Isomorphism between partially ordered sets - What is wrong with my argument?

Question

I'm trying to show

Isomorphism between two partially ordered sets is an equivalence relation.

Suppose $M$ and $M^{\prime}$ are two partially ordered set and $f:M\to M^{\prime}$ is isomorphism between them. To show reflexivity, let $a\in M$ then since $M$ is partially ordered $a\leq a$, so $f(a)\leq f(a)$. If $f(a)\leq f(b)$ and $f(b) \leq f(c)$ imply $a \leq b$ and $b \leq c$. $M$ is partially ordered set so, $a\leq c$, hence, $f(a)\leq f(c)$.

I have problems to show symmetry. If $f(a)\leq f(b)$ I can't show $f(b)\leq f(a)$. Either my argument is completely wrong or I am missing something important. Thanks for any help!

Answer 1

I assume what you mean by isomorphism is that $f$ preserves the structure of $M$ when mapping to $M^{\prime}$ in the following sense:

1. $f$ is a bijection
2. $x\leq_{M}y$ iff $f\left(x\right)\leq_{M^{\prime}}f\left(y\right)$

I also assume you define equivalence as follows: $M$ and $M^{\prime}$ are equivalent (written $M\sim M^{\prime}$) if there exists an isomorphism (as defined above) between them. Then

1. Taking $f$ to be the identity map, $M\sim M$
2. Suppose $M\sim M^{\prime}$. Then there exists an isomorphism $f_{1}\colon M\rightarrow M^{\prime}$. Taking $f\equiv f_{1}^{-1}$ ($f_{1}$ is a bijection), we get $M^{\prime}\sim M$.
3. Suppose $M\sim M^{\prime}$ and $M^{\prime}\sim M^{\prime\prime}$. Then there exist isomorphisms $f_{1}\colon M\rightarrow M^{\prime}$ and $f_{2}\colon M^{\prime}\rightarrow M^{\prime\prime}$. Taking $f\equiv f_{2}\circ f_{1}$, we can show that $f$ is an isomorphism. Suppose $a,b\in M$ with $a\leq_{M}b$. Then $f_{1}\left(a\right)\leq_{M^{\prime}}f_{1}\left(b\right)$, and as such $f_{2}\left(f_{1}\left(a\right)\right)\leq_{M^{\prime\prime}}f_{2}\left(f_{1}\left(b\right)\right)$.

Edit: The notation $a \leq_M b$ refers to the partial order on $M$ explicitly. The subscript is often omitted.