Is injectivity of $2^{(\ldots)}$ weaker than $\mathsf{GCH}$? [duplicate]

Question

The following statement cannot be proven in $\mathsf{ZFC}$:

(S) : If $A, B$ are sets with $|A| < |B|$, then $2^{|A|} = |{\cal P}(A)| < |{\cal P}(B)| = 2^{|B|}$.

Obviously, $\mathsf{GCH}$ implies (S). Does (S) imply $\mathsf{GCH}$ too?

Answer 1

No, take Merimovich's model in which $2^\kappa=\kappa^{+3}$ for all cardinals $\kappa.$

Merimovich, Carmi A power function with a fixed finite gap everywhere. J. Symbolic Logic 72 (2007), no. 2, 361–417.

Let me add that the statement "the power function is injective" is known as weak GCH, which has important applications. See Why the weak GCH is true! .

Answer 2

One can answer this easily with Easton's theorem, which shows that we can have $2^{\aleph_n}=\aleph_{n+2}$ for all finite $n$, and $2^\kappa=\kappa^+$ for all other infinite cardinals $\kappa$. In this model, the continuum function is still one-to-one, but the GCH fails. So the answer is no.