Show that V is a strong limit in second-order terms Let $X, Y$ be two proper classes. We say that the power of $X$ is equinumerous to $Y$ iff there is a relation $R$ with $domain(R) = Y$ and $range(R) = X$ such that (i) $\forall Z \subseteq X \exists y \in Y R_y = Z$ (where $R_y$ is $\{x : <y ,x> \in R\}$), and (ii) $\forall y_1 , y_2 \in Y \ (y_1 \neq y_2 \rightarrow R_{y_1} \neq R_{y_2})$. (So R codes a ''bijection'' from Y to the ''powerclass'' of X)
My question is: does KM - Global Choice prove the following statement (*) (which basically says that $V$ cannot be reached by taking ''power'' of any class smaller than it)?
(*) If there is no injection from $V$ to $X$, then the power of $X$ is not equinumerous to $V$.
(Since Global Choice implies that all proper classes are equinumerous, this is not an interesting statement given GC.)
 A: No, this is not the case.
The counterargument is exactly along the lines Asaf suggested. If $M\models\mathsf{ZFC}$ then there is a definable-over-$M$ class surjection from the class of sets of ordinals to $M$ itself. Basically, with a set of ordinals we can "code" arbitrary initial segments of the cumulative hierarchy. In detail, for a set $u$ say that a code for $u$ is a triple $(\alpha,R,\beta)$ where $\alpha$ is some ordinal, $R$ is a binary relation on $\alpha$ such that $(\alpha;R)\cong V_\gamma$ for some $\gamma$ with $u\in V_\gamma$, and the image of $\beta$ under the unique isomorphism $(\alpha;R)\cong V_\gamma$ is $u$ itself. By appropriate tuple-juggling, we can replace codes with sets of ordinals.
The point is that if global choice fails, there is no injection from $V$ to $Ord$. However, running the above idea with $M=V$ shows - informally speaking - that the powerclass of $Ord$ is at least as big as $V$.
Note that there's some subtlety around the notion of "powerclass" here: are we talking about the collection of all subclasses of a given class, or the collection of all subsets of a given class? Call these the "big powerclass" and "small powerclass" respectively. The argument above gives a strong negative answer to your question: although $V$ doesn't inject into $Ord$, already the small powerclass of $Ord$ is as big as $V$.
