one-to-one function inverse cardinality

Question

Proof:

Prove that
$f : A → B$ is one-to-one if and only if $|f^{-1}(b)| ≤ 1$ for all $b ∈ B$

I realize you must construct two proofs to prove both sides of the biconditional. I started first by assuming $f : A → B$ and trying to prove $|f^{-1}| ≤ 1$ for all $b ∈ B$. However, I thought it must be true that a function is bijective in order for the function to be invertible, so I got stumped immediately. Any tips?

Answer 1

You are missing a critical portion of the problem statement. It should be something like

Problem: Prove that $f: A \to B$ is one-to-one if and only if $|f^{-1}(\{b\})| \le 1$ for all $b\in B$.

You also have some confusion about the notation. The notation $f^{-1}$ is used to indicate preimage not inverse, which, as you note, may not exist. If $S$ is a subset of $B$, then $f^{-1}(S)$ is the subset of $A$ defined by $\{x\in A \mid f(x) \in S\}$, which may be empty in the case that $f$ does not map any points of $A$ into $S$.

Now, consult your definition of one-to-one and try to deduce that if $f$ is one-to-one, then there is at most $1$ point in the preimage of a given element $b\in B$. The converse direction is similar.