Let $f : A rightarrow B$ be an arbitrary function. [closed]
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(a) Prove that if $f$ is a bijection (and hence invertible), then $f^{−1}(f(x))=x forall x in A$, and $f(f^{−1}(x))=x ::forall:: x in B$.
(b) Conversely, show that if there is a function $g : B rightarrow A$, satisfying $g(f(x)) = x ::forall :: x in A$, and $f(g(x)) = x :: forall ::x in B$ then $f$ is a bijection, and $f^{−1} = g$.
Apologizes for the poor formatting.
This makes sense logically, the composition of $f$ and $f^{-1}$ is always going to be the input but I am not sure how to formally prove this .
functions elementary-set-theory
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
asked Nov 21 at 18:30
smith sam
184
closed as off-topic by Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy Nov 22 at 7:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
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(a) Prove that if $f$ is a bijection (and hence invertible), then $f^{−1}(f(x))=x forall x in A$, and $f(f^{−1}(x))=x ::forall:: x in B$.
(b) Conversely, show that if there is a function $g : B rightarrow A$, satisfying $g(f(x)) = x ::forall :: x in A$, and $f(g(x)) = x :: forall ::x in B$ then $f$ is a bijection, and $f^{−1} = g$.
Apologizes for the poor formatting.
This makes sense logically, the composition of $f$ and $f^{-1}$ is always going to be the input but I am not sure how to formally prove this .
functions elementary-set-theory
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
asked Nov 21 at 18:30
smith sam
184
closed as off-topic by Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy Nov 22 at 7:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
Regarding (a), isn't that the definition of $f^{-1}$?
– Federico
Nov 21 at 18:31
To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this
– smith sam
Nov 21 at 18:34
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question.
– Yanko
Nov 21 at 18:36
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website.
– smith sam
Nov 21 at 18:41
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$.
– fleablood
Nov 21 at 18:44
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show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
(a) Prove that if $f$ is a bijection (and hence invertible), then $f^{−1}(f(x))=x forall x in A$, and $f(f^{−1}(x))=x ::forall:: x in B$.
(b) Conversely, show that if there is a function $g : B rightarrow A$, satisfying $g(f(x)) = x ::forall :: x in A$, and $f(g(x)) = x :: forall ::x in B$ then $f$ is a bijection, and $f^{−1} = g$.
Apologizes for the poor formatting.
This makes sense logically, the composition of $f$ and $f^{-1}$ is always going to be the input but I am not sure how to formally prove this .
functions elementary-set-theory
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
asked Nov 21 at 18:30
smith sam
184
(a) Prove that if $f$ is a bijection (and hence invertible), then $f^{−1}(f(x))=x forall x in A$, and $f(f^{−1}(x))=x ::forall:: x in B$.
(b) Conversely, show that if there is a function $g : B rightarrow A$, satisfying $g(f(x)) = x ::forall :: x in A$, and $f(g(x)) = x :: forall ::x in B$ then $f$ is a bijection, and $f^{−1} = g$.
Apologizes for the poor formatting.
This makes sense logically, the composition of $f$ and $f^{-1}$ is always going to be the input but I am not sure how to formally prove this .
functions elementary-set-theory
functions elementary-set-theory
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
asked Nov 21 at 18:30
smith sam
184
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
asked Nov 21 at 18:30
smith sam
184
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
edited Nov 25 at 12:46
Martin Sleziak
44.6k7115269
44.6k7115269
asked Nov 21 at 18:30
smith sam
184
asked Nov 21 at 18:30
smith sam
184
asked Nov 21 at 18:30
smith sam
184
184
closed as off-topic by Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy Nov 22 at 7:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy Nov 22 at 7:49
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Yanko, Henrik, Chinnapparaj R, user10354138, Kavi Rama Murthy
If this question can be reworded to fit the rules in the help center, please edit the question.
Regarding (a), isn't that the definition of $f^{-1}$?
– Federico
Nov 21 at 18:31
To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this
– smith sam
Nov 21 at 18:34
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question.
– Yanko
Nov 21 at 18:36
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website.
– smith sam
Nov 21 at 18:41
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$.
– fleablood
Nov 21 at 18:44
|
show 1 more comment
Regarding (a), isn't that the definition of $f^{-1}$?
