Injection from $\aleph_2$ into the power set of $\mathbb{R}$

Question

Assume $\mathsf{ZF} + \mathsf{DC}$. Must there exist an injection from $\aleph_2$ to $\mathcal{P}(\mathbb{R})$? If not, what is the consistency strength of the nonexistence of such an injection?

I hope I'm not missing something obvious. Here are my thoughts so far:

1. There is always a surjection from $\mathcal{P}(\mathbb{R})$ to $\aleph_2$.
2. $\mathsf{ZFC}$, or more generally the existence of a wellordering of $\mathcal{P}(\mathbb{R})$, implies there is such an injection.
3. $\mathsf{ZF} + \mathsf{AD}$ implies the existence of a surjection from $\mathbb{R}$ to $\aleph_2$, which implies there is such an injection.
4. If there is an inner model of $\mathsf{ZFC}$ satisfying $(\aleph_1^V)^+ = \aleph_2^V$, for example in $V(\mathbb{R}^{V[G]})$ where $\kappa$ is inaccessible and $G \subset \operatorname{Col}(\omega,\mathord{<}\kappa)$ is a $V$-generic filter, then there is such an injection.

Answer 1

The consistency strength is just that of ZFC.

Start with GCH, and do a symmetric collapse of $\aleph_{\omega_1}$ to be $\aleph_2$. Since everything is $\sigma$-closed, you get DC for free, and the reals can even be well ordered.

But any injection from $\aleph_2$ into $\cal P(\Bbb R)$ would have had to be adjoined by some bounded part of the iteration (by homogeneity), so none exist, as with the usual arguments for $L(\Bbb R)$.