$\endgroup$ â Brendan W. Sullivan Nov 27 at 1:01 Fix any . Then we may define the inverse sine function $\sin^{-1}:[-1,1]\to[-\pi/2,\pi/2]$, since the sine function is bijective when the domain and codomain are restricted. Give an example of a function \(f : A \rightarrow B\) that is neither injective nor surjective. If f is given as a formula, we may be able to find a by solving the equation \(f(a) = b\) for a. Show that the function \(g : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}\) defined by the formula \(g(m, n) = (m+n, m+2n)\), is both injective and surjective. 1. Show that the function \(f : \mathbb{R}-\{0\} \rightarrow \mathbb{R}\) defined as \(f(x) = \frac{1}{x}+1\) is injective but not surjective. https://math.stackexchange.com/questions/3285806/do-injective-yet-not-bijective-functions-have-an-inverse/3285822#3285822, https://math.stackexchange.com/questions/3285806/do-injective-yet-not-bijective-functions-have-an-inverse/3285817#3285817, $\sin(x) : [0,\pi) \rightarrow \mathbb{R}$, https://math.stackexchange.com/questions/3285806/do-injective-yet-not-bijective-functions-have-an-inverse/3285818#3285818. Verify whether this function is injective and whether it is surjective. The previous example shows f is injective. A function \(f : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}\) is defined as \(f(m,n) = (m+n,2m+n)\). For this, just finding an example of such an a would suffice. Consider function \(h : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Q}\) defined as \(h(m,n)= \frac{m}{|n|+1}\). For all common algebraic structures, and, in particular for vector spaces, an injective homomorphism is also called a â¦ Functions in the first column are injective, those in the second column are not injective. f is not onto i.e. Often it is necessary to prove that a particular function \(f : A \rightarrow B\) is injective. Verify whether this function is injective and whether it is surjective. \sin: \mathbb{R} \to \mathbb{R} In other words, weâve seen that we can have functions that are injective and not surjective (if there are more girls than boys), and we can have functions that are surjective but not injective (if there are more boys than girls, then we had to send more than one boy to at least one of the girls). Injective vs. Surjective: A function is injective if for every element in the domain there is a unique corresponding element in the codomain. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. Injective and surjective functions There are two types of special properties of functions which are important in many di erent mathematical theories, and which you may have seen. Second, as you note, the restriction function Injective, Surjective, and Bijective Functions. How many are surjective? Notice that whether or not f is surjective depends on its codomain. Bijective? Is \(\theta\) injective? $$ A function is a way of matching all members of a set A to a set B. Thus, the map is injective. A function \(f : \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}\) is defined as \(f(n)=(2n, n+3)\). How many of these functions are injective? The function \(f(x) = x^2\) is not injective because \(-2 \ne 2\), but \(f(-2) = f(2)\). The formal definition I was given in my analysis papers was that in order for a function $f(x)$ to have an inverse, $f(x)$ is required to be bijective. Is it surjective? Is it surjective? Please Subscribe here, thank you!!! Note: One can make a non-injective function into an injective function by eliminating part of the domain. (max 2 MiB). How many are bijective? In advanced mathematics, the word injective is often used instead of one-to-one, and surjective is used instead of onto. Prove the function \(f : \mathbb{R}-\{1\} \rightarrow \mathbb{R}-\{1\}\) defined by \(f(x) = (\frac{x+1}{x-1})^{3}\) is bijective. Injective function: | | ||| | An injective non-surjective function (not a |bije... World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and â¦ How many are bijective? You may recall from algebra and calculus that a function may be one-to-one and onto, and these properties are related to whether or not the function is invertible. We note in passing that, according to the definitions, a function is surjective if and only if its codomain equals its range. We need to show that there is some \((x, y) \in \mathbb{Z} \times \mathbb{Z}\) for which \(g(x, y) = (b, c)\). A one-one function is also called an Injective