– Federico
Nov 21 at 18:31
To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this
– smith sam
Nov 21 at 18:34
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question.
– Yanko
Nov 21 at 18:36
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website.
– smith sam
Nov 21 at 18:41
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$.
– fleablood
Nov 21 at 18:44
Regarding (a), isn't that the definition of $f^{-1}$?
– Federico
Nov 21 at 18:31
Regarding (a), isn't that the definition of $f^{-1}$?
– Federico
Nov 21 at 18:31
To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this
– smith sam
Nov 21 at 18:34
To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this
– smith sam
Nov 21 at 18:34
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question.
– Yanko
Nov 21 at 18:36
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question.
– Yanko
Nov 21 at 18:36
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website.
– smith sam
Nov 21 at 18:41
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website.
– smith sam
Nov 21 at 18:41
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$.
– fleablood
Nov 21 at 18:44
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$.
– fleablood
Nov 21 at 18:44
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2 Answers
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a) is almost a definition.
$f:Ato B$ is a function. That means for every $x in A$ then there is a $w_x in B$ so that $f(x) = w_x$. That's .... not anything deep. That's just what a function is. (But my notation may seem strange to you.)
$f$ is a bijection so it is invertible. That means by definition that for any $yin B$ that there is exactly one $x_y in A$ so that $f(x_y) =y$.
So for any $x in A$ there is a unique $w in B$ so that $f(x) = w$. And for $w in B$ there is a unique $z= f^{-1}(w)in A$ so that $f(z) = w = f(x)$. But if that value is unique and $f(x)$ is also equal to $w$ that means $z=f^{-1}(w) = x$.
So $f^{-1}(f(x)) = f^{-1}(w) = x$.
answered Nov 21 at 18:56
fleablood
67.4k22684
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up vote
1
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To make sense of the question, note that for $f: A to B$, $f^{-1}(P)$ represents the inverse image of a set $Psubseteq B$ under $f$. If $f(x)$ is treated as the set ${f(x)}$, then, by definition of inverse image,
$$f^{-1}({f(x)}) = { a : f(a)=f(x), a in A }$$
Since $f$ is a bijection, it is injective. Therefore,
$$f(a)=f(x)implies a=x\ implies f^{-1}({f(x)})={x}, forall x in A\$$
You can prove that $f(f^{-1}(x))=x, forall x in B$, similarly.
For the second part, observe that $g(f(x))=x, forall x in A,$ is a bijective function. $g(f(x))$ is injective $implies f$ is injective and $g(f(x))$ is surjective $implies g$ is surjective. Similarly, $f(g(x))=x, forall x in B,$ is a bijective function. $f(g(x))$ is injective $implies g$ is injective and $f(g(x))$ is surjective $implies f$ is surjective.
Thus, if such a $g$ exists, then both $f$ and $g$ are bijective. Let the inverse of $f$ be denoted by $f^{-1}$,
$$f(g(x))=x, forall x in B\
implies f^{-1}(f(g(x))=f^{-1}(x), forall x in B\
implies g(x)=f^{-1}(x), forall x in B$$
answered Nov 21 at 20:17
Shubham Johri
2,063412
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
a) is almost a definition.
$f:Ato B$ is a function. That means for every $x in A$ then there is a $w_x in B$ so that $f(x) = w_x$. That's .... not anything deep. That's just what a function is. (But my notation may seem strange to you.)
$f$ is a bijection so it is invertible. That means by definition that for any $yin B$ that there is exactly one $x_y in A$ so that $f(x_y) =y$.
So for any $x in A$ there is a unique $w in B$ so that $f(x) = w$. And for $w in B$ there is a unique $z= f^{-1}(w)in A$ so that $f(z) = w = f(x)$. But if that value is unique and $f(x)$ is also equal to $w$ that means $z=f^{-1}(w) = x$.
So $f^{-1}(f(x)) = f^{-1}(w) = x$.
answered Nov 21 at 18:56
fleablood
67.4k22684
add a comment |
up vote
1
down vote
a) is almost a definition.
$f:Ato B$ is a function. That means for every $x in A$ then there is a $w_x in B$ so that $f(x) = w_x$. That's .... not anything deep. That's just what a function is. (But my notation may seem strange to you.)
$f$ is a bijection so it is invertible. That means by definition that for any $yin B$ that there is exactly one $x_y in A$ so that $f(x_y) =y$.
