A function is invertible if and only if it is bijective Proof Let f : A → B be a function, and assume first that f is invertible Then it has a unique inverse function f-1 : B
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[PDF] A function is bijective if and only if has an inverse
30 nov 2015 · Since f is surjective, there exists a ∈ A such that f(a) = b Let f−1(b) = a Since f is injective, this a is unique, so f−1 is well-defined Now we much check that f−1 is the inverse of f
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The function f : A → B has an inverse if and only if it is a bijection Proof There are two things to prove here Firstly we must show that if f has an inverse then it is
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We say that two functions f and g are equal if they have the same domain and codomain A function f : A → B is called bijective if it is both injective and surjective Proof Suppose that the function f : A → B is invertible and let f−1 be its inverse First Prove that if the composition g ◦ f is surjective, then g is surjective 2
[PDF] Lecture 6: Functions : Injectivity, Surjectivity, and Bijectivity
iii) Function f has a inverse iff f is bijective Proof Let A and B be non-empty sets and f : A → B a function i) ⇒ Suppose f has a right inverse g, then f ◦g = 1B
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If f has an inverse function, then f is both injective and surjective Proof: Suppose f has an inverse function g : B → A Then for all s, t in A we have
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22 jan 2015 · Prove that if f has a left inverse, then f is injective Another weaker notion of an inverse function is called a right inverse function Suppose that f : A
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14 mar 2013 · if f and g are bijective, then the sum f " g is bijective (in particular, if f is bijective ), then f C * D iff C * fL D Proof By part (a), f C # D iff C # fL Hint: you can use inverse trig functions, but be careful of where they exist
[PDF] Inverse Functions
A function is invertible if and only if it is bijective Proof Let f : A → B be a function, and assume first that f is invertible Then it has a unique inverse function f-1 : B
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Math 300 Introduction to Mathematical Reasoning Autumn 2017
Inverse Functions
Please read this pdf in place of Section 6.5 in the text. The text uses the term \inverse of a function"
and the notationf1in the most general possible way, and this can be confusing. We will use this pdf in place of Section 6.5 to avoid some causes of confusion in working with the notationf1. To refer to results in this pdf, label them as \InF Theorem 1," \InF Lemma 2," etc.Solving Equations, Revisited
When we discussed surjectivity of functions, we noted that determining whether a functionfis surjective often amounts to solving the equationf(x) =yfor an arbitraryyin the codomain; and determining whether it's injective amounts to determining if the solution is unique when it exists. In such cases, we were often able to derive an explicit formula for the value ofxin terms ofy. For example, Exercise 4(a) on page 318 of the textbook asks you to consider the function F:R!Rdened byF(x) = 5x+ 3. As the solution in the back of the book explains,Fis bijective, and for eachy2R, the numberx= (y3)=5 is theuniquesolution toF(x) =y. This formula actually denes a new functionG:R!R:G(y) =y35
This function has the feature that for eachy2R,G(y) is the unique numberxsuch thatF(x) =y.The following denition generalizes this.Denition.SupposeAandBare nonempty sets andf:A!Bis a function. A function
g:B!Ais called aninverse function forfif it satises the following condition:For everya2Aandb2B,f(a) =bif and only ifg(b) =a.Thus, in the example above,Gis an inverse function forF.
