Defining a bijective function that inputs and outputs a function The 2019 Stack Overflow Developer Survey Results Are InProve that the function $Phi :mathcalF(X,Y)longrightarrow Y$, is not injective.Bijective functionIs this a function and injective/surjective questionDefining a bijective function from $2mathbbN$ to $3mathbbZ-1$?show that $f(x)=-3x+4$ is bijectiveProve that if $f:Ato B$ is bijective then $f^-1:Bto A$ is bijective.Quick Clarification: Definition of Bijective FunctionProve that $exists k in mathbbN$ such that $kleq m$. And such that exists Bijective function $g:A rightarrow mathbbN^<k$.show that f is bijective functionProving a multi variable function bijective
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Defining a bijective function that inputs and outputs a function
The 2019 Stack Overflow Developer Survey Results Are InProve that the function $Phi :mathcalF(X,Y)longrightarrow Y$, is not injective.Bijective functionIs this a function and injective/surjective questionDefining a bijective function from $2mathbbN$ to $3mathbbZ-1$?show that $f(x)=-3x+4$ is bijectiveProve that if $f:Ato B$ is bijective then $f^-1:Bto A$ is bijective.Quick Clarification: Definition of Bijective FunctionProve that $exists k in mathbbN$ such that $kleq m$. And such that exists Bijective function $g:A rightarrow mathbbN^<k$.show that f is bijective functionProving a multi variable function bijective
$begingroup$
Given $f:A to B$ and $g: C to D$ bijective.
I want to define a bijective function $F: C^A to D^B$ where $C^A= h: A to C$.
My attempt: We know the function $F$ takes in a function that goes from $A to C$.
So I tried defining it by $F(phi)=g(phi(f^-1(t)))$ and this is indeed a function that goes from $B to D$.
Then I tried to show that it's injective & surjective.
Because I don't know anything about $phi$ in particular I do not know whether if it is injective or surjective I am doubting my approach. Thanks for any help in advance.
functions
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 2:39
javacoderjavacoder
848
$endgroup$
|
show 1 more comment
$begingroup$
Given $f:A to B$ and $g: C to D$ bijective.
I want to define a bijective function $F: C^A to D^B$ where $C^A= h: A to C$.
My attempt: We know the function $F$ takes in a function that goes from $A to C$.
So I tried defining it by $F(phi)=g(phi(f^-1(t)))$ and this is indeed a function that goes from $B to D$.
Then I tried to show that it's injective & surjective.
Because I don't know anything about $phi$ in particular I do not know whether if it is injective or surjective I am doubting my approach. Thanks for any help in advance.
functions
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 2:39
javacoderjavacoder
848
$endgroup$
$begingroup$
$phi$ will often be neither injective or surjective but you don't care. It's $F$ you want to show is bijective. Try to write down its inverse - just think of how you would have tackled the problem if you were told the inverses of $f$ and $g$ instead of $f$ and $g$.
$endgroup$
– Ethan Bolker
Mar 23 at 2:43
$begingroup$
Hey, thanks for the response. Even if I were told the inverse of $f$ and $g$ I believe that it does not help me or I can't see it. Like the sets $A,B,C,D$ are 4 separate sets not related to each other in any way. The inverses would just go $B to A$ and $D to C$ and $phi$ can't be replaced by any of the inverses because $phi$ goes from $A$ to $C$ and there is no functions that takes it from $A$ to $C$ at least from the given functions... I would appreciate a little more hint! Thanks. I do not see a way to compose a function that takes it from $A$ to $C$..
$endgroup$
– javacoder
Mar 23 at 2:50
$begingroup$
Anyone...? It seems the guy above doesn't know either.
$endgroup$
– javacoder
Mar 23 at 3:48
$begingroup$
Ethan Bolker did know, and told you, but you didn't understand him. You don't need to know anything about $phi$ other than it is a function from $A to C$. It is $F$ itself you need to prove is bijective, not $phi$ or $F(phi)$. Having defined $F : C^A to D^B$, try now defining $G: D^B to C^A$ in the same manner. How are $F$ and $G$ related?
