This is essentially Nakayama's lemma (not literally, but the same proof). More generally, the result is that a graded module map $\phi\colon N\to M$ is surjective if and if and only if the induced map $N\to M/A_{>0}M$ is surjective. "Only if" is obvious; consider the "if" direction. The induced map being surjective is the same as saying that $M=A_{>0}M+\operatorname{im}\phi$, so reducing both sides modulo $\operatorname{im}\phi$, we see that $M/\operatorname{im}\phi=A_{>0}(M/\operatorname{im}\phi)$. This is impossible, since $M/\operatorname{im}\phi$ has an element of minimal degree (since the grading is non-negative) and that's obviously not in $A_{>0}(M/\operatorname{im}\phi)$.

Note if $M$ is allowed to have elements in arbitrary degrees the result is false: consider $k[x,x^{-1}]$ as a graded module over $k[x]$.

This is essentially Nakayama's lemma (not literally, but the same proof). More generally, the result is that a graded module map $\phi\colon N\to M$ is surjective if and only if the induced map $N\to M/A_{>0}M$ is surjective. "Only if" is obvious; consider the "if" direction. The induced map being surjective is the same as saying that $M=A_{>0}M+\operatorname{im}\phi$, so reducing both sides modulo $\operatorname{im}\phi$, we see that $M/\operatorname{im}\phi=A_{>0}(M/\operatorname{im}\phi)$. This is impossible, since $M/\operatorname{im}\phi$ has an element of minimal degree (since the grading is non-negative) and that's obviously not in $A_{>0}(M/\operatorname{im}\phi)$.

Note if $M$ is allowed to have elements in arbitrary degrees the result is false: consider $k[x,x^{-1}]$ as a graded module over $k[x]$.