Let $\lambda$ be an infinite cardinal. Does the full support product of $\lambda$ copies of $Add(\omega, 1)$ collapse $2^\lambda$ to $\aleph\_0$?


For $\lambda = \omega$, it is known to be true (it is an exercise in Kunen's book), but what about uncountable $\lambda$? 



Products of Cohen forcings
The answer is yes.


Lemma: There is a function $F: 2^{\lambda} \rightarrow 2^{\lambda}$ such that for every $S \in [\lambda]^{\lambda}$ and function $g: \lambda \setminus S \rightarrow 2,$ $F$ restricted to extensions of $g$ is surjective.


Fix a well-ordering of all ordered pairs of the form $(g, h),$ $g$ as above and $h \in 2^{\lambda},$ of length $2^{\lambda}.$ We construct $F$ in $2^{\lambda}$ stages, each stage determining $F$ at one input. Let $(g,h)$ be the $\alpha$th pair in the ordering. At stage $\alpha,$ we have determined $F$ at less than $2^{\lambda}$ inputs, so there is some extension $g'$ of $g$ for which $F(g')$ has not been determined. We declare $F(g')=h.$ This completes the construction.


Let's consider the case where $\text{cf}(\lambda)>\omega.$ Notice that for every $h \in 2^{\lambda}$ and $p \in \mathbb{P}=\Pi\_{\alpha<\lambda} Add(\omega,1),$ there are $q \le p$ and $n<\omega$ such that $n \in \text{dom}(q\_{\alpha})$ for every $\alpha,$ and $F(\alpha \mapsto q\_{\alpha}(n))=h.$ Here $n$ is any number such that $|\{\alpha<\lambda: n \not \in \text{dom}(p\_{\alpha})\}|=\lambda.$ Thus, in $V[G],$ there is a surjection from $\omega$ to $(2^{\lambda})^V$ definable from $F$ and $G.$


Now suppose $\text{cf}(\lambda)=\omega$ and $\lambda>\omega.$ Here we use a generalization of the matrix argument in Hamkins' answer. Let $\kappa\_0=0,$ and fix a cofinal $\omega$-sequence $\langle \kappa\_n: 0<n<\omega\rangle$ of uncountable regular cardinals below $\lambda.$ In $V[G],$ $G$ determines an $\omega \times \lambda$ matrix of bits, each column a Cohen real. We construct a $\lambda$-sequence $z\_n$ as follows: Let $n\_0=n,$ and recursively define $n\_{k+1}$ to be the number of consecutive 0's above the $n\_k$th entry in the $\kappa\_k$th column. For $\alpha \in [\kappa\_k, \kappa\_{k+1}),$ let $z\_n(\alpha)$ be the $n\_k$th entry in the $\alpha$th column.


Fix $h \in (2^{\lambda})^V$ and $p \in \mathbb{P}.$ It suffices to show there are $q \le p$ and $n<\omega$ such that $q$ forces $h$ to equal $F(z\_n).$ Let $n\_k>\max(\text{dom}(p\_{\kappa\_k}))$ be such that $|\{\alpha \in [\kappa\_k, \kappa\_{k+1}): n\_k \not \in \text{dom}(p\_{\alpha})\}|=\kappa\_{k+1},$ and let $n=n\_0.$ It is easy to construct $q$ as desired.


