Bijective proof check The Next CEO of Stack Overflow(Revisited$_2$) Injectivity Relies on The Existence of an Onto Function Mapping Back to Its PreimageProof a function is continous.Examine if function $f:mathbbR^2rightarrow mathbbR^2$ which is defined as $f(x,y)=(2x-y,x-4y)$ is bijective. If bijective, find $f^-1$.Is this function bijective, surjective and injective?Injective and Surjective in composite functionsProve distance function is bijectiveBijective compositions implies each function is bijective.Given $f$ and $g$, prove/disprove whether $g circ f$ is injective/surjectiveLet $X$ a infinite set and $Y$ a finite set. Prove exists a function $f:Xrightarrow Y$ surjective and a function $g:Yrightarrow X$ injective.Let $f:Xrightarrow Y$ and $g:Yrightarrow X$ satisfice $gcirc f=1_x$. Prove $f$ is injective and $g$ is surjective.
Is there a way to save my career from absolute disaster?
How do I fit a non linear curve?
Is fine stranded wire ok for main supply line?
What is the process for purifying your home if you believe it may have been previously used for pagan worship?
Is there such a thing as a proper verb, like a proper noun?
Is it correct to say moon starry nights?
Getting Stale Gas Out of a Gas Tank w/out Dropping the Tank
Calculate the Mean mean of two numbers
Won the lottery - how do I keep the money?
Yu-Gi-Oh cards in Python 3
Can someone explain this formula for calculating Manhattan distance?
Easy to read palindrome checker
Is it convenient to ask the journal's editor for two additional days to complete a review?
What day is it again?
It is correct to match light sources with the same color temperature?
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Graph of the history of databases
what's the use of '% to gdp' type of variables?
Film where the government was corrupt with aliens, people sent to kill aliens are given rigged visors not showing the right aliens
Defamation due to breach of confidentiality
Scary film where a woman has vaginal teeth
Aggressive Under-Indexing and no data for missing index
TikZ: How to fill area with a special pattern?
From jafe to El-Guest
Bijective proof check
The Next CEO of Stack Overflow(Revisited$_2$) Injectivity Relies on The Existence of an Onto Function Mapping Back to Its PreimageProof a function is continous.Examine if function $f:mathbbR^2rightarrow mathbbR^2$ which is defined as $f(x,y)=(2x-y,x-4y)$ is bijective. If bijective, find $f^-1$.Is this function bijective, surjective and injective?Injective and Surjective in composite functionsProve distance function is bijectiveBijective compositions implies each function is bijective.Given $f$ and $g$, prove/disprove whether $g circ f$ is injective/surjectiveLet $X$ a infinite set and $Y$ a finite set. Prove exists a function $f:Xrightarrow Y$ surjective and a function $g:Yrightarrow X$ injective.Let $f:Xrightarrow Y$ and $g:Yrightarrow X$ satisfice $gcirc f=1_x$. Prove $f$ is injective and $g$ is surjective.
$begingroup$
I have been given the problem: "Say $$f:X→Y$$ and $$g:Y→Z$$are functions so that $$g◦f:X→Z$$ is bijective and $f$ is surjective. Can you deduce that $g$ is bijective?
My proof went as follows, I am not sure if this is valid?
$$ $$
Since $f$ is surjective then for any $yin Y$ there exists and $x in X$ such that $$f(x)=y$$
and $gf$ is bijective so it is surjective and injective so holds for:
$$g(f(x))=z$$for any $x$ and $$gf(x_1)=g(f(x_2)) Rightarrow x_1=x_2$$
Then since we know $gf$ is surjective so $gf(x)=z$, since $f(x)=y$, then $g(y)=z$, so for any $z in Z$ there exists $y in Y$ such that $g(y)=z$, so $g$ is surjective.
Finally we say $$x_1=x_2$$ and call $h=gf$. Since $h$ is injective and surjective $$h(x_1)=h(x_2), g(f(x_1))=g(f(x_2))$$. Since we know that $f$ is surjective so $f(x)=y$ then $g(y_1)=g(y_2)$ so it is injective. Then since we have proven $g$ is both injective and surjective then it is bijective.
functions function-and-relation-composition
asked Mar 19 at 20:38
James odareJames odare
738
$endgroup$
add a comment |
$begingroup$
I have been given the problem: "Say $$f:X→Y$$ and $$g:Y→Z$$are functions so that $$g◦f:X→Z$$ is bijective and $f$ is surjective. Can you deduce that $g$ is bijective?
