# if gof is surjective, then f is surjective

1) Démontrer que si f et g sont injectives alors gof est injective 2) Démontrer que si gof est surjective e In other words, every element of the function's codomain is the image of at most one element of its domain. Bonjour, je suis bloquée sur un exercice sur les fonctions injectives et surjectives. https://goo.gl/JQ8NysProof that if g o f is Injective(one-to-one) then f is Injective(one-to-one). Prove if gof is surjective then g is surjective. Show that f is surjective if and only if there exists g: B→A such that fog=iB, where i is the identity function. Problem 3.3.8. Can somebody help me? —Preceding unsigned comment added by 65.110.237.146 21:01, 22 September 2010 (UTC) No, the article is correct. and in this case if g o f is surjective g does have to be surjective. In mathematics, an injective function (also known as injection, or one-to-one function) is a function that maps distinct elements of its domain to distinct elements of its codomain. g(f(b)) certainly as f is injective and a ? uh i think u mean: f:F->H, g:H->G (we apply f first) and in this case if g o f is surjective g does have to be surjective. Maintenant supposons gof surjective. B - Show That If G And F Are Surjective Then Gof Is Surjective. Hence f is surjective. See the answer. "If g is not surjective, then gof is not surjective" Let g be not surjective. We say f is surjective or onto when the following property holds: For all y ∈ Y there is some x ∈ X such that f(x) = y. The x is only unique if f is bijective. Problem 27: Let f : B !C and g : C !D be functions. f(b) so we've f(a), f(b)?Y and f(a) ? (a) Suppose that f : X → Y and g: Y→ Z and suppose that g∘f is surjective. pleaseee help me solve this questionnn!?!? Now g(b) ∈ C. We claim that g(b) is not in the range of g f and hence g f is not surjective. December 10, 2020 by Prasanna. Thus g is surjective. At least, that's what one of the diagrams on the page illustrates. (iii) “The Set Of All Positive Rational Numbers Is Uncountable." On the other hand, $$g(x) = x^3$$ is both injective and surjective, so it is also bijective. (a) Proposition: If gof is surjective, then g is surjective. 2 Answers. Problem 27: Let f : B !C and g : C !D be functions. This is the part 03 out of four lectures on this topic. Suppose that gof is surjective. Let f : X → Y be a function. Expert Answer . (d) f : Z Z !Q; f(a;b) = ˆ a=b; if b 6= 0 0 if b = 0: Let c 2Q. This is not at all necessary. Recall that if f: X → Y is a function, then for every subset S ⊆ X we denote: f (S) := {y ∈ Y | ∃ x ∈ S such that f (x) = y}. Prove that the function g is also surjective. (ii) "If F: A + B Is Surjective, Then F Is Injective." Now, you're asking if g (the first mapping) needs to be surjective. Proof: This problem has been solved! This question hasn't been answered yet Ask an expert. Question: (i) "If F: A + B Is Injective, Then F Is Surjective." Since gf is surjective, doesn't that mean you can reach every element of H from G? See the answer. Then there is c in C so that for all b, g(b)≠c. By de nition of a rational number, there exist integers a;b such that b 6= 0 and c = a=b. Expert Answer 100% (1 rating) Previous question Next question Transcribed Image Text from this Question. a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. b, then f(a) ? If f is not surjective, then there is a b in B such that for all a in A, f(a) ≠ b. See the answer. This problem has been solved! The receptionist later notices that a room is actually supposed to cost..? Let Q be the relation on P (X) such that αQβ if and only if α ⊆ β. Please Subscribe here, thank you!!! Let f: R to S be a surjective ring homomorphism and I be an ideal of R. Then prove that the image f(I) is an ideal of S. RIng Theory Problems and Solutions. Induced surjection and induced bijection. Show transcribed image text. Please Subscribe here, thank you!!! (Hint : Consider f(x) = x and g(x) = |x|). First of all, you mean g:B→C, otherwise g f is not defined. Now g(b) ∈ C. We claim that g(b) is not in the range of g f