Let $(A,≤)$ and $(B,≤')$ be posets. Suppose the function $h:A→B$ satisfies $x≤y$ iff $h(x)≤'h(y)$ for all $x$ and $y$ in $A$. Prove that $h$ is one-to-one.
I attempted to prove by contradiction:
Assume $h$ is not one-to-one. Then there are $u,v \in dom(h)$ s.t. $h(u)=h(v)$ but $u\not=v$. Since $h(u)=h(v)$, we have $h(u)≤'h(v)$. Then $u≤v$ but $u\not=v$, so $u<v$... (where < is the strict order).
Wasn't sure where I could go from here. Perhaps another approach I could take? Thanks.
If $h(u)=h(v),$ then $h(u)\le h(v)$ and $h(v)\le h(u)$ (by reflexivity), so $u\le v$ and $v\le u$ (by assumption), so $u=v$ (by antisymmetry).