Why does right inverse implies surjection [duplicate]
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Prove if $f$ has a right inverse function: $fcirc g=id_x$ $iff$ $f$ is onto $Y$
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Basically, as it says in the title I require an answer on why is it so that if a function has a right inverse then it is surjective.
I am currently self studying abstract algebra and please take that in mind.
abstract-algebra functions
asked Dec 13 '18 at 17:14
amateramater
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marked as duplicate by José Carlos Santos, MisterRiemann, JMoravitz, rschwieb abstract-algebra
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Dec 13 '18 at 17:28
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This question already has an answer here:
Prove if $f$ has a right inverse function: $fcirc g=id_x$ $iff$ $f$ is onto $Y$
2 answers
Basically, as it says in the title I require an answer on why is it so that if a function has a right inverse then it is surjective.
I am currently self studying abstract algebra and please take that in mind.
abstract-algebra functions
asked Dec 13 '18 at 17:14
amateramater
1
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marked as duplicate by José Carlos Santos, MisterRiemann, JMoravitz, rschwieb abstract-algebra
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Dec 13 '18 at 17:28
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$begingroup$
This question already has an answer here:
Prove if $f$ has a right inverse function: $fcirc g=id_x$ $iff$ $f$ is onto $Y$
2 answers
Basically, as it says in the title I require an answer on why is it so that if a function has a right inverse then it is surjective.
I am currently self studying abstract algebra and please take that in mind.
abstract-algebra functions
asked Dec 13 '18 at 17:14
amateramater
1
$endgroup$
This question already has an answer here:
Prove if $f$ has a right inverse function: $fcirc g=id_x$ $iff$ $f$ is onto $Y$
2 answers
Basically, as it says in the title I require an answer on why is it so that if a function has a right inverse then it is surjective.
I am currently self studying abstract algebra and please take that in mind.
This question already has an answer here:
Prove if $f$ has a right inverse function: $fcirc g=id_x$ $iff$ $f$ is onto $Y$
2 answers
abstract-algebra functions
abstract-algebra functions
asked Dec 13 '18 at 17:14
amateramater
1
asked Dec 13 '18 at 17:14
amateramater
1
asked Dec 13 '18 at 17:14
amateramater
1
asked Dec 13 '18 at 17:14
amateramater
1
asked Dec 13 '18 at 17:14
amateramater
1
1
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Dec 13 '18 at 17:28
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Dec 13 '18 at 17:28
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3 Answers
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If $f:Xto Y$ is a morphism in a category whose objects are sets with additional structure (e.g. groups, rings, etc), and $g:Yto X$ satisfies $fcirc g=1_Y,$ then by a pure set theoretical argument $f$ is surjective: If $yin Y$ let $x=g(y)in X,$ then $f(x)=fcirc g(y)=y.$
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
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The right inverse of $f:Ato B$ is the function $h:Bto A$ such that $(fcirc h)(x)=x forall xin B$.
The identity function is surjective on $B$ which means $fcirc h$ is surjective on $B$. This means that $f$ has to be surjective, because the range of $fcirc h$ is a subset of the range of $f$;
$ R(fcirc h)subseteq R(f)subseteq B$
$R(fcirc h)=Bimplies R(f)=B$
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
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If $f:Ato B$ has a right inverse, $exists g:Bto A:fcirc g=id_B$. Since $fcirc g$ is onto, $f$ must be onto.
