Proving that a function composed with its inverse is the identity
$begingroup$
Let $f:X rightarrow Y$ be a bijection with inverse function $f^{-1}: Y rightarrow X.$ Show that $f^{-1}(f)$ and $f(f^{-1})$ are the identity functions on X and Y respectively.
My approach for the first part of the problem was to try to show that for $x in X$, we need to show that $f^{-1}(f(x)) = x$ I feel like I can use the fact that f is one to one here somehow, but i'm not sure how to get started.
functions elementary-set-theory
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
asked Sep 5 '16 at 3:17
12332111233211
317111
$endgroup$
add a comment |
$begingroup$
Let $f:X rightarrow Y$ be a bijection with inverse function $f^{-1}: Y rightarrow X.$ Show that $f^{-1}(f)$ and $f(f^{-1})$ are the identity functions on X and Y respectively.
My approach for the first part of the problem was to try to show that for $x in X$, we need to show that $f^{-1}(f(x)) = x$ I feel like I can use the fact that f is one to one here somehow, but i'm not sure how to get started.
functions elementary-set-theory
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
asked Sep 5 '16 at 3:17
12332111233211
317111
$endgroup$
$begingroup$
This is half of the definition of inverse function. The inverse is the function which gives the identity when composed in either side.
$endgroup$
– Tac-Tics
Sep 5 '16 at 3:29
2
$begingroup$
What's the definition of a function's inverse if not this?
$endgroup$
– fleablood
Sep 5 '16 at 3:55
$begingroup$
Write $f^{-1}circ f$ instead of $f^{-1}(f)$, because $f$ is not a point of $Y$.
$endgroup$
– Christian Blatter
Sep 5 '16 at 12:32
add a comment |
$begingroup$
Let $f:X rightarrow Y$ be a bijection with inverse function $f^{-1}: Y rightarrow X.$ Show that $f^{-1}(f)$ and $f(f^{-1})$ are the identity functions on X and Y respectively.
My approach for the first part of the problem was to try to show that for $x in X$, we need to show that $f^{-1}(f(x)) = x$ I feel like I can use the fact that f is one to one here somehow, but i'm not sure how to get started.
functions elementary-set-theory
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
asked Sep 5 '16 at 3:17
12332111233211
317111
$endgroup$
Let $f:X rightarrow Y$ be a bijection with inverse function $f^{-1}: Y rightarrow X.$ Show that $f^{-1}(f)$ and $f(f^{-1})$ are the identity functions on X and Y respectively.
My approach for the first part of the problem was to try to show that for $x in X$, we need to show that $f^{-1}(f(x)) = x$ I feel like I can use the fact that f is one to one here somehow, but i'm not sure how to get started.
functions elementary-set-theory
functions elementary-set-theory
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
asked Sep 5 '16 at 3:17
12332111233211
317111
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
asked Sep 5 '16 at 3:17
12332111233211
317111
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
edited Sep 5 '16 at 3:33
Andrés E. Caicedo
65k8158246
65k8158246
asked Sep 5 '16 at 3:17
12332111233211
317111
asked Sep 5 '16 at 3:17
12332111233211
317111
asked Sep 5 '16 at 3:17
12332111233211
317111
317111
$begingroup$
This is half of the definition of inverse function. The inverse is the function which gives the identity when composed in either side.
$endgroup$
– Tac-Tics
Sep 5 '16 at 3:29
2
$begingroup$
What's the definition of a function's inverse if not this?
$endgroup$
– fleablood
Sep 5 '16 at 3:55
$begingroup$
Write $f^{-1}circ f$ instead of $f^{-1}(f)$, because $f$ is not a point of $Y$.
$endgroup$
– Christian Blatter
Sep 5 '16 at 12:32
add a comment |
$begingroup$
This is half of the definition of inverse function. The inverse is the function which gives the identity when composed in either side.
$endgroup$
– Tac-Tics
Sep 5 '16 at 3:29
2
$begingroup$
What's the definition of a function's inverse if not this?