function. Consider the cosine function \(cos : \mathbb{R} \rightarrow \mathbb{R}\). So this function is not an injection. But g : X â¶ 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 functiâ¦ Explain. $$ A non-injective non-surjective function (also not a bijection) A homomorphism between algebraic structures is a function that is compatible with the operations of the structures. However, the function g : R â R 0 + defined by g ( x ) = x 2 (with the restricted codomain) is surjective, since for every y in the nonnegative real codomain Y , there is at least one x in the real domain X such that x 2 = y . It emphasizes the way we think about functions: the "domain" and "codomain" of a function are part of the data of the function, so a restriction is a different function because we've changed the domain (and dually, if we calculate that the range of the function is smaller than the given codomain, it means we can define a new function with the smaller set as its codomain, and that new function won't literally be the same as our old function even though its values are the same). Next we examine how to prove that \(f : A \rightarrow B\) is surjective. Is it surjective? Here is an outline: How to show a function \(f : A \rightarrow B\) is surjective: [Prove there exists \(a \in A\) for which \(f(a) = b\).]. (How to find such an example depends on how f is defined. To define an inverse sine (or cosine) function, we must also restrict the domain $A$ to $A'$ such that $\sin:A'\to B'$ is also injective. $$, $\sin|_{\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}$, https://math.stackexchange.com/questions/3285806/do-injective-yet-not-bijective-functions-have-an-inverse/3285824#3285824. But $sin(x)$ is not bijective, but only injective (when restricting its domain). Then, at last we get our required function as f : Z â Z given by. How to show a function \(f : A \rightarrow B\) is injective: \(\begin{array}{cc} {\textbf{Direct approach}}&{\textbf{Contrapositive approach}}\\ {\text{Suppose} a,a' \in A \text{and} a \ne a'}&{\text{Suppose} a,a' \in A \text{and} f(a) = f(a')}\\ {\cdots}&{\cdots}\\ {\text{Therefore} f(a) \ne f(a')}&{\text{Therefore} a=a'}\\ \nonumber \end{array}\). We now review these important ideas. This function is not injective because of the unequal elements \((1,2)\) and \((1,-2)\) in \(\mathbb{Z} \times \mathbb{Z}\) for which \(h(1, 2) = h(1, -2) = 3\). This is just like the previous example, except that the codomain has been changed. $$ Example: The quadratic function f(x) = x 2 is not an injection. If a function is $f:X\to Y$ is injective and not necessarily surjective then we "create" the function $g:X\to f(X)$ prescribed by $x\mapsto f(x)$. The point is that the authors implicitly uses the fact that every function is surjective on it's image. https://goo.gl/JQ8NysHow to Prove a Function is Surjective(Onto) Using the Definition On the other hand, g(x) = x3 is both injective and surjective, so it is also bijective. A function \(f : \mathbb{Z} \rightarrow \mathbb{Z}\) is defined as \(f(n) = 2n+1\). De nition. That is, no two or more elements of A have the same image in B. \sin|_{\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to \mathbb{R} Injective functions are also called one-to-one functions. This is because the contrapositive approach starts with the equation \(f(a) = f(a′)\) and proceeds to the equation \(a = a'\). We seek an \(a \in \mathbb{R}-\{0\}\) for which \(f(a) = b\), that is, for which \(\frac{1}{a}+1 = b\). To find \((x, y)\), note that \(g(x,y) = (b,c)\) means \((x+y, x+2y) = (b,c)\). This means \(\frac{1}{a} +1 = \frac{1}{a'} +1\). Verify whether this function is injective and whether it is surjective. $$ Consider the function \(\theta : \{0, 1\} \times \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(\theta(a, b) = (-1)^{a}b\). How many are bijective? This is the kind of thing that engineers don't do for the most part (because the distinction rarely matters and it's confusing to have to introduce a ton of symbols to describe what is, from a calculation standpoint, the same thing), logicians/computer scientists do frequently (because these distinctions always matter in those fields) and most mathematicians do only when there is cause for confusion (so we did it above, since we were clarifying exactly this point -- but in casual usage we would not speak of this $\sin^*$ function, most likely). Related Topics. If your function f: X â Y is