Mixing
Proving the inverse of a function $f$ is a function iff the function $f$ is a bijection.
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$,
proof-verification
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
add a comment |
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$,
proof-verification
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
Your first deduction is $f^{-1}circ f=i_A$ might be unclear, depending on what definitions you are working from. This would typically be part of your definition of inverse function. However, you are dealing with an inverse relation that happens to be a function, and it's not immediately obvious that this is equal to the inverse function (this might be your definition of inverse function, but then $f^{-1}circ f=i_A$ is not simply true by definition). It is false in general that $f^{-1}circ f=i_A$, when $f^{-1}$ is the inverse relation of $f$ (and $circ$ is composition of relations).
– Erick Wong
Nov 22 '18 at 7:49
Ahh ok, but what I am assuming for that part is that both f and f^-1 are both functions since that is what's stated in the theorem so I doubt that will be a problem. However, should I add the definition of compositional inverses in and prove both their domains would be equal?
– George
Nov 22 '18 at 8:55
Well not really showing their domains are equal but that f $circ$ g= the domain of g
– George
Nov 22 '18 at 9:00
I’m not sure that addresses my concern. The question assumes that $f$ and $f^{-1}$ are functions, but here $f^{-1}$ refers to the inverse relation, not the inverse function $f^{-1}$ (which has not been assumed to exist). The inverse relation is defined by simply reversing the ordered pairs defining $f$. Unlike the inverse function, the inverse relation has no a priori properties regarding composition. Of course this assumes the usual definitions and your source text may vary.
– Erick Wong
Nov 22 '18 at 17:32
ok, I decided to just rewrite that part using the relation definition for functions just to be safe.
– George
Nov 23 '18 at 3:13
add a comment |
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$,
proof-verification
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
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$,
proof-verification
proof-verification
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
edited Nov 22 '18 at 7:36
1ENİGMA1
958416
958416
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
asked Nov 22 '18 at 6:08
GeorgeGeorge
676
676
Your first deduction is $f^{-1}circ f=i_A$ might be unclear, depending on what definitions you are working from. This would typically be part of your definition of inverse function. However, you are dealing with an inverse relation that happens to be a function, and it's not immediately obvious that this is equal to the inverse function (this might be your definition of inverse function, but then $f^{-1}circ f=i_A$ is not simply true by definition). It is false in general that $f^{-1}circ f=i_A$, when $f^{-1}$ is the inverse relation of $f$ (and $circ$ is composition of relations).
– Erick Wong
Nov 22 '18 at 7:49
Ahh ok, but what I am assuming for that part is that both f and f^-1 are both functions since that is what's stated in the theorem so I doubt that will be a problem. However, should I add the definition of compositional inverses in and prove both their domains would be equal?
– George
Nov 22 '18 at 8:55
Well not really showing their domains are equal but that f $circ$ g= the domain of g
– George
Nov 22 '18 at 9:00
I’m not sure that addresses my concern. The question assumes that $f$ and $f^{-1}$ are functions, but here $f^{-1}$ refers to the inverse relation, not the inverse function $f^{-1}$ (which has not been assumed to exist). The inverse relation is defined by simply reversing the ordered pairs defining $f$. Unlike the inverse function, the inverse relation has no a priori properties regarding composition. Of course this assumes the usual definitions and your source text may vary.
– Erick Wong
Nov 22 '18 at 17:32
ok, I decided to just rewrite that part using the relation definition for functions just to be safe.
– George
Nov 23 '18 at 3:13
add a comment |
Your first deduction is $f^{-1}circ f=i_A$ might be unclear, depending on what definitions you are working from. This would typically be part of your definition of inverse function. However, you are dealing with an inverse relation that happens to be a function, and it's not immediately obvious that this is equal to the inverse function (this might be your definition of inverse function, but then $f^{-1}circ f=i_A$ is not simply true by definition). It is false in general that $f^{-1}circ f=i_A$, when $f^{-1}$ is the inverse relation of $f$ (and $circ$ is composition of relations).
– Erick Wong
Nov 22 '18 at 7:49
Ahh ok, but what I am assuming for that part is that both f and f^-1 are both functions since that is what's stated in the theorem so I doubt that will be a problem. However, should I add the definition of compositional inverses in and prove both their domains would be equal?
– George
Nov 22 '18 at 8:55
Well not really showing their domains are equal but that f $circ$ g= the domain of g
– George
Nov 22 '18 at 9:00
I’m not sure that addresses my concern. The question assumes that $f$ and $f^{-1}$ are functions, but here $f^{-1}$ refers to the inverse relation, not the inverse function $f^{-1}$ (which has not been assumed to exist). The inverse relation is defined by simply reversing the ordered pairs defining $f$. Unlike the inverse function, the inverse relation has no a priori properties regarding composition. Of course this assumes the usual definitions and your source text may vary.
– Erick Wong
Nov 22 '18 at 17:32
ok, I decided to just rewrite that part using the relation definition for functions just to be safe.
