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 $\aleph_0 \to \aleph_1$?
On
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.
No. You can never map $\omega$ onto $\omega_1$. Not without the axiom of choice, and certainly not with it.
The reason is that if $f\colon\alpha\to A$ is a surjection, and $\alpha$ is an ordinal, then there is an injection from $A$ into $\alpha$. Simply take $g(a)=\min\{\beta<\alpha\mid f(\beta)=a\}$. You should convince yourself that this is a well-defined injection.
$\omega_1$ is defined as the least uncountable ordinal, again without appealing to the axiom of choice, so by definition there is no injection from $\omega_1$ into $\omega$ and by the above there cannot be a surjection from $\omega$ onto $\omega_1$ either.
In the absence of the axiom of choice, it is consistent (read: possible, but not provable) that there is some set $A$ such that $\aleph_1\nleq|A|$ but there is a map from $A$ onto $\aleph_1$. But this $A$ cannot be countable, that is for sure.