injective but not necessarily surjective, you can say it has an inverse function defined on the image f (X), but not on all of Y. Let f : A â¶ B and g : X â¶ Y be two functions represented by the following diagrams. hello all! This function $g$ (closely related to $f$ and carrying the same prescription) is bijective so it has an inverse $g^{-1}:f(X)\to X$. In words, we must show that for any \(b \in B\), there is at least one \(a \in A\) (which may depend on b) having the property that \(f(a) = b\). Note that this definition is meaningful. BUT f(x) = 2x from the set of natural numbers to is not surjective, because, for example, no member in can be mapped to 3 by this function. By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. A function $f:A\to B$ that is injective may still not have an inverse $f^{-1}:B\to A$. Is f injective? $$, $$ Lets take two sets of numbers A and B. Then \((m+n, m+2n) = (k+l,k+2l)\). a function thats not surjective means that im (f)!=co-domain (8 votes) See 3 more replies \sin^*: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to [-1, 1]. Say we know an injective function â¦ Prove that the function \(f : \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(f (n) = \frac{(-1)^{n}(2n-1)+1}{4}\) is bijective. The function f is not surjective because there exists an element \(b = 1 \in \mathbb{R}\), for which \(f(x) = \frac{1}{x}+1 \ne 1\) for every \(x \in \mathbb{R}-\{0\}\). Decide whether this function is injective and whether it is surjective. There are no polyamorous matches like the absolute value function, there are just one-to-one matches like f(x) = x+3. Let the extended function be f. For our example let f(x) = 0 if x is a negative integer. Consider the function \(f : \mathbb{R}^2 \rightarrow \mathbb{R}^2\) defined by the formula \(f(x, y)= (xy, x^3)\). So this is how you can define the $\arcsin$ for instance (though for $\arcsin$ you may want the domain to be $[-\frac{\pi}{2},\frac{\pi}{2})$ instead I believe), Click here to upload your image
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. Functions can be one-to-one functions (injections), onto functions (surjections), or both one-to-one and onto functions (bijections). Then \((x, y) = (2b-c, c-b)\). On the other hand, \(g(x) = x^3\) is both injective and surjective, so it is also bijective. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy, 2021 Stack Exchange, Inc. user contributions under cc by-sa, not a duplicate; this is specific to the "inverse" of the $\sin$ function, $$ Thus we need to show that \(g(m, n) = g(k, l)\) implies \((m, n) = (k, l)\). In summary, for any \(b \in \mathbb{R}-\{1\}\), we have \(f(\frac{1}{b-1} =b\), so f is surjective. According to Definition12.4,we must prove the statement \(\forall b \in B, \exists a \in A, f(a)=b\). Then \(h(c, d-1) = \frac{c}{|d-1|+1} = \frac{c}{d} = b\). Can you think of a bijective function now? The notion of a function is fundamentally important in practically all areas of mathematics, so we must review some basic definitions regarding functions. Injective and surjective are not quite "opposites", since functions are DIRECTED, the domain and co-domain play asymmetrical roles (this is quite different than relations, which in â¦ For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. How many are surjective? We now have \(g(2b-c, c-b) = (b, c)\), and it follows that g is surjective. Watch the recordings here on Youtube! Functions in the first row are surjective, those in the second row are not. How many such functions are there? Bijective? Of these two approaches, the contrapositive is often the easiest to use, especially if f is defined by an algebraic formula. Consider the function \(\theta : \mathscr{P}(\mathbb{Z}) \rightarrow \mathscr{P}(\mathbb{Z})\) defined as \(\theta(X) = \bar{X}\). Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). Legal. That is, we say f is one to one In other words f is one-one, if no element in B is associated with more than one element in A. It has cleared my doubts and I'm grateful. Such an interval is $[-\pi/2,\pi/2]$. is injective. Therefore, f is one to one or injective function. Let $f:X\rightarrow Y$ be an injective map. So that logical problem goes away. A function f : A â¶ B is said to be a one-one function or an injection, if different elements of A have different images in B. This question concerns functions \(f : \{A,B,C,D,E,F,G\} \rightarrow \{1,2\}\). This is because $f^{-1}$ may not be able to take input values from $B$ if it is not also surjective: $f$ had no output to some points in $B$, so $f^{-1}$ cannot take inputs from these points in $B$. Suppose \((m,n), (k,l) \in \mathbb{Z} \times \mathbb{Z}\) and \(g(m,n)= g(k,l)\). Bijective? Explain. I also observe that computer scientists are far more comfortable with partial functions, which would permit $\mathrm{arc}\left(\left.