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Information about Composition of functions

an exposition of homework problem 3.2.24 for BSU's math 189 - discrete math for CS majors.

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3.2.24.b Here, we just want to ﬁnd a counterexample to the claim that if g ◦ f is onto, then f is onto AND g is onto. Consider the sets: A = {0, 1}. B = {3, 4, 5}. C = {6, 7}. Deﬁne g ◦ f = {(0, 6), (1, 7)}, which is onto. Deﬁne g = {(3, 6), (4, 6), (5, 7)} is onto. But f = {(0, 3), (1, 4)} is not onto. BAM. DONE. 3.2.24.c Prove that if g ◦ f is onto, then g must be onto. First, let’s consider some incorrect thought processes. Deﬁne some sets A, B, C. A = {0, 1, 2}. B = {3, 4}. C = {6, 7, 8}. Deﬁne f : A → B as f = {(0, 3), (1, 3), (2, 4)}. Note that f is onto. Deﬁne g : B → C as g = {(3, 6), (4, 7)}. Note that g is not onto. Now here’s where you have to be careful. You cannot deﬁne g◦f as {(0, 6), (1, 7), (2, 8)}. Now, this set does represent some onto function from A → C, but this function is not g ◦f. Consider what it means to be in g ◦ f. This is g(f(x)) for some x in the dom(f). So g ◦ f is {(x, g(f(x))) | x ∈ A. So g ◦ f = {(0, g(f(0)), (1, g(f(1)), (2, g(f(2))}. So, g ◦ f is {(0, g(3)), (1, g(3)), (2, g(4))} So, g ◦ f is {(0, 6), (1, 6), (2, 7)}. Now you can see that g◦f is not onto Be careful when you are composing your functions that you don’t just go from the domain of f directly to the range of g. You have to stop in the middle at the ran(f) which is the dom(g). Now, let’s get to the proof. Claim: if g ◦ f is onto, then g must be onto. Proof. Let f and g be functions and g ◦ f be an onto function. Let A, B, C be sets such that f : A → B and g : B → C. Now, since g ◦ f is onto, we know that every element in C gets hit by some element in f mapped through g. Suppose g is not onto. Then there is some element, say z ∈ C such that g(y) = z, y ∈ B. Now we need to consider what happens when f is onto and when f is not onto. 2

Case1: Suppose f is onto. If f is onto, then we have every element in B can be written as f(x) for some x ∈ A. So y = f(x). So, g(f(x) = z. So g ◦ f is not onto, a contradiction. So, if f is onto, and g ◦ f is onto, then g must be onto. Case2: Suppose f is not onto. So there is some element in B, say y, such that f(x) = y for any element x ∈ A. Since g is not onto, then there is some element, say z ∈ C such that g(y) = z, y ∈ B. So, then for some element x ∈ A, f(x) = y and g(y) = z, so not everything in C gets hit. So g ◦ f is not onto, a contradiction. So, If g ◦ f is onto, then g must be onto. 3

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