$endgroup$
– Paul Sinclair
Mar 23 at 14:18
$begingroup$
Thank you. Defining $G$ in the same manner as in the way I did it or in some other way? And if you are referring to the same manner as in "my way" then I have done so. But still does not ring a bell on how I can show that $F$ is bijective. $F$ and $G$ are related such that when you compose $F$ with $G$ or vice versa you'll end up with the identity function is that what you are referring to?
$endgroup$
– javacoder
Mar 23 at 21:51
|
show 1 more comment
$begingroup$
Given $f:A to B$ and $g: C to D$ bijective.
I want to define a bijective function $F: C^A to D^B$ where $C^A= h: A to C$.
My attempt: We know the function $F$ takes in a function that goes from $A to C$.
So I tried defining it by $F(phi)=g(phi(f^-1(t)))$ and this is indeed a function that goes from $B to D$.
Then I tried to show that it's injective & surjective.
Because I don't know anything about $phi$ in particular I do not know whether if it is injective or surjective I am doubting my approach. Thanks for any help in advance.
functions
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 2:39
javacoderjavacoder
848
$endgroup$
Given $f:A to B$ and $g: C to D$ bijective.
I want to define a bijective function $F: C^A to D^B$ where $C^A= h: A to C$.
My attempt: We know the function $F$ takes in a function that goes from $A to C$.
So I tried defining it by $F(phi)=g(phi(f^-1(t)))$ and this is indeed a function that goes from $B to D$.
Then I tried to show that it's injective & surjective.
Because I don't know anything about $phi$ in particular I do not know whether if it is injective or surjective I am doubting my approach. Thanks for any help in advance.
functions
functions
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 2:39
javacoderjavacoder
848
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
asked Mar 23 at 2:39
javacoderjavacoder
848
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
edited Mar 23 at 3:09
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Mar 23 at 2:39
javacoderjavacoder
848
asked Mar 23 at 2:39
javacoderjavacoder
848
asked Mar 23 at 2:39
javacoderjavacoder
848
848
$begingroup$
$phi$ will often be neither injective or surjective but you don't care. It's $F$ you want to show is bijective. Try to write down its inverse - just think of how you would have tackled the problem if you were told the inverses of $f$ and $g$ instead of $f$ and $g$.
$endgroup$
– Ethan Bolker
Mar 23 at 2:43
$begingroup$
Hey, thanks for the response. Even if I were told the inverse of $f$ and $g$ I believe that it does not help me or I can't see it. Like the sets $A,B,C,D$ are 4 separate sets not related to each other in any way. The inverses would just go $B to A$ and $D to C$ and $phi$ can't be replaced by any of the inverses because $phi$ goes from $A$ to $C$ and there is no functions that takes it from $A$ to $C$ at least from the given functions... I would appreciate a little more hint! Thanks. I do not see a way to compose a function that takes it from $A$ to $C$..
$endgroup$
– javacoder
Mar 23 at 2:50
$begingroup$
Anyone...? It seems the guy above doesn't know either.
$endgroup$
– javacoder
Mar 23 at 3:48
$begingroup$
Ethan Bolker did know, and told you, but you didn't understand him. You don't need to know anything about $phi$ other than it is a function from $A to C$. It is $F$ itself you need to prove is bijective, not $phi$ or $F(phi)$. Having defined $F : C^A to D^B$, try now defining $G: D^B to C^A$ in the same manner. How are $F$ and $G$ related?
$endgroup$
– Paul Sinclair
Mar 23 at 14:18
$begingroup$
Thank you. Defining $G$ in the same manner as in the way I did it or in some other way? And if you are referring to the same manner as in "my way" then I have done so. But still does not ring a bell on how I can show that $F$ is bijective. $F$ and $G$ are related such that when you compose $F$ with $G$ or vice versa you'll end up with the identity function is that what you are referring to?
