Let f:X->Y and g:Y->Z, show that g is surjective, if gof is surjective. Solution

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a) Suppose that g o f : X ? Z is onto. So, for any z in Z, we have (g o f)(x) = z for some x in X. Rewrite this as g(f(x)) = z. Since f : X ? Y, we see that f(x) is in Y. Taking y = f(x), we have g(y) = z. Hence, g is onto. --- Suppose that g o f : X ? Z is 1-1, and that there exist x, x' in X, we have f(x) = f(x'). Apply g to both sides: g(f(x)) = g(f(x')). ==> (g o f)(x) = (g o f)(x') Since g o f is 1-1, we conclude that x = x'. Hence, f(x) = f(x') ==> x = x'; that is, f is 1-1. ---------------- b) Define g : Y ? X by g(f(x)) = x for all x in X. To show that g is well-defined, suppose f(c) and f(d) are the same, but c and d are different (i.e. f isn't injective); that is, let y = f(c) = f(d), where c ? d. Then g(y) = g(f(c)) = c, and also g(y) = g(f(d)) = d. Hence, c = d, which is a contradiction by the injectivity of f. That g o f = id_x is true by the way we defined g. As for f o g = id_y: f(g(y)) = f(g(f(x)), for some x in X (since f is surjective) ..........= f(x), by g o f = id_x ..........= y, as required. ---------- Conversely, suppose there exists g : Y ? X with g o f = id_x and f o g = id_y. Since f o g = id_y is onto, part a allows us to conclude that f is surjective (onto). Since g o f = id_x is 1-1, part a allows us to conclude that f is injective (1-1). Hence, f is a bijection. ---------------- As for dropping a condition, think of the results from part a (and how we used it above). You should be able to construct simple counterexamples to bijectivity of f in either case.

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