Let $X$ and $Y$ be two ordered rings (i.e, they satisfy the order properties). Let $<_{X}$ and $<_{Y}$ be the strict partial orders on $X$ and $Y$ respectively.
Let $f:X \rightarrow Y$ be a ring homomorphism which is injective and order-preserving, i.e $$\forall x\, \forall y\, (x<_{X}y \implies f(x)<_{Y}f(y) ).$$ Then prove that this map is unique.
I don't think that statement is true. Consider any ordered ring $R$, and take $S=R[X]$, with the order such that the sign of $P\in R[X]$ is the sign of its leading coefficient.
Then the morphism $S\to S$ induced by $X\mapsto X^2$ is injective and order-preserving, but is not the identity.