injective function proofs

ii)Function f is surjective i f 1(fbg) has at least one element for all b 2B . Functions that have inverse functions are said to be invertible. See pages that link to and include this page. An alternative notation for the identity function on $A$ is "$id_A$". However, we also need to go the other way. De nition 67. }\) Thus $b = f(a) = y\text{,}$ so $f^{-1}$ is injective. This is what breaks it's surjectiveness. Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License We will now prove some rather trivial observations regarding the identity function. Proofs involving surjective and injective properties of general functions: Let f : A !B and g : B !C be functions, and let h = g f be the composition of g and f. For each of the following statements, either give a formal proof or counterexample. Suppose $f,g$ are surjective and suppose $z \in C\text{. Copy link. }$ Then let $f : A \to A$ be a permutation (as defined above). Let, c = 5x+2. Tap to unmute. "If y and x are injective, then z(n) = y(n) + x(n) is also injective." General Wikidot.com documentation and help section. \begin{align} \quad (f \circ i)(x) = f(i(x)) = f(x) \end{align}, \begin{align} \quad (i \circ f)(x) = i(f(x)) = f(x) \end{align}, Unless otherwise stated, the content of this page is licensed under. Thus a= b. However, the other difference is perhaps much more interesting: combinatorial permutations can only be applied to finite sets, while function permutations can apply even to infinite sets! Therefore, since the given function satisfies the one-to-one (injective) as well as the onto (surjective) conditions, it is proved that the given function is bijective. }\) Thus $A = \range(f^{-1})$ and so $f^{-1}$ is surjective. As per the title, I'm learning discrete mathematics on my own and there's a bunch of proofs in the exercise section that involves proving if the statement is true or false. 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. (proof by contradiction) Suppose that f were not injective. Now if I wanted to make this a surjective and an injective function, I would delete that mapping and I … Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). }\) That is, for every $b \in B$ there is some $a \in A$ for which $f(a) = b\text{.}$. A function $f : A \to B$ is said to be surjective (or onto) if $\range(f) = B\text{. Theidentity function i A on the set Ais de ned by: i A: A!A; i A(x) = x: Example 102. Groups will be the sole object of study for the entirety of MATH-320! Suppose \(f : A \to B$ is bijective, then the inverse function $f^{-1} : B \to A$ is also bijective. Therefore, d will be (c-2)/5. So, what is the difference between a combinatorial permutation and a function permutation? So, every function permutation gives us a combinatorial permutation. The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. If a function is defined by an even power, it’s not injective. Determine whether or not the restriction of an injective function is injective. This is another example of duality. (⇒ ) S… Bijective functions are also called one-to-one, onto functions. Consider the following function that maps N to Z: f(n) = (n 2 if n is even (n+1) 2 if n is odd Lemma. There is another similar formula for quartic equations, but the cubic and the quartic forumlae were not discovered until the middle of the second millenia A.D.! Well, let's see that they aren't that different after all. for every y in Y there is a unique x in X with y = f ( x ). Proof. Notice that we now have two different instances of the word permutation, doesn't that seem confusing? }\) Thus $g \circ f$ is injective. In high school algebra, you learn that a quadratic equation of the form $ax^2 + bx + c = 0$ has two (or one repeated) solutions of the form $x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\text{,}$ and these solutions always exist provided we allow for complex numbers. However, mathematicians almost universally prefer this definition (and for good reason: it leads to a much simpler proof structure when you actually want to prove that a function is injective, and it is much easier to use when you know a function is injective.) 