This is an exercise of an assignment I have:
Suppose $A$ and $B$ are finite sets and $f\colon A\to B$ is surjective. Is it true that the relation “$|A| < |B|$” is a sufficient condition for claiming that $f$ is a bijection? Justify your answer.
And this is my answer:
No. In fact, if $|A| < |B|$, then there should exist at least $1$ element of $A$ that points to more than $1$ element of $B$, since all elements of $B$ must be pointed (surjective), but this is not a function, because the same element of $A$ point to different elements of $B$.
Is my answer correct? I don't know if the question actually is asking for a proof or what, and if my answer is a proof, if correct.
If $\lvert A \rvert < \lvert B \rvert$, then you cannot have any surjective function $f\colon A\to B$ anyway, and the question is vacuous. (the image $f(A)$ of $A$ by any function $f$ must satisfy $\lvert f(A) \rvert \leq \lvert A \rvert$, with equality when $f$ is injective).