Theorems About Inverse Functions
Theorem 1.LetAandBbe nonempty sets, and letf:A!Bandg:B!Abe functions. Then gis an inverse function forfif and only if for everya2A,g(f(a)) =a, and(1) for everyb2B,f(g(b)) =b.(2) Proof.Assume rst thatgis an inverse function forf. We need to show that both (1) and (2) are satised. Leta2Abe arbitrary, and letb=f(a). Then by denition of an inverse function, f(a) =bimpliesg(b) =a, so we can compute g(f(a)) =g(b) =a: This proves (1). To prove (2), letb2Bbe arbitrary, and leta=g(b). Once again, the fact thatg is an inverse function forftells us thatg(b) =aimpliesf(a) =b, and therefore, f(g(b)) =f(a) =b; and (2) is proved. 1 Conversely, assume thatfandgsatisfy (1) and (2). Leta2Aandb2Bbe arbitrary; we need to show thatf(a) =bif and only ifg(b) =a. On the one hand, iff(a) =b, then g(b) =g(f(a)) (substitutingf(a) =b); =a(using equation (1)) as desired. Conversely, ifg(b) =a, then a similar argument using equation (2) shows thatf(a) =f(g(b)) =b, and we are done.There is a nice way to rephrase this result in terms of compositions. Before we do so, let us
note some important facts about compositions. IfAis any nonempty set, one of the simplest functions we can write down is theidentity function ofA, denoted byIA:A!A, and dened byIA(x) =xfor everyx2A. Lemma 2.SupposeA;B;C;Dare any nonempty sets, andf:A!B,g:B!C, andh:C!D are any functions. Then the following identities hold: fIA=IBf=f; (hg)f=h(gf):Proof.See Exercises 4 and 5 on page 331 of the textbook.Here is a concise reformulation of Theorem 1 in terms of compositions.
Corollary 3.LetAandBbe nonempty sets, and letf:A!Bandg:B!Abe functions. Then gis an inverse function forfif and only if fg=IBandgf=IA:(3) Proof.You can check using the denitions of composition and identity functions that (3) is true ifand only if both (1) and (2) are true, and then the result follows from Theorem 1.Another important consequence of Theorem 1 is that if an inverse function forfexists, it is
unique. Here is the proof. Theorem 4.LetAandBbe nonempty sets, and letf:A!Bbe a function. Ifg1:B!Aand g2:B!Aare inverse functions forf, theng1=g2.
Proof.Letf:A!B, and assumeg1;g2:B!Aare both inverse functions forf. By Corollary 3, they satisfy fg1=IB; g1f=IA; fg2=IB; g2f=IA: Using these formulas together with the results of Lemma 2, we compute as follows: g1=g1IB
=g1(fg2) = (g1f)g2 =IAg2 =g2; which is what we wanted to prove.We now make the following denitions: 2 Denition.LetAandBbe nonempty sets. A functionf:A!Bis said to beinvertibleif it has an inverse function.Notation:Iff:A!Bis invertible, we denote the (unique) inverse function byf1:B!A.Using this notation, we can rephrase some of our previous results as follows.
Corollary 5.Supposef:A!Bis an invertible function. Then f1(f(a)) =afor everya2A; (4)
f(f1(b)) =bfor everyb2B; (5) ff1=IBandf1f=IA:(6)Proof.These are just the results of Theorem 1 and Corollary 3 withgreplaced byf1.It is important to be careful with the notationf1: The superscript in this case doesnot
represent a multiplicative inverse; instead it represents a dierent function, one that satises (4), (5), and (6). Also,the notationf1should never be used to represent an inverse function unless you have veried thatfis invertible.Important note:We will see inx6.6 that there is a dierent way thatf1is used even whenfis not invertible. In that section we wil applyf1to a set, and it will not indicate thatfis invertible or that there is an inverse function. Here is a simple criterion for deciding which functions are invertible. Theorem 6.A function is invertible if and only if it is bijective. Proof.Letf:A!Bbe a function, and assume rst thatfis invertible. Then it has a unique inverse functionf1:B!A. To show thatfis surjective, letb2Bbe arbitrary, and let a=f1(b). Then (5) impliesf(a) =f(f1(b)) =b. To show thatfis injective, leta1;a22Aand supposef(a1) =f(a2). Applying the function f1to both sides yieldsf1(f(a1)) =f1(f(a2)), and then (4) shows thata1=a2.
