TITLE: $2^\omega$ vs $({\omega+1})^\omega$
QUESTION [2 upvotes]: It is easy to see that there is a surjective lattice homomorphism $s:({\omega+1})^\omega \to 2^\omega$ (construction see after the horizontal line below).  Is there a surjective lattice homomorphism $s:2^\omega \to ({\omega+1})^\omega$?

Assign to every $f\in ({\omega+1})^\omega$ the map $s(f)\in 2^\omega$ defined by setting for every $n\in\omega$: $\big(s(f)\big)(n) = 0$ if $f(n) = 0$ and $\big(s(f)\big)(n) = 1$ if $f(n)>0 $.

REPLY [9 votes]: No. The lattice $2^\omega$ is complemented, but the lattice $\omega+1$ is not complemented. Since a homomorphic image of a complemented lattice is complemented, there is no surjective lattice homomorphism from $2^\omega$ to $\omega+1.$ Equivalently, there is no surjective lattice homomorphism from $2^\omega$ to $(\omega+1)^\omega.$
Likewise, there is no surjective lattice homomorphism from $2^\omega$ to $3,$ although there is a surjective lattice homomorphism from $\omega+1$ to $3.$
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