So for any $x in A$ there is a unique $w in B$ so that $f(x) = w$. And for $w in B$ there is a unique $z= f^{-1}(w)in A$ so that $f(z) = w = f(x)$. But if that value is unique and $f(x)$ is also equal to $w$ that means $z=f^{-1}(w) = x$.
So $f^{-1}(f(x)) = f^{-1}(w) = x$.
answered Nov 21 at 18:56
fleablood
67.4k22684
add a comment |
up vote
1
down vote
up vote
1
down vote
a) is almost a definition.
$f:Ato B$ is a function. That means for every $x in A$ then there is a $w_x in B$ so that $f(x) = w_x$. That's .... not anything deep. That's just what a function is. (But my notation may seem strange to you.)
$f$ is a bijection so it is invertible. That means by definition that for any $yin B$ that there is exactly one $x_y in A$ so that $f(x_y) =y$.
So for any $x in A$ there is a unique $w in B$ so that $f(x) = w$. And for $w in B$ there is a unique $z= f^{-1}(w)in A$ so that $f(z) = w = f(x)$. But if that value is unique and $f(x)$ is also equal to $w$ that means $z=f^{-1}(w) = x$.
So $f^{-1}(f(x)) = f^{-1}(w) = x$.
answered Nov 21 at 18:56
fleablood
67.4k22684
a) is almost a definition.
$f:Ato B$ is a function. That means for every $x in A$ then there is a $w_x in B$ so that $f(x) = w_x$. That's .... not anything deep. That's just what a function is. (But my notation may seem strange to you.)
$f$ is a bijection so it is invertible. That means by definition that for any $yin B$ that there is exactly one $x_y in A$ so that $f(x_y) =y$.
So for any $x in A$ there is a unique $w in B$ so that $f(x) = w$. And for $w in B$ there is a unique $z= f^{-1}(w)in A$ so that $f(z) = w = f(x)$. But if that value is unique and $f(x)$ is also equal to $w$ that means $z=f^{-1}(w) = x$.
So $f^{-1}(f(x)) = f^{-1}(w) = x$.
answered Nov 21 at 18:56
fleablood
67.4k22684
answered Nov 21 at 18:56
fleablood
67.4k22684
answered Nov 21 at 18:56
fleablood
67.4k22684
answered Nov 21 at 18:56
fleablood
67.4k22684
67.4k22684
add a comment |
add a comment |
up vote
1
down vote
To make sense of the question, note that for $f: A to B$, $f^{-1}(P)$ represents the inverse image of a set $Psubseteq B$ under $f$. If $f(x)$ is treated as the set ${f(x)}$, then, by definition of inverse image,
$$f^{-1}({f(x)}) = { a : f(a)=f(x), a in A }$$
Since $f$ is a bijection, it is injective. Therefore,
$$f(a)=f(x)implies a=x\ implies f^{-1}({f(x)})={x}, forall x in A\$$
You can prove that $f(f^{-1}(x))=x, forall x in B$, similarly.
For the second part, observe that $g(f(x))=x, forall x in A,$ is a bijective function. $g(f(x))$ is injective $implies f$ is injective and $g(f(x))$ is surjective $implies g$ is surjective. Similarly, $f(g(x))=x, forall x in B,$ is a bijective function. $f(g(x))$ is injective $implies g$ is injective and $f(g(x))$ is surjective $implies f$ is surjective.
Thus, if such a $g$ exists, then both $f$ and $g$ are bijective. Let the inverse of $f$ be denoted by $f^{-1}$,
$$f(g(x))=x, forall x in B\
implies f^{-1}(f(g(x))=f^{-1}(x), forall x in B\
implies g(x)=f^{-1}(x), forall x in B$$
answered Nov 21 at 20:17
Shubham Johri
2,063412
add a comment |
up vote
1
down vote
To make sense of the question, note that for $f: A to B$, $f^{-1}(P)$ represents the inverse image of a set $Psubseteq B$ under $f$. If $f(x)$ is treated as the set ${f(x)}$, then, by definition of inverse image,
$$f^{-1}({f(x)}) = { a : f(a)=f(x), a in A }$$
Since $f$ is a bijection, it is injective. Therefore,
$$f(a)=f(x)implies a=x\ implies f^{-1}({f(x)})={x}, forall x in A\$$
You can prove that $f(f^{-1}(x))=x, forall x in B$, similarly.