My proof went as follows, I am not sure if this is valid?
$$ $$
Since $f$ is surjective then for any $yin Y$ there exists and $x in X$ such that $$f(x)=y$$
and $gf$ is bijective so it is surjective and injective so holds for:
$$g(f(x))=z$$for any $x$ and $$gf(x_1)=g(f(x_2)) Rightarrow x_1=x_2$$
Then since we know $gf$ is surjective so $gf(x)=z$, since $f(x)=y$, then $g(y)=z$, so for any $z in Z$ there exists $y in Y$ such that $g(y)=z$, so $g$ is surjective.
Finally we say $$x_1=x_2$$ and call $h=gf$. Since $h$ is injective and surjective $$h(x_1)=h(x_2), g(f(x_1))=g(f(x_2))$$. Since we know that $f$ is surjective so $f(x)=y$ then $g(y_1)=g(y_2)$ so it is injective. Then since we have proven $g$ is both injective and surjective then it is bijective.
functions function-and-relation-composition
asked Mar 19 at 20:38
James odareJames odare
738
$endgroup$
1
$begingroup$
you can use single dollar symbols$to enclose inline math, like this$z in Z$: $zin Z$. Please update formatting.
$endgroup$
– Lærne
Mar 19 at 20:44
$begingroup$
Will change now thank you
$endgroup$
– James odare
Mar 19 at 20:47
add a comment |
$begingroup$
I have been given the problem: "Say $$f:X→Y$$ and $$g:Y→Z$$are functions so that $$g◦f:X→Z$$ is bijective and $f$ is surjective. Can you deduce that $g$ is bijective?
My proof went as follows, I am not sure if this is valid?
$$ $$
Since $f$ is surjective then for any $yin Y$ there exists and $x in X$ such that $$f(x)=y$$
and $gf$ is bijective so it is surjective and injective so holds for:
$$g(f(x))=z$$for any $x$ and $$gf(x_1)=g(f(x_2)) Rightarrow x_1=x_2$$
Then since we know $gf$ is surjective so $gf(x)=z$, since $f(x)=y$, then $g(y)=z$, so for any $z in Z$ there exists $y in Y$ such that $g(y)=z$, so $g$ is surjective.
Finally we say $$x_1=x_2$$ and call $h=gf$. Since $h$ is injective and surjective $$h(x_1)=h(x_2), g(f(x_1))=g(f(x_2))$$. Since we know that $f$ is surjective so $f(x)=y$ then $g(y_1)=g(y_2)$ so it is injective. Then since we have proven $g$ is both injective and surjective then it is bijective.
functions function-and-relation-composition
asked Mar 19 at 20:38
James odareJames odare
738
$endgroup$
I have been given the problem: "Say $$f:X→Y$$ and $$g:Y→Z$$are functions so that $$g◦f:X→Z$$ is bijective and $f$ is surjective. Can you deduce that $g$ is bijective?
My proof went as follows, I am not sure if this is valid?
$$ $$
Since $f$ is surjective then for any $yin Y$ there exists and $x in X$ such that $$f(x)=y$$
and $gf$ is bijective so it is surjective and injective so holds for:
$$g(f(x))=z$$for any $x$ and $$gf(x_1)=g(f(x_2)) Rightarrow x_1=x_2$$
Then since we know $gf$ is surjective so $gf(x)=z$, since $f(x)=y$, then $g(y)=z$, so for any $z in Z$ there exists $y in Y$ such that $g(y)=z$, so $g$ is surjective.
Finally we say $$x_1=x_2$$ and call $h=gf$. Since $h$ is injective and surjective $$h(x_1)=h(x_2), g(f(x_1))=g(f(x_2))$$. Since we know that $f$ is surjective so $f(x)=y$ then $g(y_1)=g(y_2)$ so it is injective. Then since we have proven $g$ is both injective and surjective then it is bijective.
functions function-and-relation-composition
functions function-and-relation-composition
asked Mar 19 at 20:38
James odareJames odare
738
asked Mar 19 at 20:38
James odareJames odare
738
edited Mar 19 at 20:48
James odare
asked Mar 19 at 20:38
James odareJames odare
738
asked Mar 19 at 20:38
James odareJames odare
738
asked Mar 19 at 20:38
James odareJames odare
738
738
1
$begingroup$
you can use single dollar symbols$to enclose inline math, like this$z in Z$: $zin Z$. Please update formatting.