$endgroup$
– fleablood
Sep 5 '16 at 3:55
$begingroup$
Write $f^{-1}circ f$ instead of $f^{-1}(f)$, because $f$ is not a point of $Y$.
$endgroup$
– Christian Blatter
Sep 5 '16 at 12:32
$begingroup$
This is half of the definition of inverse function. The inverse is the function which gives the identity when composed in either side.
$endgroup$
– Tac-Tics
Sep 5 '16 at 3:29
$begingroup$
This is half of the definition of inverse function. The inverse is the function which gives the identity when composed in either side.
$endgroup$
– Tac-Tics
Sep 5 '16 at 3:29
2
2
$begingroup$
What's the definition of a function's inverse if not this?
$endgroup$
– fleablood
Sep 5 '16 at 3:55
$begingroup$
What's the definition of a function's inverse if not this?
$endgroup$
– fleablood
Sep 5 '16 at 3:55
$begingroup$
Write $f^{-1}circ f$ instead of $f^{-1}(f)$, because $f$ is not a point of $Y$.
$endgroup$
– Christian Blatter
Sep 5 '16 at 12:32
$begingroup$
Write $f^{-1}circ f$ instead of $f^{-1}(f)$, because $f$ is not a point of $Y$.
$endgroup$
– Christian Blatter
Sep 5 '16 at 12:32
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
To show that $f^{-1}circ f$ is the identity on $X$, let $x in X$. Since $f$ is a bijection, $f$ is an injection, so $f(x)$ is the unique element in $Y$ such that $f^{-1}(f(x)) = x$. Hence $f^{-1}circ f = text{id}_X$. The other direction is similar. Can you see how to show $fcirc f^{-1} = text{id}_Y$?
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
To show that $f^{-1}circ f$ is the identity on $X$, let $x in X$. Since $f$ is a bijection, $f$ is an injection, so $f(x)$ is the unique element in $Y$ such that $f^{-1}(f(x)) = x$. Hence $f^{-1}circ f = text{id}_X$. The other direction is similar. Can you see how to show $fcirc f^{-1} = text{id}_Y$?
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
$endgroup$
add a comment |
$begingroup$
To show that $f^{-1}circ f$ is the identity on $X$, let $x in X$. Since $f$ is a bijection, $f$ is an injection, so $f(x)$ is the unique element in $Y$ such that $f^{-1}(f(x)) = x$. Hence $f^{-1}circ f = text{id}_X$. The other direction is similar. Can you see how to show $fcirc f^{-1} = text{id}_Y$?
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
$endgroup$
add a comment |
$begingroup$
To show that $f^{-1}circ f$ is the identity on $X$, let $x in X$. Since $f$ is a bijection, $f$ is an injection, so $f(x)$ is the unique element in $Y$ such that $f^{-1}(f(x)) = x$. Hence $f^{-1}circ f = text{id}_X$. The other direction is similar. Can you see how to show $fcirc f^{-1} = text{id}_Y$?
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
$endgroup$
To show that $f^{-1}circ f$ is the identity on $X$, let $x in X$. Since $f$ is a bijection, $f$ is an injection, so $f(x)$ is the unique element in $Y$ such that $f^{-1}(f(x)) = x$. Hence $f^{-1}circ f = text{id}_X$. The other direction is similar. Can you see how to show $fcirc f^{-1} = text{id}_Y$?
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
answered Sep 5 '16 at 3:25
AOrtizAOrtiz
10.5k21341
10.5k21341
add a comment |
add a comment |
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$begingroup$
This is half of the definition of inverse function. The inverse is the function which gives the identity when composed in either side.
$endgroup$
– Tac-Tics
Sep 5 '16 at 3:29
2
$begingroup$
What's the definition of a function's inverse if not this?
$endgroup$
– fleablood
Sep 5 '16 at 3:55
$begingroup$
Write $f^{-1}circ f$ instead of $f^{-1}(f)$, because $f$ is not a point of $Y$.
$endgroup$
– Christian Blatter
Sep 5 '16 at 12:32