If F is a homomorphism function, i want to ask the inverse of F is also homomorphism? [closed]












0












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In abstract algebra i have a problem to prove or disprove this statemant : "If F is a homomorphism function then the inverse of F is also homomorphism function" ?!






abstract-algebra inverse-function







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closed as off-topic by Crostul, José Carlos Santos, user593746, Lord Shark the Unknown, Shailesh Dec 20 '18 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, José Carlos Santos, Community, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    This is obviously true.
    $endgroup$
    – Crostul
    Dec 19 '18 at 22:25
















0












$begingroup$


In abstract algebra i have a problem to prove or disprove this statemant : "If F is a homomorphism function then the inverse of F is also homomorphism function" ?!






abstract-algebra inverse-function







share|cite|improve this question











$endgroup$



closed as off-topic by Crostul, José Carlos Santos, user593746, Lord Shark the Unknown, Shailesh Dec 20 '18 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, José Carlos Santos, Community, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    This is obviously true.
    $endgroup$
    – Crostul
    Dec 19 '18 at 22:25














0












0








0





$begingroup$


In abstract algebra i have a problem to prove or disprove this statemant : "If F is a homomorphism function then the inverse of F is also homomorphism function" ?!






abstract-algebra inverse-function







share|cite|improve this question











$endgroup$




In abstract algebra i have a problem to prove or disprove this statemant : "If F is a homomorphism function then the inverse of F is also homomorphism function" ?!






abstract-algebra inverse-function




abstract-algebra inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 23:06









egreg

185k1486207




185k1486207










asked Dec 19 '18 at 22:21









Osaid Ibrahim Bani SalmanOsaid Ibrahim Bani Salman

244




244




closed as off-topic by Crostul, José Carlos Santos, user593746, Lord Shark the Unknown, Shailesh Dec 20 '18 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, José Carlos Santos, Community, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Crostul, José Carlos Santos, user593746, Lord Shark the Unknown, Shailesh Dec 20 '18 at 2:49


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Crostul, José Carlos Santos, Community, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    This is obviously true.
    $endgroup$
    – Crostul
    Dec 19 '18 at 22:25


















  • $begingroup$
    This is obviously true.
    $endgroup$
    – Crostul
    Dec 19 '18 at 22:25
















$begingroup$
This is obviously true.
$endgroup$
– Crostul
Dec 19 '18 at 22:25




$begingroup$
This is obviously true.
$endgroup$
– Crostul
Dec 19 '18 at 22:25










2 Answers
2






active

oldest

votes


















1












$begingroup$

Let $f: A to B$ be a bijective homomorphism (isomorphism) of rings (you can change the proof to other algebraic structures).



Note that you need it to be bijective to be sure that the inverse exists.



Then $f(1_A) = 1_B$, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for $a,b in A$.



Now
$$f^{-1}(1_B) = f^{-1}(f(1_A)) = 1_A.$$



Now take $c,d in B$. Then since $f$ has an inverse (it is bijective), there exists $a,b in A$ such that $f(a)=c$ and $f(b)=d$. In particular this gives that $a = f^{-1}(c)$ and $b = f^{-1}(d)$. It follows that



$$f^{-1}(c+d) = f^{-1}(f(a) + f(b)) = f^{-1}(f(a+b)) = a+b = f^{-1}(c) + f^{-1}(d) $$



and lastly



$$f^{-1}(cd) = f^{-1}(f(a)f(b)) = f^{-1}(f(ab)) = ab = f^{-1}(c)f^{-1}(d).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for helping me Niall
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 3:44



















1












$begingroup$

A function needs to be injective to have an inverse.



By the first isomorphism theorem if $f$ is injective, then $f$ is an isomorphism onto its image.



Because of this you can define an isomorphism as a bijective homomorphism.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this means the statemant not true?
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 19 '18 at 22:46










  • $begingroup$
    It's true, in case there's an inverse at all.
    $endgroup$
    – Chris Custer
    Dec 19 '18 at 22:47


















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $f: A to B$ be a bijective homomorphism (isomorphism) of rings (you can change the proof to other algebraic structures).



Note that you need it to be bijective to be sure that the inverse exists.



Then $f(1_A) = 1_B$, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for $a,b in A$.



Now
$$f^{-1}(1_B) = f^{-1}(f(1_A)) = 1_A.$$



Now take $c,d in B$. Then since $f$ has an inverse (it is bijective), there exists $a,b in A$ such that $f(a)=c$ and $f(b)=d$. In particular this gives that $a = f^{-1}(c)$ and $b = f^{-1}(d)$. It follows that



$$f^{-1}(c+d) = f^{-1}(f(a) + f(b)) = f^{-1}(f(a+b)) = a+b = f^{-1}(c) + f^{-1}(d) $$



and lastly



$$f^{-1}(cd) = f^{-1}(f(a)f(b)) = f^{-1}(f(ab)) = ab = f^{-1}(c)f^{-1}(d).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for helping me Niall
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 3:44
















1












$begingroup$

Let $f: A to B$ be a bijective homomorphism (isomorphism) of rings (you can change the proof to other algebraic structures).



Note that you need it to be bijective to be sure that the inverse exists.



Then $f(1_A) = 1_B$, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for $a,b in A$.



