TITLE: The function $f(x)=(x\vee a)\wedge b$ in a lattice.
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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
$$f(x) \vee f(y) = a < b = f(x\vee y).$$