\sin\right|_{[-\pi/2,\pi/2]}\right)$. (For the first example, note that the set \(\mathbb{R}-\{0\}\) is \(\mathbb{R}\) with the number 0 removed.). Verify whether this function is injective and whether it is surjective. Now, letâs see an example of how we prove surjectivity or injectivity in a given functional equation. This similarity may contribute to the swirl of confusion in students' minds and, as others have pointed out, this may just be an inherent, perennial difficulty for all students,. It is also surjective , which means that every element of the range is paired with at least one member of the domain (this is obvious because both the range and domain are the same, and each point maps to itself). This leads to the following system of equations: Solving gives \(x = 2b-c\) and \(y = c -b\). The four possible combinations of injective and surjective features are illustrated in the adjacent diagrams. Hope this will be helpful Prove that the function \(f : \mathbb{R}-\{2\} \rightarrow \mathbb{R}-\{5\}\) defined by \(f(x)= \frac{5x+1}{x-2}\) is bijective. Do injective, yet not bijective, functions have an inverse. The second line involves proving the existence of an a for which \(f(a) = b\). Show that the function \(f : \mathbb{R}-\{0\} \rightarrow \mathbb{R}-\{1\}\) defined as \(f(x) = \frac{1}{x}+1\) is injective and surjective. Clearly, f : A â¶ B is a one-one function. A function \(f : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) is defined as \(f(m,n) = 2n-4m\). Explain. (Also, it is not a surjection.) Now this function is bijective and can be inverted. Notice that at each step, we gave the function a new name, $\sin|_{\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}$ and then $\sin^*$ (the former convention is standard in math and the latter was made up for this exposition). Nevertheless, further on on the papers, I was introduced to the inverse of trigonometric functions, such as the inverse of $sin(x)$. However the image is $[-1,1]$ and therefore it is surjective on it's image. However the image is $[-1,1]$ and therefore it is surjective on it's image. We will use the contrapositive approach to show that f is injective. \sin|_{\big[-\frac{\pi}{2}, \frac{\pi}{2}\big]}: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to \mathbb{R} The function g : R â R defined by g(x) = x 2 is not surjective, since there is no real number x such that x 2 = â1. Then \(b = \frac{c}{d}\) for some \(c, d \in \mathbb{Z}\). As you can see the topics I'm studying are probably very basic, so excuse me if my question is silly, but ultimately does a function need to be bijective in order to have an inverse? Is \(\theta\) injective? a non injective/surjective function doesnt have a special name and if a function is injective doesnt say anything about im (f). You can also provide a link from the web. In algebra, as you know, it is usually easier to work with equations than inequalities. How many of these functions are injective? For this it suffices to find example of two elements \(a, a′ \in A\) for which \(a \ne a′\) and \(f(a)=f(a′)\). Let \(A= \{1,2,3,4\}\) and \(B = \{a,b,c\}\). $$ Therefore f is injective. (I'm just following your convenction for preferring $\mathrm{arc}f$ to $f^{-1}$. Subtracting the first equation from the second gives \(n = l\). a â b â f(a) â f(b) for all a, b â A âº f(a) = f(b) â a = b for all a, b â A. e.g. The following examples illustrate these ideas. There are four possible injective/surjective combinations that a function may possess. Missed the LibreFest? Is it surjective? Some people tend to call a bijection a one-to-one correspondence, but not me. Formally, to have an inverse you have to be both injective and surjective. A function \(f : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}\) is defined as \(f(m,n) = 3n-4m\). https://goo.gl/JQ8NysHow to prove a function is injective. How many such functions are there? A function f from a set X to a set Y is injective (also called one-to-one) $$ This means a function f is injective if a1â a2 implies