$endgroup$
– javacoder
Mar 23 at 21:51
|
show 1 more comment
$begingroup$
$phi$ will often be neither injective or surjective but you don't care. It's $F$ you want to show is bijective. Try to write down its inverse - just think of how you would have tackled the problem if you were told the inverses of $f$ and $g$ instead of $f$ and $g$.
$endgroup$
– Ethan Bolker
Mar 23 at 2:43
$begingroup$
Hey, thanks for the response. Even if I were told the inverse of $f$ and $g$ I believe that it does not help me or I can't see it. Like the sets $A,B,C,D$ are 4 separate sets not related to each other in any way. The inverses would just go $B to A$ and $D to C$ and $phi$ can't be replaced by any of the inverses because $phi$ goes from $A$ to $C$ and there is no functions that takes it from $A$ to $C$ at least from the given functions... I would appreciate a little more hint! Thanks. I do not see a way to compose a function that takes it from $A$ to $C$..
$endgroup$
– javacoder
Mar 23 at 2:50
$begingroup$
Anyone...? It seems the guy above doesn't know either.
$endgroup$
– javacoder
Mar 23 at 3:48
$begingroup$
Ethan Bolker did know, and told you, but you didn't understand him. You don't need to know anything about $phi$ other than it is a function from $A to C$. It is $F$ itself you need to prove is bijective, not $phi$ or $F(phi)$. Having defined $F : C^A to D^B$, try now defining $G: D^B to C^A$ in the same manner. How are $F$ and $G$ related?
$endgroup$
– Paul Sinclair
Mar 23 at 14:18
$begingroup$
Thank you. Defining $G$ in the same manner as in the way I did it or in some other way? And if you are referring to the same manner as in "my way" then I have done so. But still does not ring a bell on how I can show that $F$ is bijective. $F$ and $G$ are related such that when you compose $F$ with $G$ or vice versa you'll end up with the identity function is that what you are referring to?
$endgroup$
– javacoder
Mar 23 at 21:51
$begingroup$
$phi$ will often be neither injective or surjective but you don't care. It's $F$ you want to show is bijective. Try to write down its inverse - just think of how you would have tackled the problem if you were told the inverses of $f$ and $g$ instead of $f$ and $g$.
$endgroup$
– Ethan Bolker
Mar 23 at 2:43
$begingroup$
$phi$ will often be neither injective or surjective but you don't care. It's $F$ you want to show is bijective. Try to write down its inverse - just think of how you would have tackled the problem if you were told the inverses of $f$ and $g$ instead of $f$ and $g$.
$endgroup$
– Ethan Bolker
Mar 23 at 2:43
$begingroup$
Hey, thanks for the response. Even if I were told the inverse of $f$ and $g$ I believe that it does not help me or I can't see it. Like the sets $A,B,C,D$ are 4 separate sets not related to each other in any way. The inverses would just go $B to A$ and $D to C$ and $phi$ can't be replaced by any of the inverses because $phi$ goes from $A$ to $C$ and there is no functions that takes it from $A$ to $C$ at least from the given functions... I would appreciate a little more hint! Thanks. I do not see a way to compose a function that takes it from $A$ to $C$..
$endgroup$
– javacoder
Mar 23 at 2:50
$begingroup$
Hey, thanks for the response. Even if I were told the inverse of $f$ and $g$ I believe that it does not help me or I can't see it. Like the sets $A,B,C,D$ are 4 separate sets not related to each other in any way. The inverses would just go $B to A$ and $D to C$ and $phi$ can't be replaced by any of the inverses because $phi$ goes from $A$ to $C$ and there is no functions that takes it from $A$ to $C$ at least from the given functions... I would appreciate a little more hint! Thanks. I do not see a way to compose a function that takes it from $A$ to $C$..
