If $2^\mu$ and $2^\lambda$ are isomorphic as boolean algebras, is it true that $\mu = \lambda$? The content of this question is:

Is what I am going to say correct?

It is well known that given two cardinals $\lambda < \mu$ it might be the case that $2^{\lambda} = 2^{\mu}$. Even stronger, the Easton theorem says that it is consistent with ZFC to force some of these equalities.
This means that there might be a bijection between the set $2^\lambda$ and the set $2^\mu$. On the other hand I claim that this correspondence cannot be a morphism of boolean algebras. In fact, the following seems to hold.
Lemma Let If $2^\mu$ and $2^\lambda$ are isomorphic as boolean algebras then $\mu = \lambda$.
By boolean algebra morphism I mean a function that preserve sups and inf.
The proof is very simple. Such a map $\phi$ would preserve completely join-irreducible elements. In a boolean algebra those correspond to the atoms of the boolean algebra. Thus $\phi$ restrict to a bijection of the atoms, which is the thesis.
 A: Yes, this is correct.  More strongly, in fact, you can say that if $\lambda<\mu$, then there is no injective homomorphism of Boolean algebras $2^\mu\to 2^\lambda$, since such a homomorphism would have to send the atoms of $2^\mu$ to a collection of $\mu$ pairwise disjoint nonzero elements of $2^\lambda$, which is not possible if $\lambda<\mu$.
Another strengthening of your statement is that the category of Boolean algebras of the form $2^X$, with complete (i.e., preserving infinite joins) Boolean homomorphisms as morphisms, is equivalent to the opposite category of sets.  That is, any complete homomorphism $2^X\to 2^Y$ must be the map induced by a unique map $Y\to X$.  The proof is straightforward: just look at where each atom of $2^X$ gets sent, and use the fact that every element of $2^X$ is a join of atoms.  It follows that if $2^X$ and $2^Y$ are isomorphic in this category (i.e., isomorphic as Boolean algebras, since any isomorphism of Boolean algebras is complete), then $X$ and $Y$ are isomorphic as sets.
