Does $\mathsf{ZF}$ prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals below $\lambda$?

Question

Let $\lambda<\kappa$ be cardinals and consider the forcing $\operatorname{Col}(\lambda,\kappa)$ adding a generic surjection $\lambda\to\kappa$. More formally, $\operatorname{Col}(\lambda,\kappa)$ is the set of all partial functions $p\colon\lambda\to\kappa$ such that $|p|<\lambda$. It is well-known that $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals less than equal to $\lambda$, but its proof uses choice. (More specifically, we need $\lambda$-$\mathsf{DC}$.)

Question. Can we prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals $\le\lambda$ from $\mathsf{ZF}$ alone?

The question is trivial if $\lambda=\omega$, but I am interested in the case when $\lambda$ is a singular cardinal of cofinality $\omega$, and $\kappa$ is inaccessible. (We say $\kappa$ is inaccessible if $V_\kappa$ is a model of the second-order $\mathsf{ZF}$.)

Also, is there any forcing adding a surjection $\lambda\to \kappa$ that also preserves cardinals below $\lambda$ in a choiceless context?

Answer 1

Re the first question: No. Suppose $\lambda$ is a singular cardinal of cofinality $\omega$. Then the forcing collapses all cardinals $\leq\lambda$. For let $\left<A_n\right>_{n<\omega}\in V$ be a partition of $\lambda$ into sets $A_n$, each of cardinality ${<\lambda}$, and with $\mathrm{ordertype}(A_n)<\mathrm{ordertype}(A_{n+1})$. In $V[G]$, where $G:\lambda\to\kappa$ is the generic surjection, let $F:\omega\to\lambda$ be the function computed as follows: Let $X_n=(G``A_n)\cap\lambda$. Then $G\upharpoonright A_n\in V$, so $X_n\in V$ and $X_n$ has cardinality $<\lambda$ (in $V$). Let $\gamma_n$ be the least ordinal $\notin X_n$, so $\gamma<\lambda$. If $X_n=\gamma_n$ let $F(n)=0$. If $X_n\neq\gamma_n$, let $\delta_n$ be the least $\delta>\gamma_n$ such that $\delta\in X_n$. So $[\gamma_n,\delta_n)\cap X_n=\emptyset$ and $\gamma_n\cup\{\delta_n\}\subseteq X_n$. Now define $F(n)$ to be the ordinal $\xi$ such that $\gamma_n+\xi=\delta_n$.

Now a genericity argument shows that $F$ is surjective. For fix $0<\xi<\lambda$, and let $p\in\mathbb{P}$. Let $D_n=A_n\cap\mathrm{dom}(p)$ and $C_n=A_n\backslash\mathrm{dom}(p)$. Note that for eventually all $n$, we have $\xi\leq\mathrm{ordertype}(C_n)$ (otherwise $\mathrm{dom}(p)$ has cardinality $\lambda$). Similarly, for eventually all $n$ we have $\mathrm{card}(D_n)\leq\mathrm{ordertype}(C_n)$. So fix $n$ such that $\eta=\max(\mathrm{card}(D_n),\xi)\leq\mathrm{ordertype}(C_n)$. Note that because both $\xi$ and $(p``D_n)\cap\lambda$ have cardinality $\leq\eta$ (in $V$), there is an ordinal $\alpha<\eta^+$ such that $[\alpha,\alpha+\xi)\cap((p``D_n)\cap\lambda)=\emptyset$. Now just extend $p$ to some $q$ with $A_n\subseteq\mathrm{dom}(q)$ and $\alpha\subseteq q``A_n$ and $[\alpha,\alpha+\xi)\cap (q``A_n)=\emptyset$ and $\alpha+\xi\in q``A_n$ (recall $C_n$ has ordertype $\geq\eta$, so this is possible). Note that $F(n)=\xi$, which suffices.

Answer 2

First of all, note that if $\lambda$ is a singular limit cardinal, then the forcing is terribly behaved already in $\sf ZFC$. Even if $\lambda$ is a successor, you'd still find that the same argument works. Divide $\lambda$ into a small amount of intervals, since any small set, cardinality wise, must be bounded in almost all of them, we are guaranteed that generically the collapsing function is constant on a tail of cofinally many of the intervals, which gives us a surjection from $\operatorname{cf}(\lambda)$ onto $\kappa$, and therefore you've collapsed much more that you've bargained for.

In the more general case. Forcing with "plain ol' sequences of ordinals" is a bad idea when working in $\sf ZF$, and will rarely work. The reason is simple, if we are just concatenating sequences of ordinals without any restrictions, we are pretty much guaranteed to introduce some well-orderings into the universe. This is very easy to see in the case of $\kappa=\lambda$ which is just adding a Cohen subset to $\lambda$, in which case we force a bijection between $\mathcal P(\lambda)^V$ and $\lambda^+$, which will add well-orderings for $\mathcal P(\alpha)^V$ below $\lambda$ as well, and in that case bounded subsets could be added below $\lambda$.

More to the point, if $\kappa^{<\lambda}$ can be well-ordered, then as a well-ordered forcing things are more or less reasonably behaved. So, for example, if we assume $\sf AX_4$, which states that for any infinite ordinal $\kappa$, $[\kappa]^{\omega}$ is well-orderable. So, in this case $\operatorname{Col}(\omega_1,\kappa)$ is going to be a reasonably behaved forcing.

In the general case, it's not at all clear that there is a way in $\sf ZF$ to add generic sets of ordinals without "damage" (with the exception of Cohen reals, or more generally, $\operatorname{Col}(\omega,\alpha)$ which is likely to not cause more damage than intended). This is more of a situation that will need to be tailored to your current model. For example, if you're working in a symmetric extension, ground model forcings are well-orderable and will commute with the symmetric extension, so they can sometimes be useful and preserve nice things (but not always!). I have always seen that as one of the biggest problems in the theory of forcing without choice, and we have yet to even discuss non well-orderable sets.