TITLE: Can Assumptions about forcing produce Mice?
QUESTION [5 upvotes]: This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident: 
For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements 
$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$
(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names) 
and
$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma) $$
(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))
Now, here comes the question: Does the following conjunction:
$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

$\lambda$ is a Regular cardinal.
$\vert \mathbb{P} \vert = \lambda^{+}$
$\forall \mu \  (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$ 
$\neg Cof(\lambda, \mathbb{P})$

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)
(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)
Edit:
(It was not my intention to scare a lot of nice mice)
(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)

REPLY [10 votes]: Here's an argument for an affirmative answer to Joel's modified version of the question. Suppose we have a forcing that preserves cardinals but singularizes some cardinal $\lambda$ that was regular in the ground model.  Note that $\lambda$ had to be a limit cardinal, since otherwise singularizing it would collapse it down to its immediate predecessor cardinal (if not even lower).  Now let $C$ in the forcing extension be a cofinal subset of $\lambda$ of smaller cardinality $\kappa$.  I claim that C is not included in any set $D$ in the ground model of cardinality $\leq\max\{\kappa,\aleph_1\}$; in other words, I claim that $C$ is a counterexample to the assertion that the forcing extension satisfies the covering lemma over the ground model.  Indeed, suppose we had such a $D$.  Intersecting it with $\lambda$, we'd have a cofinal subset of $\lambda$ strictly smaller than $\lambda$ in the ground model, contrary to the assumption that $\lambda$ is regular in the ground model.  ("Strictly smaller" in the preceding sentence uses that $\lambda>\aleph_1$, which is why I pointed out earlier that $\lambda$ has to be a limit cardinal.)  So the forcing extension doesn't satisfy the covering lemma over the ground model.  That implies the existence of an inner model with a measurable cardinal, by an ancient result of mine --- "Small extensions of models of set theory" in "Axiomatic Set Theory" (Proc. of 1983 Boulder Conference, edited by Baumgartner, Martin, and Shelah) Contemporary Math. 31 (1984) pp. 35-39.