The inverse function in Set is Boolean Ring homomorphism

Preliminary

We already know that $(P(S),\mathrm{\Delta },\cap )$$(P(S),\Delta,\cap)$ is a ring. The most efficient way to show it is to consider ${\mathbb{F}}_2$$\mathbb F_2$

Firstly, we must prove it is an Abelian Group with $\mathrm{\Delta }$$\Delta$.

Obviously, it is close; the identity is $\mathrm{\varnothing }$$\empty$ $\mathrm{\forall }A\in P(S),A\mathrm{\Delta }A=\mathrm{\varnothing }$$\forall A\in P(S),A\Delta A=\empty$

To prove that it is associative,

consider ${\mathbb{F}}_2$$\mathbb F_2$ , for $A\mathrm{\Delta }B,x\in A,x\in B$$A\Delta B,x\in A,x\in B$, then $x\notin A\mathrm{\Delta }B$$x\notin A\Delta B$

It is just like $1+1=0$$1+1=0$ in ${\mathbb{F}}_2$$\mathbb F_2$ and so on.

Because + in ${\mathbb{F}}_2$$\mathbb F_2$ is associative, $\mathrm{\Delta }$$\Delta$ is also associative

And $\cap$$\cap$ it is just like $\times$$\times$ in ${\mathbb{F}}_2$$\mathbb F_2$ , if $x\in A,x\in B$$x\in A,x\in B$, then $x\in A\cap B$$x\in A\cap B$. Just like $1\times 1=1$$1\times 1=1$ and so on.

The 1 in this ring is $S$$S$, the biggest set. F2 proves the distributive law,

Actually this Ring is isomorphic to ${\mathbb{F}}_2^n$$\mathbb F_2^n$

And The ideal of this ring is for $T\subseteq S$$T\subseteq S$, all the subsets of $T$$T$, $P(T)$$P(T)$

The duality between $\mathrm{Set}$$\mathrm{Set}$ and $\mathrm{Bool}$$\mathrm{Bool}$

This article aims to prove that $f:X\to Y$$f:X\to Y$, its inverse, defined as a set value function

$f^{-1}:P(Y)\to P(X)$$f^{-1}:P(Y)\to\ P(X)$ is a Boolean Ring homomorphism

So we need to prove that $f^{-1}$$f^{-1}$ preserves symmetry difference $\mathrm{\Delta }$$\Delta$ and intersection

And to prove $f^{-1}$$f^{-1}$ preserves symmetry difference $\mathrm{\Delta }$$\Delta$ ,

we need to prove that $f^{-1}$$f^{-1}$ preserves $\cup$$\cup$ $\cap$$\cap$ and complement

To prove that $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$$f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$

We just need to consider the universal property

$f=\bar{f}\circ \pi$$f=\overline f\circ\pi$... Then it will be obviously

or consider $f(x)\in A\cup B\iff f(x)\in A\vee f(x)\in B$$f(x)\in A\cup B\Leftrightarrow f(x)\in A\vee f(x)\in B$

To prove that $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$$f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}( B)$

$f(x)\in A\cap B\iff f(x)\in A\wedge f(x)\in B$$f(x)\in A\cap B\Leftrightarrow f(x)\in A\wedge f(x)\in B$

As a corollary, $A\subseteq B\Rightarrow f^{-1}(A)\subseteq f^{-1}(B)$$A\subseteq B\Rightarrow f^{-1}(A)\subseteq f^{-1}(B)$

Because $A\subseteq B\iff A\cap B=A\iff A\cup B=B$$A\subseteq B\Leftrightarrow A\cap B= A\Leftrightarrow A\cup B= B$

Consider $f:X\to Y$$f:X\to Y$, we need to prove that $f^{-1}(A^c)=f^{-1}(A{)}^c$$f^{-1}(A^c)=f^{-1}(A)^c$

Because $Y=A\cup A^c,A\cap A^c=\mathrm{\varnothing }$$Y=A\cup A^c, A\cap A^c=\empty$

So $X=f^{-1}(A)\cup f^{-1}(A^c),f^{-1}(A)\cap f^{-1}(A^c)=\mathrm{\varnothing }$$X=f^{-1}(A)\cup f^{-1}(A^c),f^{-1}(A)\cap f^{-1}(A^c)=\empty$

Thus $f^{-1}(A^c)=f^{-1}(A{)}^c$$f^{-1}(A^c)=f^{-1}(A)^c$

Or we can prove $f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$$f^{-1}(A- B)=f^{-1}(A)- f^{-1}(B)$

Because $f(x)\in A-B\iff f(x)\in A\wedge f(x)\notin B$$f(x)\in A- B\Leftrightarrow f(x)\in A\wedge f(x)\notin B$

Thus $f^{-1}$$f^{-1}$ preserves $\mathrm{\Delta }$$\Delta$ and $\cap$$\cap$

Because $A\mathrm{\Delta }B=(A\cup B)\cap (A\cap B{)}^c$$A\Delta B=(A\cup B)\cap(A\cap B)^c$

$f^{-1}(A\mathrm{\Delta }B)=f^{-1}(A)\mathrm{\Delta }{f}^{-1}(B),f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$$f^{-1}(A\Delta B)=f^{-1}(A)\Delta f^{-1}(B), f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$, $f^{-1}(Y)=X$$f^{-1}(Y)=X$

The kernel of this ring homomorphism is $P(Y-\mathrm{im}f)$$P(Y-\mathrm{im} f)$

Because $f^{-1}$$f^{-1}$ send all the element of $Y-\mathrm{im}f$$Y-\mathrm{im} f$ to $\mathrm{\varnothing }$$\empty$, $\begin{array}{c}\frac{P(Y)}{P(Y-\mathrm{im}f)}\cong \mathrm{im}{f}^{-1}\end{array}$$\frac{P(Y)}{P(Y-\mathrm{im} f)}\cong\mathrm{im}f^{-1}\\$