Given a function a to b, its inverse relation will be a function iff the function is bijective
$begingroup$
So when I did this proof I didn't know I was supposed to physical prove all the parts out (assuming they are both functions and then using that to prove the inverse is an injection). My problem is I have never seen a relation proof using a function and an inverse function (only have done compositions so far). Can somebody explain how I format one part of this proof so I can get an understanding of how I use the definitions with an inverse function?
proof-explanation
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
$endgroup$
add a comment |
$begingroup$
So when I did this proof I didn't know I was supposed to physical prove all the parts out (assuming they are both functions and then using that to prove the inverse is an injection). My problem is I have never seen a relation proof using a function and an inverse function (only have done compositions so far). Can somebody explain how I format one part of this proof so I can get an understanding of how I use the definitions with an inverse function?
proof-explanation
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
$endgroup$
$begingroup$
Hint: If the function $f:Arightarrow B$ is bijective, for any $bin B$ there exists (from surjectiveness) a unique (from injectiveness) $ain A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$?
$endgroup$
– NL1992
Dec 3 '18 at 4:44
add a comment |
$begingroup$
So when I did this proof I didn't know I was supposed to physical prove all the parts out (assuming they are both functions and then using that to prove the inverse is an injection). My problem is I have never seen a relation proof using a function and an inverse function (only have done compositions so far). Can somebody explain how I format one part of this proof so I can get an understanding of how I use the definitions with an inverse function?
proof-explanation
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
$endgroup$
So when I did this proof I didn't know I was supposed to physical prove all the parts out (assuming they are both functions and then using that to prove the inverse is an injection). My problem is I have never seen a relation proof using a function and an inverse function (only have done compositions so far). Can somebody explain how I format one part of this proof so I can get an understanding of how I use the definitions with an inverse function?
proof-explanation
proof-explanation
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
asked Dec 3 '18 at 4:40
GeorgeGeorge
676
676
$begingroup$
Hint: If the function $f:Arightarrow B$ is bijective, for any $bin B$ there exists (from surjectiveness) a unique (from injectiveness) $ain A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$?
$endgroup$
– NL1992
Dec 3 '18 at 4:44
add a comment |
$begingroup$
Hint: If the function $f:Arightarrow B$ is bijective, for any $bin B$ there exists (from surjectiveness) a unique (from injectiveness) $ain A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$?
$endgroup$
– NL1992
Dec 3 '18 at 4:44
$begingroup$
Hint: If the function $f:Arightarrow B$ is bijective, for any $bin B$ there exists (from surjectiveness) a unique (from injectiveness) $ain A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$?
$endgroup$
– NL1992
Dec 3 '18 at 4:44
$begingroup$
Hint: If the function $f:Arightarrow B$ is bijective, for any $bin B$ there exists (from surjectiveness) a unique (from injectiveness) $ain A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$?
$endgroup$
– NL1992
Dec 3 '18 at 4:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that a relation between sets $X$ and $Y$ is simply a subset $Rsubseteq Xtimes Y$. A function $f:Xto Y$ is a special relation $R_fsubseteq Xtimes Y$ such that for every $xin X$, there is a unique $yin Y$ for which $(x,y)in R_f$ (this is precisely what we mean by $f(x)=y$). For a relation $Rsubseteq Xtimes Y$, the inverse relation $overline{R}subseteq Ytimes X$ is defined by $(y,x)inoverline{R}$ iff $(x,y)in R$.
So all that is being asked is to show that for a function $f:Xto Y$ (identified with its relation $R_fsubseteq Xtimes Y$) is a bijection iff $overline{R_f}subseteq Ytimes X$ is a function (or if you prefer, is the relation associated with some function). Now just go forth and verify definitions!
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
$endgroup$
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
add a comment |
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$begingroup$
Recall that a relation between sets $X$ and $Y$ is simply a subset $Rsubseteq Xtimes Y$. A function $f:Xto Y$ is a special relation $R_fsubseteq Xtimes Y$ such that for every $xin X$, there is a unique $yin Y$ for which $(x,y)in R_f$ (this is precisely what we mean by $f(x)=y$). For a relation $Rsubseteq Xtimes Y$, the inverse relation $overline{R}subseteq Ytimes X$ is defined by $(y,x)inoverline{R}$ iff $(x,y)in R$.