$endgroup$
– Lærne
Mar 19 at 20:44
$begingroup$
Will change now thank you
$endgroup$
– James odare
Mar 19 at 20:47
add a comment |
1
$begingroup$
you can use single dollar symbols$to enclose inline math, like this$z in Z$: $zin Z$. Please update formatting.
$endgroup$
– Lærne
Mar 19 at 20:44
$begingroup$
Will change now thank you
$endgroup$
– James odare
Mar 19 at 20:47
1
1
$begingroup$
you can use single dollar symbols
$ to enclose inline math, like this $z in Z$ : $zin Z$. Please update formatting.$endgroup$
– Lærne
Mar 19 at 20:44
$begingroup$
you can use single dollar symbols
$ to enclose inline math, like this $z in Z$ : $zin Z$. Please update formatting.$endgroup$
– Lærne
Mar 19 at 20:44
$begingroup$
Will change now thank you
$endgroup$
– James odare
Mar 19 at 20:47
$begingroup$
Will change now thank you
$endgroup$
– James odare
Mar 19 at 20:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I don't believe you've proven injectivity of $g$. What you need to prove is that if for $y_1, y_2 in Y$ you have $g(y_1) = g(y_2)$, then you have $y_1 = y_2$.
What you've proven is something weird starting from $x_1 = x_2$, and deduced $f(x_1) = y_1 = y_2 = f(x_2)$ and $gcirc f(x_1) = z_1 = z_2 = g circ f(x_2)$ or something ? Which are obvious since both $f$ and $h$ are pure function, hence must send the same value to the same value regardless of their label in the latin or greek alphabet.
What I would do, though, is use the fact than bijective functions are also inversible. That is, there is a $h^-1$ such that $h circ h^-1(z) = z$ and $h^-1 circ h (x) = x$ (prove it if you don't know that yet). Then if you have $g(y_1) = g(y_2)$ then you have $h^-1circ g(y_1) = h^-1circ g(y_2)$. I'll let you figure out the proper and complete proof.
answered Mar 19 at 20:57
LærneLærne
2,189518
$endgroup$
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
add a comment |
$begingroup$
You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154574%2fbijective-proof-check%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't believe you've proven injectivity of $g$. What you need to prove is that if for $y_1, y_2 in Y$ you have $g(y_1) = g(y_2)$, then you have $y_1 = y_2$.
What you've proven is something weird starting from $x_1 = x_2$, and deduced $f(x_1) = y_1 = y_2 = f(x_2)$ and $gcirc f(x_1) = z_1 = z_2 = g circ f(x_2)$ or something ? Which are obvious since both $f$ and $h$ are pure function, hence must send the same value to the same value regardless of their label in the latin or greek alphabet.
What I would do, though, is use the fact than bijective functions are also inversible. That is, there is a $h^-1$ such that $h circ h^-1(z) = z$ and $h^-1 circ h (x) = x$ (prove it if you don't know that yet). Then if you have $g(y_1) = g(y_2)$ then you have $h^-1circ g(y_1) = h^-1circ g(y_2)$. I'll let you figure out the proper and complete proof.
answered Mar 19 at 20:57
LærneLærne
2,189518
$endgroup$
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
add a comment |
$begingroup$
I don't believe you've proven injectivity of $g$. What you need to prove is that if for $y_1, y_2 in Y$ you have $g(y_1) = g(y_2)$, then you have $y_1 = y_2$.
What you've proven is something weird starting from $x_1 = x_2$, and deduced $f(x_1) = y_1 = y_2 = f(x_2)$ and $gcirc f(x_1) = z_1 = z_2 = g circ f(x_2)$ or something ? Which are obvious since both $f$ and $h$ are pure function, hence must send the same value to the same value regardless of their label in the latin or greek alphabet.
