(a) If f and g f are onto,then g is onto. (b) If g and g f are onto,then f is on
ID: 2942881 • Letter: #
Question
(a) If f and g f are onto,then g is onto.
(b) If g and g f are onto,then f is onto.
Explanation / Answer
Definition. A function f:A->B is onto if for every b in B, there is some a in A such that f(a) = b. In other words, every element in the range gets mapped to by some element in the domain. (a) True. We have that f:A->B is onto and that gof:A->C is onto. That means: For every b in B, there is some a in A such that f(a) = b. For every c in C, there is some a in A such that g(f(a)) = c. We want to prove that g is onto. Choose some arbitrary c in C. Then by the above, there is some a in A such that g(f(a)) = c. Let f(a) = b. Then we have found b in B such that g(b) = c, and therefore g is onto. (b) False. Counterexample: Let f:R->R be the function f(x) = x^2. Let g:[0,inf)->[0,inf] be the function g(x) = x. The function g is onto because for any nonnegative real number y, we can find the real number z = y*y = y^2, which must be in the domain [0,inf). And g(z) = z = (y^2) = y again (since y was the positive square root of y^2). So g is onto. The function g(f(x)) is onto because for any y in [0,inf), the same number y in R is such that g(f(y)) = y, because g(f(y)) = g(y^2) = (y^2) = y (again, this must be the same number y because we started with a nonnegative y). So gof is onto. But f is not onto: for any negative number y, there is no z in R such that f(z) = y, because no real number z squares to a negative. Therefore this proposition is false.
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