For the second part, observe that $g(f(x))=x, forall x in A,$ is a bijective function. $g(f(x))$ is injective $implies f$ is injective and $g(f(x))$ is surjective $implies g$ is surjective. Similarly, $f(g(x))=x, forall x in B,$ is a bijective function. $f(g(x))$ is injective $implies g$ is injective and $f(g(x))$ is surjective $implies f$ is surjective.
Thus, if such a $g$ exists, then both $f$ and $g$ are bijective. Let the inverse of $f$ be denoted by $f^{-1}$,
$$f(g(x))=x, forall x in B\
implies f^{-1}(f(g(x))=f^{-1}(x), forall x in B\
implies g(x)=f^{-1}(x), forall x in B$$
answered Nov 21 at 20:17
Shubham Johri
2,063412
add a comment |
up vote
1
down vote
up vote
1
down vote
To make sense of the question, note that for $f: A to B$, $f^{-1}(P)$ represents the inverse image of a set $Psubseteq B$ under $f$. If $f(x)$ is treated as the set ${f(x)}$, then, by definition of inverse image,
$$f^{-1}({f(x)}) = { a : f(a)=f(x), a in A }$$
Since $f$ is a bijection, it is injective. Therefore,
$$f(a)=f(x)implies a=x\ implies f^{-1}({f(x)})={x}, forall x in A\$$
You can prove that $f(f^{-1}(x))=x, forall x in B$, similarly.
For the second part, observe that $g(f(x))=x, forall x in A,$ is a bijective function. $g(f(x))$ is injective $implies f$ is injective and $g(f(x))$ is surjective $implies g$ is surjective. Similarly, $f(g(x))=x, forall x in B,$ is a bijective function. $f(g(x))$ is injective $implies g$ is injective and $f(g(x))$ is surjective $implies f$ is surjective.
Thus, if such a $g$ exists, then both $f$ and $g$ are bijective. Let the inverse of $f$ be denoted by $f^{-1}$,
$$f(g(x))=x, forall x in B\
implies f^{-1}(f(g(x))=f^{-1}(x), forall x in B\
implies g(x)=f^{-1}(x), forall x in B$$
answered Nov 21 at 20:17
Shubham Johri
2,063412
To make sense of the question, note that for $f: A to B$, $f^{-1}(P)$ represents the inverse image of a set $Psubseteq B$ under $f$. If $f(x)$ is treated as the set ${f(x)}$, then, by definition of inverse image,
$$f^{-1}({f(x)}) = { a : f(a)=f(x), a in A }$$
Since $f$ is a bijection, it is injective. Therefore,
$$f(a)=f(x)implies a=x\ implies f^{-1}({f(x)})={x}, forall x in A\$$
You can prove that $f(f^{-1}(x))=x, forall x in B$, similarly.
For the second part, observe that $g(f(x))=x, forall x in A,$ is a bijective function. $g(f(x))$ is injective $implies f$ is injective and $g(f(x))$ is surjective $implies g$ is surjective. Similarly, $f(g(x))=x, forall x in B,$ is a bijective function. $f(g(x))$ is injective $implies g$ is injective and $f(g(x))$ is surjective $implies f$ is surjective.
Thus, if such a $g$ exists, then both $f$ and $g$ are bijective. Let the inverse of $f$ be denoted by $f^{-1}$,
$$f(g(x))=x, forall x in B\
implies f^{-1}(f(g(x))=f^{-1}(x), forall x in B\
implies g(x)=f^{-1}(x), forall x in B$$
answered Nov 21 at 20:17
Shubham Johri
2,063412
edited Nov 23 at 11:04
answered Nov 21 at 20:17
Shubham Johri
2,063412
answered Nov 21 at 20:17
Shubham Johri
2,063412
answered Nov 21 at 20:17
Shubham Johri
2,063412
2,063412
add a comment |
add a comment |
Regarding (a), isn't that the definition of $f^{-1}$?
– Federico
Nov 21 at 18:31
To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this
– smith sam
Nov 21 at 18:34
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question.
– Yanko
Nov 21 at 18:36
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website.
– smith sam
Nov 21 at 18:41
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$.
– fleablood
Nov 21 at 18:44