Jtdylktuy

Can someone give me their opinion on how I wrote this proof also all my definitions are at the end.

Given a function $f:Ato B$, its inverse relation $f^{-1}$ is a function from $B$ to $A$ if and only if $f$ is a bijection from $A$ onto $B$.

$textbf{My attempt:}(Longrightarrow)$ Assume the function $f:Ato B$ has an inverse relation $f^{-1}$ which is a function from $B$ to $A$. Then $f^{-1}circ f=i_A$. Fix $a_1,a_2in A$ and $f(a_1)=f(a_2)$.Then

$i_A(a_1)=a_1$ and $i_A(a_2)=a_2$

$f^{-1}circ f(a_1)=i_A(a_1)=a_1$

$f^{-1}circ f(a_2)=i_A(a_2)=a_2$ by assumption $a_1=a_2$.

WLOG the opposite will also be true and hence $a_2=a_1$ and therefore the function $f$ is injective.
Fix $bin B$

$a=f^{-1}(b)$

$f(a)=f(f^{-1}(b))$

$=fcirc f^{-1}(b)$

$=i_B$

$=b$

Thus $f(a)=b$ so $f$ is surjective and therefore is bijective.

$(Longleftarrow)$ Assume $f$ is a bijection from $A$ onto $B$. Since $f$ is bijective, then it will be surjective and by definition 2, the domain of $f^{-1}$ would then be $B$ thus $dom(f^{-1})=B$. Also since $f$ is bijective, then it will be injective and by definition 1, $f^{-1}$ will only have one element of $A$ for each element of $B$. So by definition 4, $f:Ato B$, its inverse relation $f^{-1}$ is a function from $B$ to $A$.

Definitions

$textbf{(1)}$ A relation $Rsubseteq Atimes B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$

$textbf{(2)}$ We say a function $f:Ato B$, is a surjection from A onto B if $forall yin B$ $exists x in A (f(x)=y).$ In this case we will also call $f$ a surjection from $A$ onto $B.$

$textbf{(3)}$ We say a function $f:Ato B$, is said to be a bijection from $A$ to $B$ if it is injective and surjective.

$textbf{(4)}$ A function $f$ from a set $A$ to a set $B$ is a relation from $A$ to $B$ for which $(((x,y_1) in f) land ((x,y_2) in $f$)) rightarrow y_1=y_2$,

It follows directly from the definitions. For this, view mappings as relations.

A mapping $f:Arightarrow B$ is a relation $fsubseteq Atimes B$ which is
left-total
$$forall xin Aexists yin B[xfy]$$
and right-unique
$$forall xin Aforall y,y'in B[xAywedge xAy'rightarrow y=y'].$$
A mapping $f$ is injective (left-unique) if
$$forall x,x'in Aforall yin B[xAywedge x'Ayrightarrow x=x'].$$
A mapping $f$ is surjective (right-total) if
$$forall yin Bexists xin A[xAy].$$

So $f$ is bijective means that $f$ is left-total, right-unique, right-total, and left-unique.

Note that inverse relation $f^{-1}subseteq Btimes A$ always exists. (Going from $f$ to $f^{-1}$ the left-properties become right-properties and vice versa.)

This means that $f^{-1}$ is right-total, left-unique, left-total, and right-unique, i.e.,
$f^{-1}$ is bijective.