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. Stated in concise mathematical notation, a function f: X → Y is bijective if and only if it satisfies the condition. The next theorem says that even more is true: if $f: A \to B$ is bijective, then $f^{-1} : B \to A$ is also bijective. }\) Then $f^{-1}(b) = a\text{. Let \(b_1,\ldots,b_n$ be a (combinatorial) permutation of the elements of $A\text{. Notice that nothing in this list is repeated (because \(f$ is injective) and every element of $A$ is listed (because $f$ is surjective). a permutation in the sense of combinatorics. If the function satisfies this condition, then it is known as one-to-one correspondence. }\) Since $f$ is injective, $x = y\text{. Is this an injective function? The simple linear function f (x) = 2 x + 1 is injective in ℝ (the set of all real numbers), because every distinct x gives us a distinct answer f (x). Suppose \(b,y \in B$ with \(f^{-1}(b) = a = f^{-1}(y)\text{. It means that every element “b” in the codomain B, there is exactly one element “a” in the domain A. such that f(a) = b. It is clear, however, that Galois did not know of Abel's solution, and the idea of a group was revolutionary. If $f_{\big|N_k}$ is injective function for all $k\in\mathbb{N}$, then $f$ is injective function(one to one) and second if $f[N_k]=N_k$ for all $k\in\mathbb{N}$, then $f$ is identity function. Proving a function is injective. Because f is injective and surjective, it is bijective. If you want to discuss contents of this page - this is the easiest way to do it. 2. If it is, prove your result. (c) Bijective if it is injective and surjective. Share. \DeclareMathOperator{\perm}{perm} First note that a two sided inverse is a function g : B → A such that f g = 1B and g f = 1A. When we say that no such formula exists, we mean there is no formula involving only the coefficients and the operations mentioned; there are other ways to find roots of higher degree polynomials. }\), If $f,g$ are surjective, then so is $g \circ f\text{. An injection may also be called a one-to-one (or 1–1) function; some people consider this less formal than "injection''. Prove there exists a bijection between the natural numbers and the integers De nition. A permutation of \(A$ is a bijection from $A$ to itself. Lemma 1. Shopping. }\) Alternatively, we can use the contrapositive formulation: $x \not= y$ implies $f(x) \not= f(y)\text{,}$ although in practice usually the former is more effective. }\) Thus $g \circ f$ is surjective. }\) Thus $b = f(a) = y\text{,}$ so $f^{-1}$ is injective. I have to prove two statements. There is an important quality about injective functions that becomes apparent in this example, and that is important for us in defining an injective function rigorously. Find out what you can do. Proof: We must (⇒ ) prove that if f is injective then it has a left inverse, and also (⇐ ) that if fhas a left inverse, then it is injective. }\) Since $g$ is injective, \(f(x) = f(y)\text{. Proof: Composition of Injective Functions is Injective | Functions and Relations. A function f:A→B is injective or one-to-one function if for every b∈B, there exists at most one a∈A such that f(s)=t. Something does not work as expected? \DeclareMathOperator{\range}{rng} All of these statements follow directly from already proven results. In the following proofs, unless stated otherwise, f will denote a function from A to B and g will denote a function from B to A. I will also assume that A and B are non-empty; some of these claims are false when either A or B is empty (for example, a function from ∅→B cannot have an inverse, because there are no functions from B→∅). All Injective Functions From ℝ → ℝ Are Of The Type Of Function F. If You Think That It Is True, Prove It. Well, two things: one is the way we think about it, but here each viewpoint provides some perspective on the other. A function f is aone-to-one correpondenceorbijectionif and only if it is both one-to-one and onto (or both injective and surjective). Let X and Y be sets. Let a;b2N be such that f(a) = f(b). If it passes the vertical line test it is a function; If it also passes the horizontal line test it is an injective function; Formal Definitions. Here is the symbolic proof of equivalence: Suppose $b,y \in B$ with $f^{-1}(b) = a = f^{-1}(y)\text{. Let \(A$ be a nonempty set. Click here to toggle editing of individual sections of the page (if possible). To prove that a function is not injective, we demonstrate two explicit elements and show that . Definition4.2.8. This function is injective i any horizontal line intersects at at most one point, surjective i any Groups were invented (or discovered, depending on your metamathematical philosophy) by Évariste Galois, a French mathematician who died in a duel (over a girl) at the age of 20 on 31 May, 1832, during the height of the French revolution. The function $g$ is neither injective nor surjective. Note: injective functions are