f(a1)â f(a2). Let f : A ----> B be a function. Please Subscribe here, thank you!!! â´ 5 x 1 = 5 x 2 â x 1 = x 2 â´ f is one-one i.e. :D i have a question here..its an exercise question from the usingz book. It follows that \(m+n=k+l\) and \(m+2n=k+2l\). Thus g is injective. Otherwise I would use standard notation.). Notice we may assume d is positive by making c negative, if necessary. In my old calc book, the restricted sine function was labelled Sin$(x)$. Next, subtract \(n = l\) from \(m+n = k+l\) to get \(m = k\). Consider the function \(\theta : \{0, 1\} \times \mathbb{N} \rightarrow \mathbb{Z}\) defined as \(\theta(a, b) = a-2ab+b\). The rst property we require is the notion of an injective function. To see that g is surjective, consider an arbitrary element \((b, c) \in \mathbb{Z} \times \mathbb{Z}\). Nor is it surjective, for if \(b = -1\) (or if b is any negative number), then there is no \(a \in \mathbb{R}\) with \(f(a)=b\). Here is a brief overview of surjective, injective and bijective functions: Surjective: If f: P â Q is a surjective function, for every element in Q, there is at least one element in P, that is, f (p) = q. To prove that a function is surjective, we proceed as follows: . This is illustrated below for four functions \(A \rightarrow B\). f(x) = 0 if x â¤ 0 = x/2 if x > 0 & x is even = -(x+1)/2 if x > 0 & x is odd. This question concerns functions \(f : \{A,B,C,D,E\} \rightarrow \{1,2,3,4,5,6,7\}\). Determine whether this is injective and whether it is surjective. (hence bijective). The figure given below represents a one-one function. How many of these functions are injective? Every identity function is an injective function, or a one-to-one function, since it always maps distinct values of its domain to distinct members of its range. First, as you say, there's no way the normal $\sin$ function When we speak of a function being surjective, we always have in mind a particular codomain. This is a reasonable thing to be confused about since the terminology reveals an inconsistency between the way computer-scientists talk about functions, pure mathematicians talk about functions, and engineers talk about functions. Solving for a gives \(a = \frac{1}{b-1}\), which is defined because \(b \ne 1\). Is \(\theta\) injective? Verify whether this function is injective and whether it is surjective. whose graph is the wave could ever have an inverse. Please Subscribe here, thank you!!! It's not injective and so there would be no logical way to define the inverse; should $\sin^{-1}(0) = 0$ or $2\pi$? Discussion: Any horizontal line y=c where c>0 intersects the graph in two points. But a function is injective when it is one-to-one, NOT many-to-one. How many are surjective? Suppose \(a, a′ \in \mathbb{R}-\{0\}\) and \(f (a) = f (a′)\). the question is: We may categorise functions of {0; 1} -> {0; 1} according to whether they are injective, surjective both. Note that is not surjective because, for example, the vector cannot be obtained as a linear combination of the first two vectors of the standard basis (hence there is at least one element of the codomain that does not belong to the range of). \sin^*: \big[-\frac{\pi}{2}, \frac{\pi}{2}\big] \to [-1, 1]. Bijective? If this is the case, how can we talk about the inverse of trigonometric functions such as $sin$ or $cosine$? Injective, Surjective, and Bijective tells us about how a function behaves. â¢ A function that is both injective and surjective is called a bijective function or a bijection. A function $f:X\to Y$ has an inverse if and only if it is bijective. Have questions or comments? Subtracting 1 from both sides and inverting produces \(a =a'\). Nor is it surjective, for if b = â 1 (or if b is any negative number), then there is no a â R with f(a) = b. https://goo.gl/JQ8NysProof that if g o f is Injective(one-to-one) then f is Injective(one-to-one). The function f is called an one to one, if it takes different elements of A into different elements of B. This question concerns functions \(f : \{A,B,C,D,E,F,G\} \rightarrow \{1,2,3,4,5,6,7\}\). However, sometimes papers speaks about inverses of injective functions that are not necessarily surjective on the natural domain. We will use the contrapositive approach to show that g is injective. Here are the exact definitions: 1. injective (or one-to-one) if for all \(a, a′ \in A, a \ne a′\) implies \(f(a) \ne f(a')\); 2. surjective (or onto B) if for every \(b \in B\) there is an \(a \in A\) with \(f(a)=b\); 3. bijective if f is both injective and surjective. The two main approaches for this are summarized below. So, f is a function. What if it had been defined as \(cos : \mathbb{R} \rightarrow [-1, 1]\)? