In general, we know that adding a subset to a regular cardinal $\kappa$ can collapse cardinals. If, for example, there is $\gamma < \kappa$ with $2^\gamma >\kappa$, then $Add(\kappa,1)$ will collapse $2^\gamma$ to $\kappa$, since every subset of $\gamma$ will appear as a block in the generic.


However, this argument requires that we use the poset defined in $V$. We can easily construct situations in which $2^\gamma > \kappa$ but forcing with $Add(\kappa,1)^L$ does not collapse cardinals, by the same techniques that allow us to achieve Easton's Theorem. If we force over $L$ with $Add(\gamma, \kappa^+) \times Add(\kappa,1)$, for example, then cardinals are preserved. I am interested to know if this holds in general.



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> Does $Add(\kappa,1)^L$ ever collapse cardinals?
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Does $Add(\kappa,1)^L$ ever collapse cardinals?
Yes, it follows from a theorem of Mack Stanley (for this special case, in fact from a theorem of Foreman-Magidor-Shelah) that if $0^\sharp$ exists, then forcing with $Add(\kappa, 1)^L$ collapses $\kappa$ into $\omega.$


See Stanley's paper "Forcing disabled" (and Foreman-Magidor-Shelah paper "$0^\sharp$ and some forcing principles", J Symbolic Logic 51 (1986), 39-46). See also [Can the Cohen forcing collapse cardinals?](https://mathoverflow.net/questions/184495/can-the-cohen-forcing-collapse-cardinals) and the answer given there.


