Final answer:
If g o f is one-to-one and f is onto, then g is one-to-one. If g o f is onto and g is one-to-one, then f is onto. The function g is not the inverse of f.
Step-by-step explanation:
(a) If g o f is one-to-one and f is onto, then g is one-to-one:
To prove this statement, let's assume that g is not one-to-one. This means that there exist two elements b1 and b2 in B such that b1 != b2, but g(b1) = g(b2). Since f is onto, there exist elements a1 and a2 in A such that f(a1) = b1 and f(a2) = b2. Now, since g o f is one-to-one, it implies that (g o f)(a1) != (g o f)(a2), which means that g(f(a1)) != g(f(a2)). But we know that g(f(a1)) = g(b1) and g(f(a2)) = g(b2), which contradicts our assumption that g is not one-to-one. Therefore, we can conclude that if g o f is one-to-one and f is onto, then g is one-to-one.
(b) If g o f is onto and g is one-to-one, then f is onto:
Let's assume that f is not onto. This means that there exists an element b in B such that for every element a in A, f(a) != b. Now, since g o f is onto, it implies that for every element c in C, there exists an element a in A such that (g o f)(a) = g(f(a)) = c. But we know that f(a) != b for every a in A. Therefore, we can conclude that f is onto.
(c) Verification and explanation of g o f = l_A and why g is not the inverse of f:
The function g o f is given by (g o f)(1) = g(f(1)) = g(a) = 1 and (g o f)(2) = g(f(2)) = g(b) = 2. Therefore, we have (g o f) = {(1, 1), (2, 2)}, which is the identity function on A, denoted as l_A. Now, to check if g is the inverse of f, we need to verify if g o f = f o g = l_A. But we know that (g o f) = l_A, whereas (f o g) = {(a, 1), (b, 2), (c, 1)}, which is not equal to l_A. Therefore, we can conclude that g is not the inverse of f.