What I would do, though, is use the fact than bijective functions are also inversible. That is, there is a $h^-1$ such that $h circ h^-1(z) = z$ and $h^-1 circ h (x) = x$ (prove it if you don't know that yet). Then if you have $g(y_1) = g(y_2)$ then you have $h^-1circ g(y_1) = h^-1circ g(y_2)$. I'll let you figure out the proper and complete proof.
answered Mar 19 at 20:57
LærneLærne
2,189518
$endgroup$
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
add a comment |
$begingroup$
I don't believe you've proven injectivity of $g$. What you need to prove is that if for $y_1, y_2 in Y$ you have $g(y_1) = g(y_2)$, then you have $y_1 = y_2$.
What you've proven is something weird starting from $x_1 = x_2$, and deduced $f(x_1) = y_1 = y_2 = f(x_2)$ and $gcirc f(x_1) = z_1 = z_2 = g circ f(x_2)$ or something ? Which are obvious since both $f$ and $h$ are pure function, hence must send the same value to the same value regardless of their label in the latin or greek alphabet.
What I would do, though, is use the fact than bijective functions are also inversible. That is, there is a $h^-1$ such that $h circ h^-1(z) = z$ and $h^-1 circ h (x) = x$ (prove it if you don't know that yet). Then if you have $g(y_1) = g(y_2)$ then you have $h^-1circ g(y_1) = h^-1circ g(y_2)$. I'll let you figure out the proper and complete proof.
answered Mar 19 at 20:57
LærneLærne
2,189518
$endgroup$
I don't believe you've proven injectivity of $g$. What you need to prove is that if for $y_1, y_2 in Y$ you have $g(y_1) = g(y_2)$, then you have $y_1 = y_2$.
What you've proven is something weird starting from $x_1 = x_2$, and deduced $f(x_1) = y_1 = y_2 = f(x_2)$ and $gcirc f(x_1) = z_1 = z_2 = g circ f(x_2)$ or something ? Which are obvious since both $f$ and $h$ are pure function, hence must send the same value to the same value regardless of their label in the latin or greek alphabet.
What I would do, though, is use the fact than bijective functions are also inversible. That is, there is a $h^-1$ such that $h circ h^-1(z) = z$ and $h^-1 circ h (x) = x$ (prove it if you don't know that yet). Then if you have $g(y_1) = g(y_2)$ then you have $h^-1circ g(y_1) = h^-1circ g(y_2)$. I'll let you figure out the proper and complete proof.
answered Mar 19 at 20:57
LærneLærne
2,189518
edited Mar 20 at 15:19
answered Mar 19 at 20:57
LærneLærne
2,189518
answered Mar 19 at 20:57
LærneLærne
2,189518
answered Mar 19 at 20:57
LærneLærne
2,189518
2,189518
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
add a comment |
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
Thank you, i will look at it now!
$endgroup$
– James odare
Mar 19 at 21:02
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
How did you get $h^-1circ g(y_1) = h^-1circ g(y_2)$?
$endgroup$
– James odare
Mar 19 at 21:06
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
would you mind completing the proof?
$endgroup$
– James odare
Mar 19 at 21:44
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
$begingroup$
Hum... if $z_1 = z_2$, then obviously $h^-1(z_1) = h^-1(z_1)$ else $h^-1$ would be a ill-defined function. If the values are the same, then the function must send them to the same image value ! In this case we have $z_1$ to be $g(y_1)$... and likewise $z_2$ is $g(y_2)$. I still believe you can complete the proof on your own.
$endgroup$
– Lærne
Mar 20 at 15:18
add a comment |
$begingroup$
You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
$endgroup$
add a comment |
$begingroup$
You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
$endgroup$
add a comment |
$begingroup$
You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
$endgroup$
You can prove it more easily(I think much more) by using RAA argument. That is, first suppose that gf is not bijective and lead a contradiction from the supposition and the fact that f is a "surjective" "function".
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
answered Mar 19 at 22:21
Ki Yoon EumKi Yoon Eum
277
277
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154574%2fbijective-proof-check%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
you can use single dollar symbols
$to enclose inline math, like this$z in Z$: $zin Z$. Please update formatting.$endgroup$
– Lærne
Mar 19 at 20:44
$begingroup$
Will change now thank you
$endgroup$
– James odare
Mar 19 at 20:47