– George
Nov 23 '18 at 3:13
Your first deduction is $f^{-1}circ f=i_A$ might be unclear, depending on what definitions you are working from. This would typically be part of your definition of inverse function. However, you are dealing with an inverse relation that happens to be a function, and it's not immediately obvious that this is equal to the inverse function (this might be your definition of inverse function, but then $f^{-1}circ f=i_A$ is not simply true by definition). It is false in general that $f^{-1}circ f=i_A$, when $f^{-1}$ is the inverse relation of $f$ (and $circ$ is composition of relations).
– Erick Wong
Nov 22 '18 at 7:49
Your first deduction is $f^{-1}circ f=i_A$ might be unclear, depending on what definitions you are working from. This would typically be part of your definition of inverse function. However, you are dealing with an inverse relation that happens to be a function, and it's not immediately obvious that this is equal to the inverse function (this might be your definition of inverse function, but then $f^{-1}circ f=i_A$ is not simply true by definition). It is false in general that $f^{-1}circ f=i_A$, when $f^{-1}$ is the inverse relation of $f$ (and $circ$ is composition of relations).
– Erick Wong
Nov 22 '18 at 7:49
Ahh ok, but what I am assuming for that part is that both f and f^-1 are both functions since that is what's stated in the theorem so I doubt that will be a problem. However, should I add the definition of compositional inverses in and prove both their domains would be equal?
– George
Nov 22 '18 at 8:55
Ahh ok, but what I am assuming for that part is that both f and f^-1 are both functions since that is what's stated in the theorem so I doubt that will be a problem. However, should I add the definition of compositional inverses in and prove both their domains would be equal?
– George
Nov 22 '18 at 8:55
Well not really showing their domains are equal but that f $circ$ g= the domain of g
– George
Nov 22 '18 at 9:00
Well not really showing their domains are equal but that f $circ$ g= the domain of g
– George
Nov 22 '18 at 9:00
I’m not sure that addresses my concern. The question assumes that $f$ and $f^{-1}$ are functions, but here $f^{-1}$ refers to the inverse relation, not the inverse function $f^{-1}$ (which has not been assumed to exist). The inverse relation is defined by simply reversing the ordered pairs defining $f$. Unlike the inverse function, the inverse relation has no a priori properties regarding composition. Of course this assumes the usual definitions and your source text may vary.
– Erick Wong
Nov 22 '18 at 17:32
I’m not sure that addresses my concern. The question assumes that $f$ and $f^{-1}$ are functions, but here $f^{-1}$ refers to the inverse relation, not the inverse function $f^{-1}$ (which has not been assumed to exist). The inverse relation is defined by simply reversing the ordered pairs defining $f$. Unlike the inverse function, the inverse relation has no a priori properties regarding composition. Of course this assumes the usual definitions and your source text may vary.
– Erick Wong
Nov 22 '18 at 17:32
ok, I decided to just rewrite that part using the relation definition for functions just to be safe.
– George
Nov 23 '18 at 3:13
ok, I decided to just rewrite that part using the relation definition for functions just to be safe.
– George
Nov 23 '18 at 3:13
add a comment |
1 Answer
1
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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.
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
add a comment |
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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.
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
add a comment |
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.
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
add a comment |
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.
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
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.
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
edited Nov 22 '18 at 16:34
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
answered Nov 22 '18 at 9:34
WuestenfuxWuestenfux
3,7361411
3,7361411
add a comment |
add a comment |
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Your first deduction is $f^{-1}circ f=i_A$ might be unclear, depending on what definitions you are working from. This would typically be part of your definition of inverse function. However, you are dealing with an inverse relation that happens to be a function, and it's not immediately obvious that this is equal to the inverse function (this might be your definition of inverse function, but then $f^{-1}circ f=i_A$ is not simply true by definition). It is false in general that $f^{-1}circ f=i_A$, when $f^{-1}$ is the inverse relation of $f$ (and $circ$ is composition of relations).
– Erick Wong
Nov 22 '18 at 7:49
Ahh ok, but what I am assuming for that part is that both f and f^-1 are both functions since that is what's stated in the theorem so I doubt that will be a problem. However, should I add the definition of compositional inverses in and prove both their domains would be equal?
– George
Nov 22 '18 at 8:55
Well not really showing their domains are equal but that f $circ$ g= the domain of g
– George
Nov 22 '18 at 9:00
I’m not sure that addresses my concern. The question assumes that $f$ and $f^{-1}$ are functions, but here $f^{-1}$ refers to the inverse relation, not the inverse function $f^{-1}$ (which has not been assumed to exist). The inverse relation is defined by simply reversing the ordered pairs defining $f$. Unlike the inverse function, the inverse relation has no a priori properties regarding composition. Of course this assumes the usual definitions and your source text may vary.
– Erick Wong
Nov 22 '18 at 17:32
ok, I decided to just rewrite that part using the relation definition for functions just to be safe.
– George
Nov 23 '18 at 3:13