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Onto or Surjective function. Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. Below is a visual description of Definition 12.4. The inverse is conventionally called $\arcsin$. However, h is surjective: Take any element \(b \in \mathbb{Q}\). So we can calculate the range of the sine function, namely the interval $[-1, 1]$, and then define a third function: However, if you restrict the codomain of $f$ to some $B'\subset B$ such that $f:A\to B'$ is bijective, then you can define an inverse $f^{-1}:B'\to A$, since $f^{-1}$ can take inputs from every point in $B'$. Then you can consider the same map, with the range $Y':=\text{range}(f)$. Functions may be "injective" (or "one-to-one") An injective function is a matchmaker that is not from Utah. Thus, f : A â¶ B is one-one. In other words the map $\sin(x):[0,\pi)\rightarrow [-1,1]$ is now a bijection and therefore it has an inverse. To prove that a function is not injective, you must disprove the statement \((a \ne a') \Rightarrow f(a) \ne f(a')\). Whatever we do the extended function will be a surjective one but not injective. Decide whether this function is injective and whether it is surjective. If for instance you consider the functions $\sin(x) : [0,\pi) \rightarrow \mathbb{R}$ then it is injective but not surjective. (Scrap work: look at the equation .Try to express in terms of .). To show that it is surjective, take an arbitrary \(b \in \mathbb{R}-\{1\}\). Some people call the inverse $\sin^{-1}$, but this convention is confusing and should be dropped (both because it falsely implies the usual sine function is invertible and because of the inconsistency with the notation $\sin^2(x)$). For example, \(f(x) = x^2\) is not surjective as a function \(\mathbb{R} \rightarrow \mathbb{R}\), but it is surjective as a function \(R \rightarrow [0, \infty)\). Sometimes you can find a by just plain common sense.) By assigning arbitrary values on Y â f (X), you get a left inverse for your function. Consider the logarithm function \(ln : (0, \infty) \rightarrow \mathbb{R}\). But there's still the problem that it fails to be surjective, e.g. even after we restrict, it doesn't make sense to ask what the inverse value is at $17$ since no value of the domain maps to $17$. surjective as for 1 â N, there docs not exist any in N such that f (x) = 5 x = 1 200 Views A function f:AâB is injective or one-to-one function if for every bâB, there exists at most one aâA such that f(s)=t. In essence, injective means that unequal elements in A always get sent to unequal elements in B. Surjective means that every element of B has an arrow pointing to it, that is, it equals f(a) for some a in the domain of f. For more concrete examples, consider the following functions \(f , g : \mathbb{R} \rightarrow \mathbb{R}\). Moreover, the above mapping is one to one and onto or bijective function. How many such functions are there? Example: The function f(x) = 2x from the set of natural numbers to the set of non-negative even numbers is a surjective function. Every element of A has a different image in B. For this, Definition 12.4 says we must prove that for any two elements \(a, a′ \in A\), the conditional statement \((a \ne a′) \Rightarrow f(a) \ne f(a′)\) is true. This is something that, if we were being extremely literal (for example, maybe if we were writing code that tried to compare two different functions), we would always do. An injective function need not be surjective (not all elements of the codomain may be associated with arguments), and a surjective function need not be injective (some images may be associated with more than one argument). The function f(x) = x2 is not injective because â 2 â 2, but f( â 2) = f(2). \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\), [ "article:topic", "showtoc:no", "authorname:rhammack", "license:ccbynd" ], https://math.libretexts.org/@app/auth/2/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FMathematical_Logic_and_Proof%2FBook%253A_Book_of_Proof_(Hammack)%2F12%253A_Functions%2F12.02%253A_Injective_and_Surjective_Functions, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\). 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