Is it true that the relation |A| < |B| is a sufficient condition for claiming that $f$ is a bijection? 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).