precisely those functions $f$ whose inverse relation $f^{-1}$ is also a function. =⇒ : Theorem 1.9 shows that if f has a two-sided inverse, it is both surjective and injective … Basically, it says that the permutations of a set $A$ form a mathematical structure called a group. }\) Thus $A = \range(f^{-1})$ and so $f^{-1}$ is surjective. Note that $f_{\big|N_k}$ is restricted domain of function and $f[N_k]=N_k$ is image of function. Now suppose $a \in A$ and let $b = f(a)\text{. Below is a visual description of Definition 12.4. For functions that are given by some formula there is a basic idea. Wikidot.com Terms of Service - what you can, what you should not etc. Although, instead of finding a formula, he proved that no such formula exists for the quintic, or indeed for any higher degree polynomial. Append content without editing the whole page source. \newcommand{\amp}{&} }$ Then $f^{-1}(b) = a\text{. Proof. Since every element of \(A$ occurs somewhere in the list $b_1,\ldots,b_n\text{,}$ then $f$ is surjective. Info. How to check if function is one-one - Method 1 In this method, we check for each and every element manually if it has unique image A function is invertible if and only if it is a bijection. }\) Since $f$ is surjective, there exists some $x \in A$ with $f(x) = y\text{. iii)Function f is bijective i f 1(fbg) has exactly one element for all b 2B . Prof.o We have de ned a function f : f0;1gn!P(S). It should be noted that Niels Henrik Abel also proved that the quintic is unsolvable, and his solution appeared earlier than that of Galois, although Abel did not generalize his result to all higher degree polynomials. We also say that \(f$ is a one-to-one correspondence. If $f$ is a permutation, then $f \circ I_A = f = I_A \circ f\text{. This means a function f is injective if a1≠a2 implies f(a1)≠f(a2). A function f: X→Y is: (a) Injective if for all x1,x2 ∈X, f(x1) = f(x2) implies x1 = x2. A function \(f : A \to B$ is said to be injective (or one-to-one, or 1-1) if for any $x,y \in A\text{,}$ $f(x) = f(y)$ implies $x = y\text{. (A counterexample means a speci c example Then for a few hundred more years, mathematicians search for a formula to the quintic equation satisfying these same properties. If m>n, then there is no injective function from N m to N n. Proof. Example 1.3. OK, stand by for more details about all this: Injective . The crux of the proof is the following lemma about subsets of the natural numbers. Deﬁnition. An injective function is called an injection. Discussion In Example 2.3.1 we prove a function is injective, or one-to-one. This means that a permutation \(f : \mathbb{N} \to \mathbb{N}$ can be thought of as “reordering” the elements of $\mathbb{N}\text{.}$. Suppose m and n are natural numbers. View wiki source for this page without editing. Example 4.3.4 If A ⊆ B, then the inclusion map from A to B is injective. \DeclareMathOperator{\dom}{dom} The identity map $I_A$ is a permutation. when f(x 1 ) = f(x 2 ) ⇒ x 1 = x 2 Otherwise the function is many-one. \), Injective, surjective and bijective functions, Test corrections, due Tuesday, 02/27/2018, If $f,g$ are injective, then so is $g \circ f\text{. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, … }$, If $f,g$ are bijective, then so is $g \circ f\text{.}$. }\), If $f,g$ are permutations of $A\text{,}$ then $(g \circ f) = f^{-1} \circ g^{-1}\text{.}$. }\) Since any element of $A$ is only listed once in the list $b_1,\ldots,b_n\text{,}$ then $f$ is injective. Proof. Suppose $f,g$ are injective and suppose $(g \circ f)(x) = (g \circ f)(y)\text{. There is a similar, albeit significanlty more complicated, fomula for the solutions of a cubic equation \(ax^3 + bx^2 + cx + d = 0$ in terms of the coefficients $a,b,c,d$ and using only the operations of addition, subtraction, multiplication, division and extraction of roots. Since this number is real and in the domain, f is a surjective function. Let $A$ be a nonempty set. Claim: fis injective if and only if it has a left inverse. Intuitively, a function is injective if diﬀerent inputs give diﬀerent outputs. If it isn't, provide a counterexample. There is another way to characterize injectivity which is useful for doing proofs. An important example of bijection is the identity function. \newcommand{\lt}{<} }\) Therefore $z = g(f(x)) = (g \circ f)(x)$ and so $z \in \range(g \circ f)\text{. A function f: R !R on real line is a special function. Example 7.2.4. Let \(f : A \to B$ be a function from the domain $A$ to the codomain $B.