The function $f(x)=(x\vee a)\wedge b$ in a lattice.

Question

Is there an algebraic modular lattice $(X,\vee,\wedge)$ and $a,b\in X$ with $a\le b$ such that the function $$f:X\to X$$ $$f(x)=(x\vee a)\wedge b$$ is not $\vee$-homomorphism?

Answer 1

The following is an example of a finite (hence algebraic) modular lattice $L$ for which your $f$ is not a join homomorphism:

Consider $L \cong M_3$, that is, the five element modular lattice of height 2. Label the three atoms $x$, $b$, and $y$. Label the bottom element $a$ and the top element $1$. Then $x\vee y = 1$, so

$$f(x\vee y) = (1\vee a) \wedge b = 1\wedge b = b,$$ while $$f(x) = (x\vee a) \wedge b = x \wedge b = a,$$ and $$f(y) = (y\vee a) \wedge b = y \wedge b = a,$$ so $$f(x) \vee f(y) = a < b = f(x\vee y).$$