and hence g f is not surjective. Now, you're asking if g (the first mapping) needs to be surjective. Moreover, f is the composition of the canonical projection from f to the quotient set, and the bijection between the quotient set and the codomain of f. The composition of two surjections is again a surjection, but if g o f is surjective, then it can only be concluded that g is surjective (see figure). But x in f^(-1)(H) implies that f(x) is in H, by definition of inverse functions. Show that if f: A→B is surjective and and H is a subset of B, then f(f^(-1)(H)) = H. Homework Equations The Attempt at a Solution Let y be an element of f(f^(-1)(H)). Suppose that h is bijective and that f is surjective. Prove that the function g is also surjective. (g o f)(x) = g(f(x)), so you want f:F->G, g:G->H. Thanks. (a) Prove that if f and g are surjective, then gf is surjective. Finding an inversion for this function is easy. Any function induces a surjection by restricting its codomain to its range. Notice that whether or not f is surjective depends on its codomain. In fact you also need to assume that f is surjective to have g necessarily injective (think about it, gof tells you nothing about what g does to things that are not in the range of f). Then g(f(a)) = g(b). i believe the direct proof is easiest: assuming fof is surjective: for all b in A there exists at least one a in A st f(f(a))=b however, since f(a) is in A, there exists at least one f(a) st f(a)=b therefore f is surjective Am i correct in saying this? See the answer. Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License As a counterexample, let f: R->{0} defined by f(x)=0. Similarly, in the case of b) you assume that g is not surjective (i.e. I'll just point out that as you've written it, that composition is impossible. Let f(x) = x and g(x) = |x| where f: N → Z and g: Z → Z g(x) = ﷯ = , ≥0 ﷮− , <0﷯﷯ Checking g(x) injective(one-one) Nor is it surjective, for if $$b = -1$$ (or if b is any negative number), then there is no $$a \in \mathbb{R}$$ with $$f(a)=b$$. Now that I get it, it seems trivial. https://goo.gl/JQ8Nys The Composition of Surjective(Onto) Functions is Surjective Proof. Problem 3.3.8. Let f: A B and g: B C be functions. (ii) "If F: A + B Is Surjective, Then F Is Injective." Homework Equations 3. Proof: This problem has been solved! Math I - CPGEI - P2 Correction DM 2 Exercice 13 Soit E et F deux ensembles non vides et f : E !F. Favourite answer. Expert Answer . In each part of the exercise, give examples of sets A;B;C and functions f : A !B and g : B !C satisfying the indicated properties. Then isn't g surjective to f(x) in H? Hi, I've proved this directly but as an exercise I'm trying to do it by contrapositive. that there exists an element y of the codomain of g which is not part of the range of g) and then show that this (same) element y cannot be part of the range of $$\displaystyle g\circ f$$ either, which is to say that $$\displaystyle g\circ f$$ is not surjective. Problem. (b) Prove that if f and g are injective, then gf is injective. Homework Equations 3. le but : f croissante et surjective de [a,b] sur [f(a),f(b)] implique f continue sur [a,b]. Nor is it surjective, for if $$b = -1$$ (or if b is any negative number), then there is no $$a \in \mathbb{R}$$ with $$f(a)=b$$. It's both. If fog is injective, then g is injective. Sorry if this is a dumb question, but this has been stumping me for a week. Let f: A B and g: B C be functions. Suppose that x and y are in B and g(x) = g(y). Edit: Woops sorry, I was writing about why f doesn't need to be a surjection, not g. Further answer here. uh i think u mean: f:F->H, g:H->G (we apply f first). "g could map every point in G to a single point to F, and f could take that single point in F to every point in H.", Thanks! If, for some $x,y\in\mathbb{R}$, we have $f(x)=f(y)$, that means $x|x|=y|y|$. i believe the direct proof is easiest: assuming fof is surjective: for all b in A there exists at least one a in A st f(f(a))=b however, since f(a) is in A, there exists at least one f(a) st f(a)=b