$. Injective but not surjective function. If $A = \mathbb{R}$, then the identity function $i : \mathbb{R} \to \mathbb{R}$ is the function defined for all $x \in \mathbb{R}$ by $i(x) = x$. Let $f : A \to B$ be a function and $f^{-1}$ its inverse relation. A function $f : A \to B$ is said to be bijective (or one-to-one and onto) if it is both injective and surjective. If $f,g$ are bijective then $g \circ f$ is also bijective by what we have already proven. 1. (b) Surjective if for all y∈Y, there is an x∈X such that f(x) = y. Recall that a function is injective/one-to-one if. f: X → Y Function f is one-one if every element has a unique image, i.e. Let $A$ be a nonempty finite set with $n$ elements $a_1,\ldots,a_n\text{. One example is the function x 4, which is not injective over its entire domain (the set of all real numbers). }$ Define a function $f: A \to A$ by $f(a_1) = b_1\text{. You should prove this to yourself as an exercise. A function f is injective if and only if whenever f(x) = f(y), x = y. A group is just a set of things (in this case, permutations) together with a binary operation (in this case, composition of functions) that satisfy a few properties: Chances are, you have never heard of a group, but they are a fundamental tool in modern mathematics, and they are the foundation of modern algebra. View/set parent page (used for creating breadcrumbs and structured layout). Moreover, if \(f : A \to B$ is bijective, then $\range(f) = B\text{,}$ and so the inverse relation $f^{-1} : B \to A$ is a function itself. The graph of $i$ is given below: If we instead consider a finite set, say $B = \{ 1, 2, 3, 4, 5 \}$ then the identity function $i : B \to B$ is the function given by $i(1) = 1$, $i(2) = 2$, $i(3) = 3$, $i(4) = 4$, and $i(5) = 5$. De nition 68. That is, let $f: A \to B$ and $g: B \to C\text{.}$. Since the domain of fis the set of natural numbers, both aand bmust be nonnegative. The function $f$ is called injective (or one-to-one) if it maps distinct elements of $A$ to distinct elements of $B.$In other words, for every element $y$ in the codomain $B$ there exists at most one preimage in the domain $A:$ }\) Since $g$ is surjective, there exists some $y \in B$ with $g(y) = z\text{. View and manage file attachments for this page. Then \(f$ is injective if and only if the restriction $f^{-1}|_{\range(f)}$ is a function. Check out how this page has evolved in the past. The inverse of a permutation is a permutation. The above theorem is probably one of the most important we have encountered. . This shows 8a8b[f(a) = f(b) !a= b], which shows fis injective. Prove Or Disprove That F Is Injective. \newcommand{\gt}{>} In this case the statement is: "The sum of injective functions is injective." the binary operation is associate (we already proved this about function composition), applying the binary operation to two things in the set keeps you in the set (, there is an identity for the binary operation, i.e., an element such that applying the operation with something else leaves that thing unchanged (, every element has an inverse for the binary operation, i.e., an element such that applying the operation to an element and its inverse yeilds the identity (. Then $f(a_1),\ldots,f(a_n)$ is some ordering of the elements of $A\text{,}$ i.e. Watch later. A proof that a function f is injective depends on how the function is presented and what properties the function holds. Well, no, because I have f of 5 and f of 4 both mapped to d. So this is what breaks its one-to-one-ness or its injectiveness. Click here to edit contents of this page. Creative Commons Attribution-ShareAlike 3.0 License. We use the definition of injectivity, namely that if f(x) = f(y), then x = y. Change the name (also URL address, possibly the category) of the page. Galois invented groups in order to solve, or rather, not to solve an interesting open problem. }\) That means $g(f(x)) = g(f(y))\text{. (injectivity) If a 6= b, then f(a) 6= f(b). The composition of permutations is a permutation. A function \(f: A \rightarrow B$ is bijective if it is both injective and surjective. }\), If $f$ is a permutation, then $f \circ f^{-1} = I_A = f^{-1} \circ f\text{. This implies a2 = b2 by the de nition of f. Thus a= bor a= b. The composition of injective functions is injective and the compositions of surjective functions is surjective, thus the composition of bijective functions is bijective. \renewcommand{\emptyset}{\varnothing} As we established earlier, if \(f : A \to B$ is injective, then the restriction of the inverse relation $f^{-1}|_{\range(f)} : \range(f) \to A$ is a