how to find $\alpha(a\wedge b)=\alpha(a)\wedge\alpha(b)$

Question

for $\alpha(a\wedge b)=\alpha(a)\wedge\alpha(b)$ how to do? answer: $\alpha $ and $ \alpha^{-1}$ are order-preserving and $a \leqslant b$ and $a=a\wedge b$ so $\alpha(a) = \alpha(a\wedge b)$ so $\alpha(a)=\alpha(a\wedge b) \leqslant \alpha(b)$ and how to continue?

Answer 1

How to continue... Suppose that $\alpha$ and $\alpha^{-1}$ are order-preserving. For $a, b$ in $L_1$ we have $a \wedge b \leqslant a$ and $a \wedge b \leqslant b$, so $\alpha(a \wedge b) \leqslant \alpha(a)$ and $\alpha(a \wedge b) \leqslant \alpha(b)$, hence $\alpha(a \wedge b) \leqslant \alpha(a) \wedge \alpha(b)$.

Furthermore, if $u \leqslant \alpha(a) \wedge \alpha(b)$, then $u \leqslant \alpha(a)$ and $u \leqslant \alpha(b)$, hence $\alpha^{-1}(u) \leqslant a$ and $\alpha^{-1}(u) \leqslant b$, so $\alpha^{-1}(u) \leqslant a \wedge b$, and thus $u \leqslant \alpha(a \wedge b)$. This implies that $\alpha(a) \wedge \alpha(b) = \alpha(a \wedge b)$.