Endoskeleton of a poset is isomorphic to skeleton of the poset proof?

Question

Let $(X, \leq)$ be a poset (note that $\leq \; \subseteq X \times X$). Show that any endoskeleton of $(X, \leq)$ is isomorphic to the skeleton of $(X, \leq)$.

Some definitions:

Clique of $(X, \leq)$: $(Y, \leq')$ is a clique of $(X, \leq)$ iff $Y \subseteq X$ and $\leq'$ is a restriction of $\leq$ and a clique.

Maximal Clique of $(X, \leq)$: A clique of $(X, \leq)$ that is maximal.

Skeleton of $(X, \leq)$: Let $(X_i, \leq_i)$ be the family of all maximal cliques of $(X, \leq)$. Then the skeleton of $(X, \leq)$ is $(\cup_i X_i, \cup_i \leq_i)$.

Endoskeleton of $(X, \leq)$: Let $(X_i, \leq_i)$ be the family of all maximal cliques of $(X, \leq)$. Then the endoskeleton of $(X, \leq)$ is $(X',\leq')$ s.t. $\forall i \; X' \cap X_i \neq \varnothing$ and $\leq'$ is a restriction of $\leq$.

Monotone function: Given two preorders $(X, \leq)$ and $(Y, \leq')$, a function $f: X \rightarrow Y$ is monotone iff $\forall x,x' \in X \; x \leq x' \Rightarrow f(x) \leq' f(x')$.

Isomorphic pair: Given two preorders $(X, \leq)$ and $(Y, \leq')$, the monotonic functions $f: X \rightarrow Y$ and $g: Y \rightarrow X$ are isomorphic if $g \circ f = id_X$ and $f \circ g = id_Y$.

Isomorphism: Two preorders are isomorphic to each other if there exists an isomorphism pair between them.

My attempt: I am unable to find two monotonic function between those sets which are inverses of each other. Have I misunderstood some definition or do the functions elude me?

Answer 1

A partial order is a preorder that has no nontrivial cliques. If you had two distinct elements, $x$ and $y$, that formed a clique, then you would have $x\leq y$ and $y\leq x$, which contradicts the antisymmetry of the poset. Hence, every element of $X$ forms a trivial maximal clique, consisting of only itself. Then the skeleton of a poset and all of its endoskeletons (of which there is only one), is isomorphic to the poset itself.