Show that an order-isomorphism is necessarily a bijective mapping but the converse is not true.
My try for the converse: let $L=M=\{0,1\}$ and $f$ is defined by $f(0)=1,f(1)=0$. This can be easily verified that $f$ is bijective. But
\begin{equation*}
f(0\wedge 1)=f(0)=1\ne 0=1\wedge 0= f(0)\wedge f(1)
\end{equation*}
and \begin{equation*}
f(0\vee 1)=f(1)=0\ne 1=1\vee 0= f(0)\vee f(1)
\end{equation*}
That is $f$ is not a homomorphism.
Since $f$ is not a homomorphism, it is not isomorphism and hence cannot be order-isomorphism.
Is it correct?