Mapping on cardinal without Axiom of Choice Define $|A|\le|B|$ iff there exists injective mapping $A \to B$.
If Axiom of Choice is assumed then this is equivalent as $|A|\le|B|$ iff there exists surjective mapping $B \to A$. But:
If Axiom of Choice is not assumed: Is it possible to have surjective mapping $\aleph_0 \to \aleph_1$?

No.
Without choice, we can say that if $A,B$ are sets and $B$ is well-orderable, then $|A|\le|B|$ iff there is a surjection $B\to A$. The Axiom of Choice is equivalent to the statement that every set is well-orderable, which is why we needn't make the distinction in that setting.
If we could map $\aleph_0\to\aleph_1$ surjectively, then we could map $\aleph_1\to\aleph_0$ injectively. But $\aleph_1$ is the cardinality of the least uncountable ordinal--that is, the least ordinal that cannot be mapped injectively into $\aleph_0$--so this is impossible.
