One-to-one and Onto: True or False

Question

1) Suppose ${f}: X\to Y$ is one-to-one and $A\subseteq X$. Then $f^{-1}({f}(A))=A$.

True

2) Suppose ${f}: X \to X$, and assume that ${f} \circ {f}$ is one-to-one and onto. Then ${f}$ is one-to-one and onto.

True

3) Suppose ${f}:X\to Y$ is a function. Then ${f}(X)=Y$ if and only if ${f}$ is a bijection.

False

The first two I 'think' I worked them out correctly and found them to be true. But, the third one was the one I was mostly unsure on.

I thought these three questions were short (only true/false) and similar enough to warrant a single post.

Answer 1

For 3) take X = [-1, 1], Y = [0, 1] and f(x) = x^2 or /x/, then f[-1, 1] = [0, 1] but neither of them is a bijection since f(-1) = f(1) = 1. So 3) is false.

For 1) It is true. If x is in A ==> f(x) is in f(A). So x must be in f^(-1)(f(A)). So A is a subset of f^(-1)(f(A)). Conversely, if x is in f^(-1)(f(A)) ==> f(x) is in f(A). So f(x) = f(a) for some a in A, since f is one to one, x = a. So x is in A also. So f^(-1)(f(A)) is a subset of A. Thus A = f^(-1)(f(A)).