Levy collapse gone bad Let $\kappa$ be strongly inaccessible, and let $\mu<\kappa$ be regular.  What is the effect on $\kappa$ of forcing with the following?
(1) The product of $Col(\mu,\alpha)$ for $\alpha<\kappa$, with supports of size $<\kappa$.  (bounded support)
(2) The product of $Col(\mu,\alpha)$ for $\alpha<\kappa$, with supports of size $\leq \kappa$.  (full support)
The failure of $\kappa$-cc is clear, but do these collapse $\kappa$?
 A: (1) In general, no. If $\diamondsuit(\lambda)$ holds stationarily often below $\kappa$ (e.g., if $V=L$ and $\kappa$ is Mahlo, or $\diamondsuit(\kappa)$ and $\kappa$ is weakly compact), then $\mathbb{P}=\mathbb{P}_{\text{bounded}}$ does not collapse $\kappa.$
For $p \in \mathbb{P},$ let $\text{supp}(p)= \sup \{\alpha<\kappa: p_{\alpha} \neq \emptyset\}.$
Lemma: If $p \Vdash \dot{\gamma} < \check{\kappa},$ there is $p' \le p$ and $\beta<\kappa$ such that $p' \restriction \text{supp}(p)=p$  and $p' \Vdash \dot{\gamma}<\check{\beta}.$
First we'll show the lemma implies the result. Suppose there is $p$ and $\dot{f}$ such that $p \Vdash$ "$\dot{f}$ is a surjection from $\mu$ to $\kappa.$" We construct a descending sequence $\langle p_{\alpha}: \alpha \le \mu\rangle$ by

*

*$p_0=p.$

*$p_{\alpha+1}$ is a $p' \le p_{\alpha}$ such that $p' \restriction \text{supp}(p_{\alpha})=p_{\alpha}$ and $p'$ bounds $\dot{f}(\check{\alpha}).$

*For limit $\alpha,$ $p_{\alpha}= \bigcup_{\xi<\alpha} p_{\xi}.$
Then $p_{\mu} \Vdash$ "$\dot{f}$ is bounded," contradiction.
Now we prove the lemma. Let $p \Vdash \dot{\gamma}<\check{\kappa}.$ Let $\langle r_{\alpha}: \alpha<\kappa \rangle$ enumerate $\mathbb{P}.$
We define an ascending sequence $\langle \xi_{\alpha}: \alpha<\kappa \rangle$ by

*

*$\xi_0=0$ and $\xi_1=\text{supp}(p).$

*For $\alpha>0,$ $\xi_{\alpha+1} \ge \sup\{\eta: \text{supp}(r_{\eta}) \le \xi_{\alpha}\}$ is minimal such that, for every $r$ supported on $\xi_{\alpha},$ if there is $r'\le r$ with $r' \restriction \xi_{\alpha} = r$ such that $r'$ decides $\dot{\gamma},$ then there is such an $r'$ with $\text{supp}(r') \le \xi_{\alpha+1}.$

*For limit $\alpha,$ $\xi_{\alpha}= \sup_{\eta<\alpha} \xi_{\eta}.$
Let $\lambda$ be least such that $\diamondsuit(\lambda)$ holds and $\xi_{\lambda}=\lambda.$ Fix a sequence $\langle f_{\alpha}: \alpha<\lambda \rangle$ of functions $f_{\alpha}: \alpha \rightarrow \alpha$ such that, for every $f: \lambda \rightarrow \lambda,$ there is $\alpha>0$ such that $f \restriction \alpha = f_{\alpha}.$
We construct a descending sequence $\langle p_{\alpha}: \alpha \le \lambda \rangle,$ each $p_{\alpha}$ supported on $\xi_{\alpha},$ and a sequence $\langle \beta_{\alpha}: \alpha < \lambda \rangle$ by

*

*$p_0=\emptyset,$ $\beta_0=0,$ and $p_1=p.$


*Let $\alpha>0.$ Suppose that for all $\eta < \alpha,$ $r_{f_{\alpha}(\eta)}$ is supported on $[\xi_{\eta}, \xi_{\eta+1}).$ Further suppose there is $(q,\beta)$ such that $q$ is supported on $[\xi_{\alpha}, \xi_{\alpha+1})$ and $(p_{\alpha} \cup q \cup \bigcup_{\eta<\alpha} r_{f_{\alpha}(\eta)} ) \Vdash \dot{\gamma}=\check{\beta}.$ Then let $p_{\alpha+1}=p_{\alpha} \cup q$ and $\beta_{\alpha}=\beta$ for some such pair.
Otherwise, let $p_{\alpha+1}=p_{\alpha}$ and $\beta_{\alpha}=0.$


*For limit $\alpha,$ $p_{\alpha}= \bigcup_{\xi<\alpha} p_{\xi}.$
I claim $p':=p_{\lambda} \Vdash \dot{\gamma}<\sup_{\alpha<\lambda} \beta_{\alpha}.$ Suppose not. Let $q \le p'$ and $\beta \ge \sup_{\alpha<\lambda} \beta_{\alpha}$ be such that $q \Vdash \dot{\gamma} = \check{\beta}.$ Define $f: \lambda \rightarrow \lambda$ by $r_{f(\eta)}=q \restriction [\xi_{\eta}, \xi_{\eta+1}).$ Let $\alpha>0$ be such that $f_{\alpha}=f \restriction \alpha.$ Then $q \le p_{\alpha+1} \cup \bigcup_{\eta<\alpha} r_{f_{\alpha}(\eta)} \Vdash \dot{\gamma}=\check{(\beta_{\alpha})},$ contradiction.
(2) Yes, and in fact $\mathbb{P}=\mathbb{P}_{\text{full}}$ collapses $(2^{\kappa})^V$ to $\mu.$ In $V[G]$ there is a $g$ generic for $\Pi_{\alpha<\kappa} Add(\mu, 1),$ and we can construct from $g$ a surjection from $\mu$ to $(2^{\kappa})^V$ by a simple generalization of the "$\text{cf}(\lambda)>\omega$" case here: https://mathoverflow.net/a/334974/109573.
