Is relation R isotone isomorphism with operation *

Question

I have follow operation $*$ defined like $MxM \rightarrow M$ where $M=\{a,b,c,d\}$

and relation $R=\{(a,a),(b,b),(c,c),(d,d),(a,b),(b,c),(a,c)\}$

Decide, if $R$ is isotone with operation $*$ on M.

Solution

I know definition of isotone isomorphism:

$(G,\leq ), (H, \preceq )$ are ordered set

$f: G \rightarrow H$

$\forall x,y \in G: x \leq y \implies f(x) \leq f(y)$

But I don't know how to apply this rule.

My guess

I think, that from $R$ implies

$(a,a) \leq (a,b) \leq (a,c) \leq (b,b) \leq (b,c) \leq (c,c)$ and $(d,d)$

If I do map to $*$ operation a get:

$(a,a)\in R \rightarrow a\in M \\ (a,b)\in R \rightarrow b\in M \\ etc..$

I get:

$a \leq b \leq c \nleq a \leq d \leq a$

So, the answer is no, $R$ is not isotone isomorphism with M.

Please,is that good solution? Or how should a solution looks like?

Answer 1

$R$ cannot be an isomorphism because it not a bijection, it is not even a function.

@amrsa $x\to x\star a $ is the identity map so it is an isotone isomorphism. Similarly for $x\to a\star x $.