function f which has a onIMG SRC="images//I> correspondence from a set A to a set B, then there is a function from B to A that "undoes" the action of f. This function is called the inverse function for f. A function f and its inverse function f -1. Onto function or Surjective function : Function f from set A to set B is onto function if each element of set B is connected with set of A elements. Prove:’ 1.’The’composition’of’two’surjective’functions’is’surjective.’ 2.’The’composition’of’two’injectivefunctionsisinjective.’ This means that given any element a in A, there is a unique corresponding element b = f(a) in B. Let $$(x,y)=(a-\frac{b}{3} ,\frac{b}{3})$$. Therefore, $$t^{-1}(\{-1\}) = \{2,3\}$$. ), and ƒ (x) = x². Proving or Disproving That Functions Are Onto. We find $x=\frac{y-11}{5}.$ (We'll need to verify $$x$$ is a real number - an element in the domain.). Let f : A !B be bijective. $$f :{\mathbb{Z}}\to{\mathbb{Z}}$$; $$f(n)=n^3+1$$, $$g :{\mathbb{Q}}\to{\mathbb{Q}}$$; $$g(x)=n^2$$, $$h :{\mathbb{R}}\to{\mathbb{R}}$$; $$h(x)=x^3-x$$, $$k :{\mathbb{R}}\to{\mathbb{R}}$$; $$k(x)=5^x$$, $$p :{\mathscr{P}(\{1,2,3,\ldots,n\})}\to{\{0,1,2,\ldots,n\}}$$; $$p(S)=|S|$$, $$q :{\mathscr{P}(\{1,2,3,\ldots,n\})}\to{\mathscr{P}(\{1,2,3,\ldots,n\})}$$; $$q(S)=\overline{S}$$, $$f_1:\{1,2,3,4,5\}\to\{a,b,c,d\}$$; $$f_1(1)=b$$, $$f_1(2)=c$$, $$f_1(3)=a$$, $$f_1(4)=a$$, $$f_1(5)=c$$, $${f_2}:{\{1,2,3,4\}}\to{\{a,b,c,d,e\}}$$; $$f_2(1)=c$$, $$f_2(2)=b$$, $$f_2(3)=a$$, $$f_2(4)=d$$, $${f_3}:{\mathbb{Z}}\to{\mathbb{Z}}$$; $$f_3(n)=-n$$, $${g_1}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$; $$g_1(1)=b$$, $$g_1(2)=b$$, $$g_1(3)=b$$, $$g_1(4)=a$$, $$g_1(5)=d$$, $${g_2}:{\{1,2,3,4,5\}}\to{\{a,b,c,d,e\}}$$; $$g_2(1)=d$$, $$g_2(2)=b$$, $$g_2(3)=e$$, $$g_2(4)=a$$, $$g_2(5)=c$$. Equivalently, a function is surjective if its image is equal to its codomain. 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