Conversely, assumefis bijective. We dene a functiong:B!Aas follows: Givenb2B, becausefis surjective there is an elementa2Asuch thatf(a) =b, and becausefis injective,a is the unique such element. Thus we can unambiguously deneg(b) to be this particular value of a. By our denition ofg, we see thatg(b) =aif and only iff(a) =b. Thus by the denition of aninverse function,gis an inverse function off, sofis invertible.These theorems yield a streamlined method that can often be used for proving that a function
is bijective and thus invertible. Given a functionf:A!B, if we can (by any convenient means) come up with a functiong:B!Aand prove that it satises bothfg=IBandgf=IA, then Corollary 3 impliesgis an inverse function forf, and thus Theorem 6 implies thatfis bijective. Moreover, since the inverse is unique, we can conclude thatg=f1. Consider for example the functionF:R!Rgiven byF(x) = 5x+3, which we studied above. Preliminary computations suggest thatG:R!Rgiven byG(y) = (y3)=5 ought to be an inverse function for it. We can verify this by showing thatFG=IRandGF=IR, which is just a couple of computations:FG(y) =Fy35
= 5y35 + 3 = (y3) + 3 =y;GF(x) =G(5x+ 3) =(5x+ 3)35
=5x5 =x: OnceGis dened, this computation is all that is needed to prove thatFis bijective and invertible, and thatG=F1. (Of course, in general, you have to take care when deningGto ensure that 3 it gives a well-dened element ofAfor every element ofB, or in other words that it is indeed a function.) Next we have a formula for the inverse of an inverse. Theorem 7.LetAandBbe nonempty sets, and supposef:A!Bis invertible. Thenf1:B!Ais also invertible, and(f1)1=f.
Proof.By Corollary 3,f1is invertible if there is a functiong:A!Bthat satisesgf1=IB andf1g=IA; and in that case the functiongis the unique inverse off1. Sinceg=fis sucha function, it follows thatf1is invertible andfis its inverse.Now we consider inverses of composite functions. Suppose we have two functionsf:A!B
andg:B!C: A f!Bg!C: Theorem 6.20 in the textbook showed that if both functionsfandgare bijective, then so is the composite functiongf:A!C. Here is a nice formula for its inverse. Theorem 8.LetA;B;Cbe nonempty sets, and letf:A!Bandg:B!Cbe functions. Iff andgare invertible, thengf:A!Cis invertible and(gf)1=f1g1. Proof.Assumefandgare invertible. Then they are both bijective by Theorem 6 above, sogf is bijective by Theorem 6.20 and thus invertible. To show that the inverse is equal tof1g1, by Corollary 3 we just need to show that the following two compositions are identity maps: (gf)(f1g1) =IC; (f1g1)(gf) =IA: To prove the rst identity, we compute as follows using the results of Lemma 2: (gf)(f1g1) =gf(f1g1) =g(ff1)g1)) =g(IBg1) =gg1 =IC:The proof of the other identity is similar:
(f1g1)(gf) =f1g1(gf) =f1(g1g)f =f1(IBf) =f1f =IA; which completes the proof.4Functions as Sets of Ordered Pairs
We end this section with a brief discussion of a way of looking at functions as certain kinds of sets.