Now
$$f^{-1}(1_B) = f^{-1}(f(1_A)) = 1_A.$$



Now take $c,d in B$. Then since $f$ has an inverse (it is bijective), there exists $a,b in A$ such that $f(a)=c$ and $f(b)=d$. In particular this gives that $a = f^{-1}(c)$ and $b = f^{-1}(d)$. It follows that



$$f^{-1}(c+d) = f^{-1}(f(a) + f(b)) = f^{-1}(f(a+b)) = a+b = f^{-1}(c) + f^{-1}(d) $$



and lastly



$$f^{-1}(cd) = f^{-1}(f(a)f(b)) = f^{-1}(f(ab)) = ab = f^{-1}(c)f^{-1}(d).$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for helping me Niall
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 3:44














1












1








1





$begingroup$

Let $f: A to B$ be a bijective homomorphism (isomorphism) of rings (you can change the proof to other algebraic structures).



Note that you need it to be bijective to be sure that the inverse exists.



Then $f(1_A) = 1_B$, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for $a,b in A$.



Now
$$f^{-1}(1_B) = f^{-1}(f(1_A)) = 1_A.$$



Now take $c,d in B$. Then since $f$ has an inverse (it is bijective), there exists $a,b in A$ such that $f(a)=c$ and $f(b)=d$. In particular this gives that $a = f^{-1}(c)$ and $b = f^{-1}(d)$. It follows that



$$f^{-1}(c+d) = f^{-1}(f(a) + f(b)) = f^{-1}(f(a+b)) = a+b = f^{-1}(c) + f^{-1}(d) $$



and lastly



$$f^{-1}(cd) = f^{-1}(f(a)f(b)) = f^{-1}(f(ab)) = ab = f^{-1}(c)f^{-1}(d).$$






share|cite|improve this answer









$endgroup$



Let $f: A to B$ be a bijective homomorphism (isomorphism) of rings (you can change the proof to other algebraic structures).



Note that you need it to be bijective to be sure that the inverse exists.



Then $f(1_A) = 1_B$, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$ for $a,b in A$.



Now
$$f^{-1}(1_B) = f^{-1}(f(1_A)) = 1_A.$$



Now take $c,d in B$. Then since $f$ has an inverse (it is bijective), there exists $a,b in A$ such that $f(a)=c$ and $f(b)=d$. In particular this gives that $a = f^{-1}(c)$ and $b = f^{-1}(d)$. It follows that



$$f^{-1}(c+d) = f^{-1}(f(a) + f(b)) = f^{-1}(f(a+b)) = a+b = f^{-1}(c) + f^{-1}(d) $$



and lastly



$$f^{-1}(cd) = f^{-1}(f(a)f(b)) = f^{-1}(f(ab)) = ab = f^{-1}(c)f^{-1}(d).$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 22:47









CharlesCharles

18818




18818












  • $begingroup$
    Thanks for helping me Niall
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 3:44


















  • $begingroup$
    Thanks for helping me Niall
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 20 '18 at 3:44
















$begingroup$
Thanks for helping me Niall
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 3:44




$begingroup$
Thanks for helping me Niall
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 20 '18 at 3:44











1












$begingroup$

A function needs to be injective to have an inverse.



By the first isomorphism theorem if $f$ is injective, then $f$ is an isomorphism onto its image.



Because of this you can define an isomorphism as a bijective homomorphism.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this means the statemant not true?
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 19 '18 at 22:46










  • $begingroup$
    It's true, in case there's an inverse at all.
    $endgroup$
    – Chris Custer
    Dec 19 '18 at 22:47
















1












$begingroup$

A function needs to be injective to have an inverse.



By the first isomorphism theorem if $f$ is injective, then $f$ is an isomorphism onto its image.



Because of this you can define an isomorphism as a bijective homomorphism.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    this means the statemant not true?
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 19 '18 at 22:46










  • $begingroup$
    It's true, in case there's an inverse at all.
    $endgroup$
    – Chris Custer
    Dec 19 '18 at 22:47














1












1








1





$begingroup$

A function needs to be injective to have an inverse.



By the first isomorphism theorem if $f$ is injective, then $f$ is an isomorphism onto its image.



Because of this you can define an isomorphism as a bijective homomorphism.






share|cite|improve this answer











$endgroup$



A function needs to be injective to have an inverse.



By the first isomorphism theorem if $f$ is injective, then $f$ is an isomorphism onto its image.



Because of this you can define an isomorphism as a bijective homomorphism.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 22:51

























answered Dec 19 '18 at 22:37









Chris CusterChris Custer

14.3k3827




14.3k3827












  • $begingroup$
    this means the statemant not true?
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 19 '18 at 22:46










  • $begingroup$
    It's true, in case there's an inverse at all.
    $endgroup$
    – Chris Custer
    Dec 19 '18 at 22:47


















  • $begingroup$
    this means the statemant not true?
    $endgroup$
    – Osaid Ibrahim Bani Salman
    Dec 19 '18 at 22:46










  • $begingroup$
    It's true, in case there's an inverse at all.
    $endgroup$
    – Chris Custer
    Dec 19 '18 at 22:47
















$begingroup$
this means the statemant not true?
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 19 '18 at 22:46




$begingroup$
this means the statemant not true?
$endgroup$
– Osaid Ibrahim Bani Salman
Dec 19 '18 at 22:46












$begingroup$
It's true, in case there's an inverse at all.
$endgroup$
– Chris Custer
Dec 19 '18 at 22:47




$begingroup$
It's true, in case there's an inverse at all.
$endgroup$
– Chris Custer
Dec 19 '18 at 22:47



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