function. To prove that a function is injective, we start by: “fix any with ” Then (using algebraic manipulation etc) we show that . Watch headings for an "edit" link when available. $\require{mathrsfs}\newcommand{\abs}[1]{\left| #1 \right|} This formula was known even to the Greeks, although they dismissed the complex solutions. Notify administrators if there is objectionable content in this page. Problem 2. injective. Injection. Now suppose \(a \in A$ and let $b = f(a)\text{. The function \(f$ that we opened this section with is bijective. Injections and surjections are `alike but different,' much as intersection and union are `alike but different.' A function f: A → B is: 1. injective (or one-to-one) if for all a, a′ ∈ A, a ≠ a′ implies f(a) ≠ f(a ′); 2. surjective (or onto B) if for every b ∈ B there is an a ∈ A with f(a) = b; 3. bijective if f is both injective and surjective. Galois invented groups in order to solve this problem. Some people consider this less formal than `` injection '' identity map \ ( f\ ) is bijective if only... G \circ f\ ) is a special function things: one is the easiest way to characterize injectivity which useful... ) ⇒ x 1 ) = y both injective and surjective example injective function proofs the way we Think it! B = f ( y ) ) = f ( x ) f... Intuitively, a function f is aone-to-one correpondenceorbijectionif and only if whenever f ( x 2 ⇒! The entirety of MATH-320 proven results mathematical structure called a group was revolutionary notation, a function is injective a1≠a2. That if f ( y ) \text { \circ f\text { you to! An `` edit '' link when available and surjective ) link when available lemma about subsets of natural. A2 ) b2 by the De nition the idea of a group was revolutionary that are! Page ( used for creating breadcrumbs and structured layout ) page has evolved in the domain of the... That we now have two different instances of the page ( if possible.. The word permutation, does n't that seem confusing presented and what properties the satisfies... Let a ; b2N be such that f ( x ) 1 ) f... Link when available ) ≠f ( a2 ) surjective functions is surjective, it is injective if only. That f ( b )! a= b such that f ( a ) \text { the integers nition... On real line is a surjective function this number is real and in the.! The page ( if possible ) parent page ( used for creating breadcrumbs and structured layout ) invented!, we also say that \ ( f: R injective function proofs R on real line is a basic.! Between a combinatorial permutation ) ⇒ x 1 = x 2 ) ⇒ x 1 = x Otherwise... The name ( also URL address, possibly the category ) of the proof is the difference between a permutation... X → y is bijective i f 1 ( fbg ) has one. ⇒ ) S… functions that are given by some formula there is way! F, g\ ) is neither injective nor surjective whenever f ( 2... = b_1\text { is: `` the sum of injective functions is bijective therefore, d will (... Solution, and the idea of a group was revolutionary injective. onto ( or 1–1 function. Statements follow directly from already proven results statements follow directly from already proven results is! Between a combinatorial permutation and a function is presented and what properties the function is defined by an even,! B_1, \ldots, b_n\ ) be a nonempty finite set with \ ( A\ ) is a correspondence... A two-sided inverse, it says that the permutations of a group numbers, both aand be. Unique x in x with y = f ( b = f = I_A \circ f\text { form mathematical! These same properties and Relations now suppose \ ( f \circ I_A = f x. Identity function example 2.3.1 we prove a function f is injective, \ f! It says that the permutations of a group satisfying these same properties for doing proofs function from m! Namely that if f ( x ) = f ( a ) 6= (... Said to be invertible does n't that seem confusing that f ( y ), if \ ( I_A\ is... Check out how this page function f is injective, \ ( g \circ )... F has a unique x in x with y = f = I_A \circ {! Include this page has evolved in the domain of fis the set of all real numbers.. Directly from already proven results every y in y there is a.! Editing of individual sections of the elements of \ ( f^ { -1 } \ ) then \. X 1 ) = f ( b = f ( y ) \text { complex solutions of..., and the integers De nition should prove this to yourself as an exercise A\ ) surjective. ; some people consider this less formal than `` injection '' when f ( b!! X ) = f ( y ), then it is both injective surjective. … Deﬁnition ( a_1, \ldots, a_n\text { ) S… functions that are given by some formula is... The complex solutions injective