therefore f is surjective Am i correct in saying this? For example, to show that a function, f, from A to B, is surjective, you must show that, if y is any member of B, then there exist x in A so that f(x)= y. Then there exist two distinct elements of A, say x and y, such that f(x)=f(y). Question: (i) "If F: A + B Is Injective, Then F Is Surjective." To show that a function, f, from A to B, is injective, you must show that if f(x1)= y and f(x2)= y, where x1 and x2 are members of A and y is a member of B, then … On the other hand, $$g(x) = x^3$$ is both injective and surjective, so it is also bijective. I think I just couldn't separate injection from surjection. (iii) “The Set Of All Positive Rational Numbers Is Uncountable." 4. Jan 18, 2011 #7 2.En d eduire que si f est surjective alors, pour tout B 2P(F), f(f 1(B)) = B. Please help with this math problem I'm desperate!? merci pour votre aide. Therefore, f is surjective. It is possible that f … [J'ai corrigé ton titre, il était trop subjectif :) AD Then there is some element of C, call it c, which is not mapped to by g. That is, for all b in B, g(b)!=c [!= means "not equal to"]. In each part of the exercise, give examples of sets A;B;C and functions f : A !B and g : B !C satisfying the indicated properties. Hence g(f (a)) = c: b) If g f is surjective, then g is surjective, but f may not be. Question: Prove If Gof Is Surjective Then G Is Surjective. E. emakarov. 9 years ago. In fact you also need to assume that f is surjective to have g necessarily injective (think about it, gof tells you nothing about what g does to things that are not in the range of f). f(b) as g is injective g(f(a)) ? Now, proof by contrapositive: (1) "If g f is surjective, then g is surjective" is the same statement as (2) "if g is not surjective, then g f is not surjective." Soit c quelconque dans C. gof étant surjective, il existe au moins un a dans A tel que gof(a) = c. Mais alors, si on pose f(a) = b, on a trouvé b dans B tel que g(b)=c : g est surjective aussi. So we assume g is not surjective. is surjective then also f is surjective b If f g is injective then also f is from SFS IT50 at Eindhoven University of Technology Join Yahoo Answers and get 100 points today. Thus g is surjective. b we ought to instruct that g(f(a)) ? This problem has been solved! Suppose that g f is surjective. gof injective does not imply that g is injective. Can somebody help me? Let F be the set of functions from X to {0, 1, 2}. New comments cannot be posted and votes cannot be cast, Press J to jump to the feed. 3.Montrer que, pour tout A ˆE, A ˆf 1(f(A)). (d) f : Z Z !Q; f(a;b) = ˆ a=b; if b 6= 0 0 if b = 0: Let c 2Q. Dcamd re : Composition, injectivité, surjectivité 09-02-09 à 22:22. Answer: If g is not surjective, then there exists c 2 C such that g(b) 6= c for all b 2 B: But then g(f (a)) 6= c for all a 2 A: Thus we have proven the contrapositive, and we –nd that if g f is surjective then g is surjective. Therefore if we let y = f(x) 2B, then g(y) = z. Sean H. Lv 5. g(f(b)) QED. 4.Montrer que si f est injective alors, pour tout A 2P(E), f 1(f(A)) = A. Get answers by asking now. b - show that if g and f are surjective then gof is surjective. The description of remaining three parts has been given below. This problem has been solved! Therefore if we let y = f(x) 2B, then g(y) = z. Exercice : Soit E,F,G trois ensembles non vides et soit f:E va dans F et g:F va dans G deux fonctions. Clearly, f is a bijection since it is both injective as well as surjective. Since f is surjective there are x' and y' in A such that f(x') = x and f(y') = y and since gof is injective gof(x') = g(x) = g(y) = gof(y') implies x' = y'. So we assume g is not surjective. "If g is not surjective, then gof is not surjective" Let g be not surjective. You just made this clear for me. Proof: (c) Proposition: If Gof Is Surjective And G Is Injective, Then F Is Surjective. But then g(f(x))=g(f(y)) [this is simply because g is a function]. Now, proof by contrapositive: (1) "If g f is surjective, then g is surjective" is the same statement as (2) "if g is not surjective, then g f is not surjective." Proof: (c) Proposition: If Gof Is Surjective And G Is Injective, Then F Is Surjective. Now, if fg is a surjective map, that means that for all elements of H, at least one element of G is mapped to it. [f]^{}[/2]Homework Statement Show that if f: A→B is surjective and and H is a subset of B, then f(f^(-1)(H)) = H. Homework Equations The Attempt at a Solution Let y be an element of f(f^(-1)(H)). This problem has been solved! Suppose a ∈ A is such that (g f)(a) = g(b). Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). (a) Suppose that f : X → Y and g: Y→ Z and suppose that g∘f is surjective. Number of one-one onto function (bijection): If A and B are finite sets and f : A B is a bijection, then A … Therefore x =f(x') = f(y') =y and so g is injective. Let f(x) = x and g(x) = |x| where f: N → Z and g: Z → Z g(x) = ﷯ = , ≥0 ﷮− , <0﷯﷯ Checking g(x) injective(one-one) For example, g could map every point in G to a single point to F, and f could take that single point in F to every point in H. The only thing that fg being surjective implies is that f (the second mapping) is surjective. Injective, Surjective and Bijective. Let X be a set. 1.Montrer que, pour tout B ˆF, f(f 1(B)) = B \f(E). Thus, f : A B is one-one. https://goo.gl/JQ8Nys Proof that if g o f is Surjective(Onto) then g is Surjective(Onto). You should probably ask in r/learnmath or r/cheatatmathhomework. Should I delete it anyway? If you are looking for something more complicated, suppose f(x) : R -> R and pushes everything besides 0 one away from origin i.e. MHF Hall of Honor. Expert Answer 100% (1 rating) Previous question Next question Get more help from Chegg. Y=7x(6/7 -1/4) is this a solution or a linear equation. Therefore, f(a;b) = a=b = c and hence f is surjective. Homework Statement Show that if ##f:A\\rightarrow B## is subjective and ##H\\subseteq B## then ##f(f^{-1}(H))=H##, give an example to show the equality need to hold if ##f## if not surjective. Then since g f is surjective, there exists x 2A such that (g f)(x) = g(f(x)) = z. you dont have to provide any answers, ill just go back to the drawing board if not. I mean if g maps f(F) surjectively to G, since f(F) is a subset of H, of course g maps H surjectively to G. (Hint : Consider f(x) = x and g(x) = |x|). (a) g is not injective but g f is injective. Merci d'avance. Shouldn't it be "g" is surjective (but "f" need not be)? So we have gof(x)=gof(y), so that gof is not injective. Merci Lafol ! Previous question Next question Transcribed Image Text from this Question. Press question mark to learn the rest of the keyboard shortcuts. If R is a Noetherian ring and f: R to R' is a surjective homomorphism, then we prove that R' is a Noetherian ring. For example, g could map every point in G to a single point to F, and f could take that single point in F to every point in H. The only thing that fg being surjective implies is that f (the second mapping) is surjective. Still have questions? (b) Show by example that even if f is not surjective, g∘f can still be surjective. 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. But since g is injective, it must be that f(a) = … Show transcribed image text. We prove that if f is a surjective group homomorphism from an abelian group G to a group G', then the group G' is also abelian group. explain. If say f(x_1) does not belong to D_g, then gof is not well-defined at all, since gof(x_1) =g(f(x_1)) is not defined. Let z 2C. Prove if gof is surjective then g is surjective. Then f(b 5 3) = 3(b 5 3) + 5 = b 5 + 5 = b. See the answer. I mean if g maps f(F) surjectively to G, since f(F) is a subset of H, of course g maps H surjectively to G. g: {1,2} -> {1} g(x) = 1 f: {1} -> {1,2} f(x) = 1. If h is surjective, then f is surjective. So assume those two hypotheses and let's say f:A->B and g:B->C. Conversely, if f o g is surjective, then f is surjective (but g need not be). If, for some $x,y\in\mathbb{R}$, we have $f(x)=f(y)$, that means $x|x|=y|y|$. Prove the following. Assuming m > 0 and m≠1, prove or disprove this equation:? then for every c in C there exists an a in A such that g(f(a))= c, where f(a) is in B so there must exist a b in B for every c in C such that g(b)= c. (b=f(a)) therefore g must also be surjective? Oct 2009 5,577 2,017. They pay 100 each. Then since g f is surjective, there exists x 2A such that (g f)(x) = g(f(x)) = z. Needs to be a surjection by restricting its codomain to its range is... Hi, i 've proved this directly but as an exercise i 'm desperate!?!!... 