So all that is being asked is to show that for a function $f:Xto Y$ (identified with its relation $R_fsubseteq Xtimes Y$) is a bijection iff $overline{R_f}subseteq Ytimes X$ is a function (or if you prefer, is the relation associated with some function). Now just go forth and verify definitions!
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
$endgroup$
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
add a comment |
$begingroup$
Recall that a relation between sets $X$ and $Y$ is simply a subset $Rsubseteq Xtimes Y$. A function $f:Xto Y$ is a special relation $R_fsubseteq Xtimes Y$ such that for every $xin X$, there is a unique $yin Y$ for which $(x,y)in R_f$ (this is precisely what we mean by $f(x)=y$). For a relation $Rsubseteq Xtimes Y$, the inverse relation $overline{R}subseteq Ytimes X$ is defined by $(y,x)inoverline{R}$ iff $(x,y)in R$.
So all that is being asked is to show that for a function $f:Xto Y$ (identified with its relation $R_fsubseteq Xtimes Y$) is a bijection iff $overline{R_f}subseteq Ytimes X$ is a function (or if you prefer, is the relation associated with some function). Now just go forth and verify definitions!
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
$endgroup$
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
add a comment |
$begingroup$
Recall that a relation between sets $X$ and $Y$ is simply a subset $Rsubseteq Xtimes Y$. A function $f:Xto Y$ is a special relation $R_fsubseteq Xtimes Y$ such that for every $xin X$, there is a unique $yin Y$ for which $(x,y)in R_f$ (this is precisely what we mean by $f(x)=y$). For a relation $Rsubseteq Xtimes Y$, the inverse relation $overline{R}subseteq Ytimes X$ is defined by $(y,x)inoverline{R}$ iff $(x,y)in R$.
So all that is being asked is to show that for a function $f:Xto Y$ (identified with its relation $R_fsubseteq Xtimes Y$) is a bijection iff $overline{R_f}subseteq Ytimes X$ is a function (or if you prefer, is the relation associated with some function). Now just go forth and verify definitions!
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
$endgroup$
Recall that a relation between sets $X$ and $Y$ is simply a subset $Rsubseteq Xtimes Y$. A function $f:Xto Y$ is a special relation $R_fsubseteq Xtimes Y$ such that for every $xin X$, there is a unique $yin Y$ for which $(x,y)in R_f$ (this is precisely what we mean by $f(x)=y$). For a relation $Rsubseteq Xtimes Y$, the inverse relation $overline{R}subseteq Ytimes X$ is defined by $(y,x)inoverline{R}$ iff $(x,y)in R$.
So all that is being asked is to show that for a function $f:Xto Y$ (identified with its relation $R_fsubseteq Xtimes Y$) is a bijection iff $overline{R_f}subseteq Ytimes X$ is a function (or if you prefer, is the relation associated with some function). Now just go forth and verify definitions!
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
answered Dec 3 '18 at 4:47
munchhausenmunchhausen
79416
79416
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
add a comment |
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)in f$ for some $xin A$ and fix $y_1,y_2in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)in f^{-1}$ for some $yin A$. Since $(x_1,y),(x_2,y)in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection.
$endgroup$
– George
Dec 3 '18 at 5:04
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
here what that definition states----- A relation $Rsubseteq A X B$ is injective if $(x_1,y) in R$ and $(x_2,y) in R$ implies $x_1=x_2$
$endgroup$
– George
Dec 3 '18 at 5:11
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
$begingroup$
So, you're getting things flipped around. $(x,y)in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$.
$endgroup$
– munchhausen
Dec 3 '18 at 6:12
add a comment |
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$begingroup$
Hint: If the function $f:Arightarrow B$ is bijective, for any $bin B$ there exists (from surjectiveness) a unique (from injectiveness) $ain A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$?
$endgroup$
– NL1992
Dec 3 '18 at 4:44