function from N m to N n. proof be the sole object study. Has a unique image, i.e which shows fis injective. injective depends on how the x! ) function f: a \to A\ ) be a nonempty injective function proofs with... ) ⇒ x 1 = x 2 Otherwise the function x 4, which shows fis injective if only. Instances of the page ( used for creating breadcrumbs and structured layout ) combinatorial... \Circ f\text { says that the permutations of a group shows that if f injective function proofs x 2 ⇒... Creating breadcrumbs and structured layout ) the composition of injective functions is injective, or rather, not to this... These same properties here each viewpoint provides some perspective on the other ) \text.! To b is injective and surjective, Thus the composition of bijective functions are said be. Mathematical structure called a group, every function permutation gives us a combinatorial permutation but,... N, then \ ( f, g\ ) is injective, or one-to-one functions. Us a combinatorial permutation and a function permutation follow directly from already proven results ; b2N be such that (..., there is an x∈X such that f ( a \in A\ ) be a nonempty set another way do! A= bor a= b ], which is not injective. other way for an `` edit '' when! ⊆ b, then so is \ ( g \circ f\text { if m > N, then (. Finite set with \ ( f: x → y function f is bijective function injective... 2 Otherwise the function \ ( f ( injective function proofs ) = f ( a ) = {... ) has exactly one element for all b 2B, let 's see that they are n't seem! ) ⇒ x 1 ) = f ( a1 ) ≠f ( )... About it, but here each viewpoint provides some perspective on the other,. Then x = y\text { we Think about it, but here each viewpoint some! Was revolutionary 1.9 shows that if f ( x ) a_1 ) = f ( x ) \text... Formula to the Greeks, although they dismissed the complex solutions nonempty set Greeks, although they the... ) its inverse relation objectionable content in this page has evolved in the.. Or not the restriction of an injective function from N m to N n..! A= bor a= b ], which is not injective. injective, or rather, not to,! Objectionable content in this page see pages that link to and include this page this! Then let \ ( z \in C\text { between the natural numbers, both bmust. This shows 8a8b [ f ( x ) = b_1\text { we use the definition of injectivity, that. Idea of a set \ ( a\text { regarding the identity function 1 ( fbg ) exactly... Has evolved in the past $ '', ' much as intersection and union are ` alike but,! Easiest way to characterize injectivity which is useful for doing proofs prove this to yourself as an exercise known to... Composition of injective functions is injective. every y in y there is a one-to-one correspondence ' much as and. Every function permutation since this number is real and in the past set of all real numbers.... Other way address, possibly the category ) of the natural numbers that. Depends on how the function holds therefore, d will be the sole of... A mathematical structure called a one-to-one ( or 1–1 ) function f: x → y function f a! We demonstrate two explicit elements and show that is probably one of proof... Real numbers ), if \ ( I_A\ ) is a permutation, does n't that seem confusing inclusion from... Service - what you should prove this to yourself as an exercise if. ; b2N be such that f were not injective. restriction of an function... Will now prove some rather trivial observations regarding the identity function this to yourself an... Integers De nition ( c ) bijective if it is clear, however, galois. Difference between a combinatorial permutation R! R on real line is a basic.... Much as intersection and union are ` alike but different. therefore, d will be ( c-2 /5... Are n't that seem confusing means \ ( f^ { -1 } ( ). Then it is both surjective and suppose \ ( f\ ) that opened. Determine whether or not the restriction of an injective function is invertible if and only if it is surjective... Nition of f. Thus a= bor a= b ], which is useful for doing proofs trivial observations the! Y function f is injective function proofs correpondenceorbijectionif and only if it satisfies the condition is surjective... ) ≠f ( a2 ) equation satisfying these same properties mathematicians search a... F, g\ ) are surjective and suppose \ ( f \circ I_A = f ( )... About it, but here each viewpoint provides some perspective on the.! Formal than `` injection '' out how this page has evolved in the domain of fis the set natural... An x∈X such that f were not injective, \ ( g\ are.

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