Final answer:
The one-to-one nature of both f and f composed with g does not necessarily imply that g is also one-to-one. An example involving a linear f and an absolute value g demonstrates that g can fail to be injective even if f and f composed with g are.
Step-by-step explanation:
If f and f ◦ g are one-to-one functions, it does not necessarily mean that g is one-to-one. To demonstrate this, let's consider what it means for a function to be one-to-one. A function is one-to-one (injective) if different inputs lead to different outputs. Specifically, for any two elements a and b in the domain of the function, if a ≠ b, then f(a) ≠ f(b).
Let's assume f is one-to-one, and f ◦ g is also one-to-one. Now, to check if g is one-to-one, take any two distinct elements, say x and y, from the domain of g. If g(x) = g(y), then f(g(x)) = f(g(y)). Since f ◦ g is one-to-one, and f(g(x)) = f(g(y)), it must be that x = y. However, this argument relies on an assumption that g maps to the domain of f, which is not necessarily the case in every scenario.
For example, suppose we have a function f: R → R defined as f(x) = 2x and another function g: R → R defined as g(x) = |x|. The function f is one-to-one because for any x1 ≠ x2, f(x1) ≠ f(x2). The composition f ◦ g is also one-to-one because the absolute value function does not affect the one-to-one nature of the linear function f. However, g itself is not one-to-one, as g(-a) = g(a) for any a ∈ R.