The new omega one is the old aleph omega

If $\alpha <\kappa$, then $\alpha <{\mathrm{\aleph }}_n^V$ for some $n<\omega$. For $\alpha =0$ it is finite. Otherwise restrict the surjection from F1 by replacing values outside $\alpha$ with $0$; this is a surjection $\omega \twoheadrightarrow \alpha$ in $N$. Thus every ordinal below $\kappa$ is countable in $N$, and consequently $\kappa \le {\omega }_1^N$.

Suppose for contradiction that some $f\in N$ is a surjection $\omega \twoheadrightarrow \kappa$. Choose an HS name $\dot{f}$ and use F2 to fix $m<\omega$ such that $H_m$ fixes it. For $k<\omega$ and $\alpha <\kappa$ let $u_{k,\alpha }=\parallel \dot{f}(\check{k})=\check{\alpha }\parallel$. These Boolean values are fixed by $H_m$, because $\dot{f}$ and the check names are fixed.

Let $P_{<m}=\{p\upharpoonright m:p\in P\}$. For each $k<\omega$ put $A_k=\{\alpha <\kappa :\mathrm{\exists }q\in P_{<m}(q{\le }_B{u}_{k,\alpha })\}$. Distinct $\alpha ,\beta \in A_k$ require incompatible witnesses, since a condition cannot force two different values of the function at $k$. Choosing the least witness in a fixed ground well-order injects $A_k$ into $P_{<m}$. Ground AC and the finite-function calculation give $\mathrm{\mid }{P}_{<m}{\mathrm{\mid }}^V\le {\mathrm{\aleph }}_m^V$, hence $\mathrm{\mid }{\bigcup }_{k<\omega }{A}_k{\mathrm{\mid }}^V\le {\mathrm{\aleph }}_m^V<\kappa$.

If $p⊩\dot{f}(\check{k})=\check{\alpha }$, then $p{\le }_B{u}_{k,\alpha }$, and F3 gives $p\upharpoonright m{\le }_B{u}_{k,\alpha }$; hence $\alpha \in A_k$. Because the alleged $f$ is surjective, F4 supplies such a $k$ and $p\in G$ for every $\alpha <\kappa$. Thus $\kappa ={\bigcup }_{k<\omega }{A}_k$, contradicting the strict bound in step 1.3. Therefore no such $f$ belongs to $N$, so $\kappa$ is uncountable in $N$.

Since ${\omega }_1^N$ is the least uncountable ordinal of $N$, step 2.1 gives ${\omega }_1^N\le \kappa$, while step 1.1 gives the reverse inequality. Hence ${\omega }_1^N=\kappa ={\mathrm{\aleph }}_{o}meg{a}^V$.