# if gof is surjective, then f is surjective

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merci pour votre aide. Conversely, if f o g is surjective, then f is surjective (but g need not be). December 10, 2020 by Prasanna. Let f: A B and g: B C be functions. Let f : X → Y be a function. They pay 100 each. 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}. Jan 18, 2011 #7 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. Hence f is surjective. Show transcribed image text. This problem has been solved! See the answer. Let Q be the relation on P (X) such that αQβ if and only if α ⊆ β. a ≠ b ⇒ f(a) ≠ f(b) for all a, b ∈ A f(a) = f(b) ⇒ a = b for all a, b ∈ A. e.g. 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. Let f : X → Y be a function. b, then f(a) ? Show transcribed image text. D emonstration. Expert Answer . Since gf is surjective, doesn't that mean you can reach every element of H from G? Then f(b 5 3) = 3(b 5 3) + 5 = b 5 + 5 = b. See the answer. 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"]. If gof is injective then (f is not surjective V g is injective) I started by assuming that gof was injective and went to show that g was injective by contradiction and just hit a wall. I think I just couldn't separate injection from surjection. Suppose that gof is surjective. Expert Answer 100% (1 rating) Previous question Next question Get more help from Chegg. Posté par . Can somebody help me? 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. 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. Dcamd re : Composition, injectivité, surjectivité 09-02-09 à 22:22. you dont have to provide any answers, ill just go back to the drawing board if not. 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'. Problem 27: Let f : B !C and g : C !D be functions. 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. Proof: This problem has been solved! Question: (i) "If F: A + B Is Injective, Then F Is Surjective." 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. 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. (c) Prove that if f and g are bijective, then gf is bijective. If, for some $x,y\in\mathbb{R}$, we have $f(x)=f(y)$, that means $x|x|=y|y|$. This problem has been solved! Injective, Surjective and Bijective. If f is not surjective, then there is a b in B such that for all a in A, f(a) ≠ b. Then since g f is surjective, there exists x 2A such that (g f)(x) = g(f(x)) = z. Then easily we see that f(1) = 1 and g(1) = 1 so g(f(1)) = 1 which is a surjection and a bijection since g(f) : {1} -> {1}. Finding an inversion for this function is easy. 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). One-one function (Injection) A function f : A B is said to be a one-one function or an injection, if different elements of A have different images in B. Homework Statement Suppose f: A → B is a function. Let X be a set. Y=7x(6/7 -1/4) is this a solution or a linear equation. 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. Then f is surjective since it is a projection map, and g is injective by definition. As eruonna pointed out, you either meant to ask about fg, or you mean to say that (g: F->H, f:G->F). By de nition of a rational number, there exist integers a;b such that b 6= 0 and c = a=b. Clearly, f is surjective, but all … 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. Then since g f is surjective, there exists x 2A such that (g f)(x) = g(f(x)) = z. Please Subscribe here, thank you!!! At least, that's what one of the diagrams on the page illustrates. This question hasn't been answered yet Ask an expert. To apply (g o f), First apply f, then g, even though it's written the other way. (iii) “The Set Of All Positive Rational Numbers Is Uncountable." Since, f is surjective, there is a unique x, such that f(x) = y. 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. 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. 3 friends go to a hotel were a room costs $300. For the answering purposes, let's assuming you meant to ask about fg. You just made this clear for me. 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). Problem. See the answer. —Preceding unsigned comment added by 65.110.237.146 21:01, 22 September 2010 (UTC) No, the article is correct. In other words, every element of the function's codomain is the image of at most one element of its domain. gof injective does not imply that g is injective. Therefore if we let y = f(x) 2B, then g(y) = z. 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? (b) Show by example that even if f is not surjective, g∘f can still be surjective. Une aide serait la bienvenue. Exercise problem and solution in ring theory. (b) Prove that if f and g are injective, then gf is injective. Therefore, f(a;b) = a=b = c and hence f is surjective. 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. 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