Excercise 1.5.1 from Blackburn, de Rijke and Venema's Modal Logic

I've recently started to learn Modal Logic using this book as a reference. At the end of Section 1.5 "Modal Consequence Relations" there are some excercises about the topic. The first one says:

Let $\mathsf{K}$ be a class of frames for the basic modal similarity type, and let $\mathsf{M}(\mathsf{K})$ denote the class of models based on a frame in $\mathsf{K}$. Prove that $p\Vdash^{g}_{\mathsf{M}(\mathsf{K})} \Diamond p$ iff $\mathsf{K}\models \forall x \exists y Ryx$ (every point has a predecessor). Does this equivalence hold as well if we work with $\Vdash^{g}_{\mathsf{K}}$ instead?

I'm a little bit confuse since if we consider the model $\mathcal{M}$ with two points $x_0$ and $y_0$ such that $R y_0 y_0$, $R y_0 x_0$ and $V(p)=\{x_0,y_0\}$ then $\mathcal{M}\models \forall x \exists y Ryx$ but $\mathcal{M}\Vdash p$ and $\mathcal{M}\not\Vdash \Diamond p$ because $\mathcal{M},x_0\not\Vdash \Diamond p$.

Is there something I misunderstood?

Solution 1:

In the online copy of the book you linked to, the statement is $\square p\Vdash^g_{\mathsf{M(K)}} p$. If you have a print copy of the book that reads $p\Vdash^g_{\mathsf{M(K)}} \lozenge p$, maybe this is a typo that was corrected in the online version?