A submodel of set theory with all reals which every set is analytic

Question

This is a continuation of my previous question. Recall that a subset $A \subseteq {}^\omega\omega$ is analytic if it is the continuous image of the Baire space. I would like to know if there exist two models $N \subseteq M$ such that:

1. $M \models \mathsf{ZFC}$ (or just $M \models \mathsf{ZF} + \text{There exists a non-principal ultrafilter}$).

2. $N \models \mathsf{ZF} + \text{Every subset of reals is analytic}$.

3. $({}^\omega\omega)^M = ({}^\omega\omega)^N$.

From the comments of previous posts, it's worth noting that:

1. Under $\mathsf{ZF} + \mathsf{DC}$, one can construct the universal analytic set and prove that it is not analytic. Thus, $N \not\models \mathsf{DC}$.

2. A (somewhat trivial) example of a model in which every subset of reals is analytic is the Feferman-Levy model, where the reals is a countable union of countable sets.

Answer 1

In fact, the principle "Every set is analytic" is not consistent with $\mathsf{ZF}$ in the first place. We don't need choice to get a surjection $h$ from Baire space to the set of continuous maps on Baire space. But once we have such an $h$, the "diagonalizing" set $\{x: x\not\in h(x)\}$ can't be analytic.

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Let me show how to get such an $h$ in $\mathsf{ZF}$ alone. Working in $\mathsf{ZF}$, let $C$ be the set of continuous maps $\omega^\omega\rightarrow\omega^\omega$. To each $\gamma\in C$ we assign the set $$S(\gamma)=\{(\sigma,\tau)\in\omega^{<\omega}\times\omega^{<\omega}: \forall f\succ \sigma(\gamma(f)\succ\tau)\}.$$ By your favorite coding mechanism we can identify $S(\gamma)$ with some $\hat{S}(\gamma)\in\omega^\omega$. But $\hat{S}(\gamma)=\hat{S}(\gamma')$ implies $\gamma=\gamma'$ for continuous $\gamma,\gamma'$, so in fact $\hat{S}$ gives an injection from $C$ to $\omega^\omega$.

Now turning a surjection into an injection without choice is hard, but the converse is trivial: for $x\in\omega^\omega$, let $h(r)=\hat{S}^{-1}(r)$ if $r\in ran(\hat{S})$, and let $h(r)$ be the always-zero map otherwise.