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $f:Xto Y$ is a morphism in a category whose objects are sets with additional structure (e.g. groups, rings, etc), and $g:Yto X$ satisfies $fcirc g=1_Y,$ then by a pure set theoretical argument $f$ is surjective: If $yin Y$ let $x=g(y)in X,$ then $f(x)=fcirc g(y)=y.$
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
$endgroup$
add a comment |
$begingroup$
If $f:Xto Y$ is a morphism in a category whose objects are sets with additional structure (e.g. groups, rings, etc), and $g:Yto X$ satisfies $fcirc g=1_Y,$ then by a pure set theoretical argument $f$ is surjective: If $yin Y$ let $x=g(y)in X,$ then $f(x)=fcirc g(y)=y.$
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
$endgroup$
add a comment |
$begingroup$
If $f:Xto Y$ is a morphism in a category whose objects are sets with additional structure (e.g. groups, rings, etc), and $g:Yto X$ satisfies $fcirc g=1_Y,$ then by a pure set theoretical argument $f$ is surjective: If $yin Y$ let $x=g(y)in X,$ then $f(x)=fcirc g(y)=y.$
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
$endgroup$
If $f:Xto Y$ is a morphism in a category whose objects are sets with additional structure (e.g. groups, rings, etc), and $g:Yto X$ satisfies $fcirc g=1_Y,$ then by a pure set theoretical argument $f$ is surjective: If $yin Y$ let $x=g(y)in X,$ then $f(x)=fcirc g(y)=y.$
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
answered Dec 13 '18 at 17:21
positrón0802positrón0802
4,283520
4,283520
add a comment |
add a comment |
$begingroup$
The right inverse of $f:Ato B$ is the function $h:Bto A$ such that $(fcirc h)(x)=x forall xin B$.
The identity function is surjective on $B$ which means $fcirc h$ is surjective on $B$. This means that $f$ has to be surjective, because the range of $fcirc h$ is a subset of the range of $f$;
$ R(fcirc h)subseteq R(f)subseteq B$
$R(fcirc h)=Bimplies R(f)=B$
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
The right inverse of $f:Ato B$ is the function $h:Bto A$ such that $(fcirc h)(x)=x forall xin B$.
The identity function is surjective on $B$ which means $fcirc h$ is surjective on $B$. This means that $f$ has to be surjective, because the range of $fcirc h$ is a subset of the range of $f$;
$ R(fcirc h)subseteq R(f)subseteq B$
$R(fcirc h)=Bimplies R(f)=B$
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
$endgroup$
add a comment |
$begingroup$
The right inverse of $f:Ato B$ is the function $h:Bto A$ such that $(fcirc h)(x)=x forall xin B$.
The identity function is surjective on $B$ which means $fcirc h$ is surjective on $B$. This means that $f$ has to be surjective, because the range of $fcirc h$ is a subset of the range of $f$;
$ R(fcirc h)subseteq R(f)subseteq B$
$R(fcirc h)=Bimplies R(f)=B$
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
$endgroup$
The right inverse of $f:Ato B$ is the function $h:Bto A$ such that $(fcirc h)(x)=x forall xin B$.
The identity function is surjective on $B$ which means $fcirc h$ is surjective on $B$. This means that $f$ has to be surjective, because the range of $fcirc h$ is a subset of the range of $f$;
$ R(fcirc h)subseteq R(f)subseteq B$
$R(fcirc h)=Bimplies R(f)=B$
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
answered Dec 13 '18 at 17:21
Shubham JohriShubham Johri
4,666717
4,666717
add a comment |
add a comment |
$begingroup$
If $f:Ato B$ has a right inverse, $exists g:Bto A:fcirc g=id_B$. Since $fcirc g$ is onto, $f$ must be onto.
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
$endgroup$
add a comment |
$begingroup$
If $f:Ato B$ has a right inverse, $exists g:Bto A:fcirc g=id_B$. Since $fcirc g$ is onto, $f$ must be onto.
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
$endgroup$
add a comment |
$begingroup$
If $f:Ato B$ has a right inverse, $exists g:Bto A:fcirc g=id_B$. Since $fcirc g$ is onto, $f$ must be onto.
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
$endgroup$
If $f:Ato B$ has a right inverse, $exists g:Bto A:fcirc g=id_B$. Since $fcirc g$ is onto, $f$ must be onto.
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
answered Dec 13 '18 at 17:25
Chris CusterChris Custer
11.2k3824
11.2k3824
add a comment |
add a comment |