Reading this discussion is optional
Recall that we dened a function as a pair of sets (the domainAand the codomainB), together with a \rule" that associates with each element ofAone and only one element ofB. You might have noticed that we were a little vague about exactly what a \rule" is supposed to be. Does it have to be a formula? Or can it be a set of formulas for dierent parts of the domain? Or can it be some other type of algorithm? Does such an algorithm need to produce an exact value after nitely many steps, or is something like an innite series acceptable? Can a function assign \random" values (whatever that might mean)? These are questions that vexed mathematicians for centuries as the foundations of calculus were being worked out. Shortly after calculus was invented, most mathematicians believed a function had to be dened by an explicit formula. Gradually, the requirements were relaxed to allow more things to be considered functions. Finally, in the mid-twentieth century, mathematicians settled on a denition that is general enough to accommodate anything anybody wanted to call a \function."It is, ultimately, a denition of a function as a kind ofset, like just about everything else in modern
mathematics. Like the denition of ordered pairs as sets (see Exercise 10 on p. 263 of the textbook), it is not very practical to work with, but it is important to know that there is such a denition, because it ensures that the theory of functions rests on the same solid logical foundation as set theory. The key idea is that once the domainAand codomainBhave been chosen, whatever techniques we might use to specify a functionf:A!B, the function is completely determined by the pairing of each element ofAwith a specic element ofB. In other words, if we know all of the ordered pairs (a;b)2AB, wherearanges over all elements ofAandb=f(a) for each sucha, then we know the function. Thus a function determines a certain subset ofAB, which we denote by f: f=f(a;b)2AB:a2Aandb=f(a)g: For example, iff:R!Ris the functionf(x) =x2, then fis the subsetf(x;y)2R2:y=x2g; it is the familiar parabola that you drew when you were rst studying quadratic equations. By analogy with examples like this, for any functionf:A!B, one often calls the set fthe graph off. As we have seen, every function determines a graph, which is a subset of the Cartesian product of its domain with its codomain. The graph of a functionf:A!Bis not just anarbitrarysubset ofAB, though. From the denition of a function that we've been working with, we see that it satises a couple of conditions: First, for everya2A, there must be an elementbsuch that (a;b)2f(namely,b=f(a)). And second, that element must beuniquely determined bya; to put this a little more precisely, if b1;b2are elements ofBsuch that (a;b1) and (a;b2) are both in f(meaning thatf(a) =b1and
f(a) =b2), thenb1=b2. Thus the subset fABalways satises these conditions: (i) For everya2A, there existsb2Bsuch that (a;b)2f; (ii) For alla2Aandb1;b22B, if (a;b1) and (a;b2) are in f, thenb1=b2. Condition (i) is sometimes summarized by saying thatfiseverywhere dened(meaning that it produces an output for every element of the domain), and the second by saying thatfisuniquely dened(meaning that it produces only one output for each input). We say thatfiswell-dened if it satises both of these conditions. Saying thatfis well-dened means nothing more nor less than \fis a function." Be careful to note the dierence between conditions (i) and (ii) above and the conditions for the functionfto be surjective or injective: (iii)fissurjectiveif for everyb2B, there existsa2Asuch that (a;b)2f; 5 (iv)fisinjectiveif for alla1;a22Aandb2B, if (a1;b) and (a2;b) are in f, thena1=a2. Every function satises (i) and (ii), but only some functions satisfy (iii) and/or (iv). Now let's try to go back the other way|suppose we are given two setsAandB, and some subset AB. Can be the graph of a function fromAtoB? Clearly it has to satisfy conditions (i) and (ii) above (with freplaced by ). But the remarkable thing is that nothing more is needed: If isanysubset ofABthat satises these two conditions, then we get a well-dened function f:A!Bby declaring that for everya2A,f(a) is the unique elementb2Bsuch that (a;b)2. Based on this observation, mathematicians in the mid-twentieth century reached consensus that a function should be formallydenedas a subset of the Cartesian product satisfying these twoproperties. Thus here is the ocial denition:Set-theoretic denition of a function:LetAandBbe nonempty sets. Afunction
fromAtoBis a subset ABthat satises the following properties: (i) For everya2A, there existsb2Bsuch that (a;b)2;(ii) For alla2Aandb1;b22B, if (a;b1) and (a;b2) are in , thenb1=b2.Because this denition is simply a description of a certain kind of set, it does not depend on
any particular method for specifying howf(a) is determined froma, thus resolving the ambiguities about what kinds of rules are legitimate ways to dene functions. Since a function by this denitionisthe subset AB, it would make sense to give the subset the same name as the function itself, and many authors do so, speaking of \a functionfAB." But in practice, mathematicians generally work with functions using the usual functional notation (b=f(a)), and think about functions in terms of rules for producing an output value from an input value, rather than as sets of ordered pairs. When it is useful to work directly with the set of ordered pairs, it is usually referred to asthe graph of the functionas we did above, not as the function itself. The ne print:If a functionf:A!Bis invertible, then its graph has a very simple description in terms of the graph off. You can check easily that in that case the graph of the inverse function f