# Proving isomorphisms from posets.

An isomorphism from a poset $(S_1,R_1)$ to a poset $(S_2,R_2)$ is a bijection $f: S_1 \rightarrow S_2$ such that, for all $x,y \in S_1$

$(x,y) \in R_1 \leftrightarrow (f(x), f(y)) \in R_2$

When such an isomorphism exists, we say that $(S_1,R_1)$ is isomorphic to $(S_2,R_2)$.

Questions:

1. Show that every poset is isomorphic to itself.

Attempt: Every poset is isomorphic to itself...it's like the reflexive definition for equivalence relations which is $(\forall x \in S)[(x,x) \in R]$

So maybe I could apply the definition?

$(\forall x \in S)[(x,x) \in (S_1,R_1)]$

$(\forall x \in S)[(x,x) \in (S_2,R_2)]$

2.Prove that if $f$ is an isomorphism then so is $f^{-1}$

Attempt: Definition 6.3.6 states that we let $R$ be any relation on a set $S$. The inverse relation of R, denoted $R^{-1}$ is defined by the condition $(x,y) \in R \leftrightarrow (y,x) R^{-1}$

Using the definition on $(x,y) \in R_1 \leftrightarrow (f(x),f(y)) \in R_2$, we have

$(x,y) \in R_1 \leftrightarrow (y,x) \in R_1^{-1}$

$(f(x),f(y)) \in R_2 \leftrightarrow f(y),f(x) \in R_2^{-1}$

$(y,x) \in R_1^{-1} \leftrightarrow f(y),f(x) \in R_2^{-1}$

It appears that they're kind of symmetrical in a way.

1. Prove that the composition of two isomorphisms is an isomorphism.

Attempt: Definition 6.3.9 states that we let $R_1$ and $R_2$ be relations on a set S. The composition of $R_2$ with $R_1$ is the relation

$R_2 \circ R_1 =[(x,y) \in S \times S:( \exists v \in S)[(x,y) \in R_1 \land (v,y) \in R_2]$

Maybe we should have a composition like $(S_1,R_1) \circ (S_2, R_2)$. Then ,

$(S_2, R_2) \circ (S_1,R_1) =[(x,y) \in S \times S:( \exists v \in S)[(x,y) \in (S_1,R_1) \land (v,y) \in (S_2, R_2)]$

I'm kind of confused and lost...maybe the transitive property would work better?

Definition: R is transitive if $(\forall x, y, z \in S)[((x,y) \in R \land (y,z) \in R) \rightarrow (x,z) \in R]$

so maybe let $x = (S_1,R_1), y = (S_2, R_2)$ and $z = (S_3,R_3)$. Then

$(\forall (S_1,R_1), y, z \in S)[(((S_1,R_1),(S_2, R_2)) \in R \land ((S_2, R_2),(S_3,R_3)) \in R) \rightarrow ((S_1,R_1),(S_3,R_3)) \in R]$...

If the original poset definition was being used here I could understand it a little bit better, but this is a different form. Any hints?

1. Isomorphism relation is reflexive. That means there exists a bijection from the poset to itself that preserves the poset-relation. There might be many such bijections, but there is one that always works: identity, obviously $$\forall x,y \in S_1.\ (x,y) \in R_1 \leftrightarrow (x, y) \in R_1$$ and so function $\mathrm{id}_{S_1} : S_1 \to S_1$ given by $\mathrm{id}_{S_1}(x) = x$ is a valid isomorhism between $(S_1,R_1)$ and itself.
2. Isomorphism relation is symmetric. That means that if $f : S_1 \to S_2$ is a valid isomorphism between $(S_1,R_1)$ and $(S_2,R_2)$, then $f^{-1} : S_2 \to S_1$ is a valid isomorphism between $(S_2,R_2)$ and $(S_1,R_1)$. Note, that we don't change or inverse the $R_1$ or $R_2$ relations as you did in your approach. Instead we inverse the isomorphism, which is one level higher. In particular we would like to show $$\Bigg(\forall x,y \in S_1.\ (x,y) \in R_1 \leftrightarrow (f(x), f(y)) \in R_2\Bigg) \!\!\iff\!\! \Bigg(\forall x',y' \in S_2.\ (x',y') \in R_2 \leftrightarrow (f^{-1}(x'), f^{-1}(y')) \in R_1\Bigg)$$ which is true because $f$ and $f^{-1}$ are both bijections and mutual inverses.
3. Isomorphism relation is transitive. That is, if $f : S_1 \to S_2$ and $g : S_2 \to S_3$ are valid isomorphisms, then so is $(g \circ f) : S_1 \to S_3$. Note that we don't compose the poset relations, but the isomorphisms. Then we would like to show that $$\Bigg(\forall x,y \in S_1.\ (x,y) \in R_1 \leftrightarrow (f(x), f(y)) \in R_2\Bigg)$$ and $$\Bigg(\forall x',y' \in S_2.\ (x,y) \in R_2 \leftrightarrow (g(x'), g(y')) \in R_3\Bigg)$$ together imply $$\Bigg(\forall x'',y'' \in S_1.\ (x'',y'') \in R_1 \leftrightarrow \big((g\circ f)(x''), (g\circ f)(y')\big) \in R_3\Bigg)$$ which is true by transitivity of $\leftrightarrow$.
I hope this helps $\ddot\smile$