$endgroup$
– javacoder
Mar 23 at 2:50
$begingroup$
Anyone...? It seems the guy above doesn't know either.
$endgroup$
– javacoder
Mar 23 at 3:48
$begingroup$
Anyone...? It seems the guy above doesn't know either.
$endgroup$
– javacoder
Mar 23 at 3:48
$begingroup$
Ethan Bolker did know, and told you, but you didn't understand him. You don't need to know anything about $phi$ other than it is a function from $A to C$. It is $F$ itself you need to prove is bijective, not $phi$ or $F(phi)$. Having defined $F : C^A to D^B$, try now defining $G: D^B to C^A$ in the same manner. How are $F$ and $G$ related?
$endgroup$
– Paul Sinclair
Mar 23 at 14:18
$begingroup$
Ethan Bolker did know, and told you, but you didn't understand him. You don't need to know anything about $phi$ other than it is a function from $A to C$. It is $F$ itself you need to prove is bijective, not $phi$ or $F(phi)$. Having defined $F : C^A to D^B$, try now defining $G: D^B to C^A$ in the same manner. How are $F$ and $G$ related?
$endgroup$
– Paul Sinclair
Mar 23 at 14:18
$begingroup$
Thank you. Defining $G$ in the same manner as in the way I did it or in some other way? And if you are referring to the same manner as in "my way" then I have done so. But still does not ring a bell on how I can show that $F$ is bijective. $F$ and $G$ are related such that when you compose $F$ with $G$ or vice versa you'll end up with the identity function is that what you are referring to?
$endgroup$
– javacoder
Mar 23 at 21:51
$begingroup$
Thank you. Defining $G$ in the same manner as in the way I did it or in some other way? And if you are referring to the same manner as in "my way" then I have done so. But still does not ring a bell on how I can show that $F$ is bijective. $F$ and $G$ are related such that when you compose $F$ with $G$ or vice versa you'll end up with the identity function is that what you are referring to?
$endgroup$
– javacoder
Mar 23 at 21:51
|
show 1 more comment
0
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$begingroup$
$phi$ will often be neither injective or surjective but you don't care. It's $F$ you want to show is bijective. Try to write down its inverse - just think of how you would have tackled the problem if you were told the inverses of $f$ and $g$ instead of $f$ and $g$.
$endgroup$
– Ethan Bolker
Mar 23 at 2:43
$begingroup$
Hey, thanks for the response. Even if I were told the inverse of $f$ and $g$ I believe that it does not help me or I can't see it. Like the sets $A,B,C,D$ are 4 separate sets not related to each other in any way. The inverses would just go $B to A$ and $D to C$ and $phi$ can't be replaced by any of the inverses because $phi$ goes from $A$ to $C$ and there is no functions that takes it from $A$ to $C$ at least from the given functions... I would appreciate a little more hint! Thanks. I do not see a way to compose a function that takes it from $A$ to $C$..
$endgroup$
– javacoder
Mar 23 at 2:50
$begingroup$
Anyone...? It seems the guy above doesn't know either.
$endgroup$
– javacoder
Mar 23 at 3:48
$begingroup$
Ethan Bolker did know, and told you, but you didn't understand him. You don't need to know anything about $phi$ other than it is a function from $A to C$. It is $F$ itself you need to prove is bijective, not $phi$ or $F(phi)$. Having defined $F : C^A to D^B$, try now defining $G: D^B to C^A$ in the same manner. How are $F$ and $G$ related?
$endgroup$
– Paul Sinclair
Mar 23 at 14:18
$begingroup$
Thank you. Defining $G$ in the same manner as in the way I did it or in some other way? And if you are referring to the same manner as in "my way" then I have done so. But still does not ring a bell on how I can show that $F$ is bijective. $F$ and $G$ are related such that when you compose $F$ with $G$ or vice versa you'll end up with the identity function is that what you are referring to?
$endgroup$
– javacoder
Mar 23 at 21:51