All definitions I found say that a functor $F:\mathcal{A}\rightarrow\mathcal{B}$ is right-exact, if for every short exact sequence $$ 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$$ gives an exact sequence $$F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow 0.$$
Somehow I imagine to remember, that it is sufficient to require that any surjection $B \rightarrow C \rightarrow 0$ is mapped to a surjection $F(B) \rightarrow F(C) \rightarrow 0$, though I was neither able to prove it, nor did I find any reference. So it appears to me that this is wrong.
I would like to see a proof, or a counterexample for the claim that $F$ is right-exact if and only if it maps surjections to surjections.