Let $L_1, L_2$ be lattices and $f: L_1\to L_2$ a bijective (suffices to demand surjectivity) mapping such that $$\forall a,b\in L_1,\quad a\leq b\iff f(a)\leq f(b) $$
Show that $f$ is an isomorphism of lattices.
It remains to verify whether $f$ is a homomorphism and for that it suffices to check whether $f$ is a $\vee$-homomorphism ($\wedge$-hom is established dually) that is $$\forall a,b\in L_1,\quad f(a\vee b) \overset{?}= f(a)\vee f(b) $$ Since $a \leq a\vee b$ and $b\leq a\vee b$, then $$f(a)\leq f(a\vee b)\quad\mbox{and}\quad f(b)\leq f(a\vee b) $$ By definition of least upper bound we have $$f(a)\vee f(b)\leq f(a\vee b) $$ But how does one establish the converse inequality i.e $$f(a\vee b)\overset{?}\leq f(a)\vee f(b) $$ Lecture notes say this is obvious :(
We could assume $f(a\vee b) > f(a)\vee f(b)\geq f(a)$, for instance. Then order reflection gives $a\vee b\geq a$, but so what? Or is it $a\vee b > a$, then $a\vee b > b$, but still, so what? I don't understand how this claim is obvious.
There is $c$ such that $f(c)=f(a)\vee f(b)$. Since $f(c)=f(a)\vee f(b)\leq f(a\vee b)$, we have that $c\leq a\vee b$. If $c<a\vee b$, then either $a\not\leq c$ or $b\not\leq c$ since $a\vee b$ was the least upper bound. Without loss of generality, assume $a\not\leq c$. Then $f(a)\not\leq f(c)=f(a)\vee f(b)$; a contradiction. Thus $c\not< a\vee b$.