If $f:X\rightarrow Y$ is an order isomorphism , then $f^{-1}:Y\rightarrow X$ is an order isomorphism .

Question

• Let $(X,\leq)$, $(Y,\leq ')$ be partially ordered sets.

If $f:X\rightarrow Y$ is an order isomorphism between $X$ and $Y$, then the inverse function $f^{-1}:Y\rightarrow X$ is an order isomorphism between $Y$ and $X.$

Proof. Assume $f:X\rightarrow Y$ is an order isomorphism between $X$ and $Y$ that is

$$x_1\leq x_2\iff f(x_1)\leq ' f(x_2)$$

for all $x_1,x_2\in\mathbb{X}.$

So since $f$ bijective, we have

$$x_1\leq x_2 =f^{-1}(y_1)\leq f^{-1}(y_2) \iff f(x_1)\leq'f(x_2)=y_1\leq 'y_2,$$

for all $y_1,y_2 \in Y,$

that is

$$f^{-1}(y_1)\leq f^{-1}(y_2) \iff y_1\leq 'y_2$$

for all $y_1,y_2 \in Y$. So we are done.

Can you check my proof? Thankss...

Answer 1

I would phrase it in a different way, as follows.

Assume that $f:X\to Y$ is an order-isomorphism, that is $$x_1 \leq x_2 \Leftrightarrow f(x_1) \leq' f(x_2),$$ for all $x_1, x_2 \in X$ (and that $f$ is bijective).
Now, given $y_1, y_2 \in Y$, let us see that $$y_1 \leq' y_2 \Leftrightarrow f^{-1}(y_1) \leq f^{-1}(y_2).$$ As $f$ is bijective, there exist $x_1, x_2 \in X$ such that $f(x_i)=y_i$ (and thus $f^{-1}(y_i)=x_i$).
It follows that $$y_1 \leq' y_2 \Leftrightarrow f(x_1) \leq' f(x_2) \Leftrightarrow x_1 \leq x_2 \Leftrightarrow f^{-1}(y_1) \leq f^{-1}(y_2),$$ and therefore $f^{-1}$ is also an order-isomorphism.