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This topic board if not ( both one-to-one and Onto ) distinct elements of a Rational number there... G is injective. does not imply that g ( y ' ) = g ( f 1 ( (... Stranger on the internet those two hypotheses and let 's assuming you meant to Ask about.! 'Ve f ( b ) so we have gof ( x ) = a=b Positive Rational is! Proof that if g and f are surjective, then f is not surjective, g. Injective. writing about why f does n't need to be a surjection, g.... Are in b and g is injective. from surjection Statement assume f: +. H, g ( the first mapping ) needs to be a surjection by restricting its codomain its... ) so we 've f ( a ) Prove that if f and g are bijective and. A is such that αQβ if and only if there exists g: c... Questionnn!?!?!?!?!?!?!?!!! Are bijective, then f is surjective then g is bijective g '' is surjective.... To the feed this questionnn!?!?!?!?!?!!! =Gof ( y ), so that for all b, g ( f ( )! 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Induces a surjection by restricting its codomain to its range gof injective not! Even if f and g o f is injective and a, i. } defined by f ( y ) = x and g is.! Je suis bloquée sur un exercice sur les fonctions injectives et surjectives understand Answer... Imply that g is surjective then g ( x ) = x and,. Bloquée sur un exercice sur les fonctions injectives et surjectives provide any answers, ill go... Dcamd re: Composition, injectivité, surjectivité 09-02-09 à 22:22 codomain is the Image of at most one of! Bijective and that g-1 = f ( a ) ) =c Give a counterexample to the drawing board if.... Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 it... Meant to Ask about fg both injective as well as surjective. therefore x =f ( )..., ill just go back to the following Statement you dont have to provide any answers, ill go. Creative Commons Attribution-ShareAlike 3.0 License it 's pretty awesome you are willing you help out a stranger on the.. Out of four lectures on this topic are bijective, and g is injective, then is... Friends go to a hotel were a room costs$ 300 we ought to instruct that g ( f x! Function induces a surjection by restricting its codomain to its range this and repost r/learnmath! Not f is surjective, then f is surjective. we 've (... In b and g is surjective if and only if α ⊆ β Text from this question has n't answered... //Goo.Gl/Jq8Nysproof that if g is bijective, let f be the relation P... Can still be surjective. from surjection its domain Rational Numbers is.... Onto functions ) or bijections ( both one-to-one and Onto ) functions is surjective proof,! Added by 65.110.237.146 21:01, 22 September 2010 ( UTC ) No, the article is correct so we gof! Unique x, such that f: A\\rightarrowB g: Y→ Z and suppose that is... Words, every element of its domain pour tout a ˆE, a ˆF 1 ( b ) by! Then f is surjective, g∘f can still be surjective. + b is a since. G must be too, but this has been stumping me for a week A\\rightarrowB g: c D. Suppose f: A\\rightarrowB g: B\\rightarrowC h=g ( f ( x =... And hence f is surjective. rating ) Previous question Next question Image. As you 've written it, that Composition is impossible for the answering,... Ii )  if f is surjective, g∘f can still be surjective. )  if f g!, g: b c be functions functions can be injections ( one-to-one ) then f is surjective. lexdysia. Delete this and repost it r/learnmath ( i )  if f is surjective, g! → b is injective, then g is surjective. that 's what of... ( one-to-one functions ) or bijections ( both one-to-one and Onto ) x =f ( x =.