Restrictions of compositions

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Considering two functions, $f,g:Xto X$ s.t. $Ysubseteq X$ is invariant for both $f$ and $g$, that is $f(Y)subseteq Y$ and similar for $g$. Do we then have $f|_Ycirc g|_Y=(fcirc g)|_Y$. It seems trivial, but I'm not sure how to show it formally. It would be nice to see a set theoretic argument via the restrictions of relations.

functions elementary-set-theory function-and-relation-composition

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edited Jul 21 at 21:54

asked Jul 21 at 21:38

zzuussee

1,512419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does invariant mean?
  – William Elliot
  Jul 21 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yes, I'm sorry, I've edited my question
  – zzuussee
  Jul 21 at 21:54
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Considering two functions, $f,g:Xto X$ s.t. $Ysubseteq X$ is invariant for both $f$ and $g$, that is $f(Y)subseteq Y$ and similar for $g$. Do we then have $f|_Ycirc g|_Y=(fcirc g)|_Y$. It seems trivial, but I'm not sure how to show it formally. It would be nice to see a set theoretic argument via the restrictions of relations.

functions elementary-set-theory function-and-relation-composition

share|cite|improve this question

edited Jul 21 at 21:54

asked Jul 21 at 21:38

zzuussee

1,512419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does invariant mean?
  – William Elliot
  Jul 21 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yes, I'm sorry, I've edited my question
  – zzuussee
  Jul 21 at 21:54
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Considering two functions, $f,g:Xto X$ s.t. $Ysubseteq X$ is invariant for both $f$ and $g$, that is $f(Y)subseteq Y$ and similar for $g$. Do we then have $f|_Ycirc g|_Y=(fcirc g)|_Y$. It seems trivial, but I'm not sure how to show it formally. It would be nice to see a set theoretic argument via the restrictions of relations.

functions elementary-set-theory function-and-relation-composition

share|cite|improve this question

edited Jul 21 at 21:54

asked Jul 21 at 21:38

zzuussee

1,512419

Considering two functions, $f,g:Xto X$ s.t. $Ysubseteq X$ is invariant for both $f$ and $g$, that is $f(Y)subseteq Y$ and similar for $g$. Do we then have $f|_Ycirc g|_Y=(fcirc g)|_Y$. It seems trivial, but I'm not sure how to show it formally. It would be nice to see a set theoretic argument via the restrictions of relations.

functions elementary-set-theory function-and-relation-composition

share|cite|improve this question

edited Jul 21 at 21:54

asked Jul 21 at 21:38

zzuussee

1,512419

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 21:54

edited Jul 21 at 21:54

edited Jul 21 at 21:54

asked Jul 21 at 21:38

zzuussee

1,512419

asked Jul 21 at 21:38

zzuussee

1,512419

asked Jul 21 at 21:38

zzuussee

1,512419

1,512419

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does invariant mean?
  – William Elliot
  Jul 21 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yes, I'm sorry, I've edited my question
  – zzuussee
  Jul 21 at 21:54
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does invariant mean?
  – William Elliot
  Jul 21 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah yes, I'm sorry, I've edited my question
  – zzuussee
  Jul 21 at 21:54
  

  
  
  
  

What does invariant mean?
– William Elliot
Jul 21 at 21:51

What does invariant mean?
– William Elliot
Jul 21 at 21:51

Ah yes, I'm sorry, I've edited my question
– zzuussee
Jul 21 at 21:54

Ah yes, I'm sorry, I've edited my question
– zzuussee
Jul 21 at 21:54

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

The composition $fmid_Ycirc gmid_Y$ is well defined if and only if $g(Y)subset Y$. If that is given, the equation is true as you can see below.

By definition of the restriction, you get $f(x)=fmid_Y(x)$, $g(x)=gmid_Y(x)$ and $fcirc g(x)=(fcirc g)mid_Y(x)$ for all $xin Y$.

For all $xin Y$ this yields
$$
(fmid_Ycirc gmid_Y)(x)=fmid_Y(gmid_Y(x))=fmid_Y(g(x))=f(g(x))=fcirc g(x)=(fcirc g)mid_Y(x).
$$
(At the third equation you use $g(x)in Y$.) Hence, $fmid_Ycirc gmid_Y=(fcirc g)mid_Y$ holds.

share|cite|improve this answer

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

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0
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It does look trivial: $text lhs=f(g(y))=textrhs forall yin Y$.

That is, we have $bigcup_yin Y(y,f(g(y)))$ for both functions.

Note: $fmid _Y=fcirc i:Yrightarrow Y$, where $i:Yhookrightarrow X$ is the inclusion of $Y$ into $X$. I have written $Y$ for the codomain by invariance...

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edited Jul 22 at 0:41

answered Jul 21 at 22:41

Chris Custer

5,4082622

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The composition $fmid_Ycirc gmid_Y$ is well defined if and only if $g(Y)subset Y$. If that is given, the equation is true as you can see below.

By definition of the restriction, you get $f(x)=fmid_Y(x)$, $g(x)=gmid_Y(x)$ and $fcirc g(x)=(fcirc g)mid_Y(x)$ for all $xin Y$.

For all $xin Y$ this yields
$$
(fmid_Ycirc gmid_Y)(x)=fmid_Y(gmid_Y(x))=fmid_Y(g(x))=f(g(x))=fcirc g(x)=(fcirc g)mid_Y(x).
$$
(At the third equation you use $g(x)in Y$.) Hence, $fmid_Ycirc gmid_Y=(fcirc g)mid_Y$ holds.

share|cite|improve this answer

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

add a comment | 

up vote
2
down vote



accepted

The composition $fmid_Ycirc gmid_Y$ is well defined if and only if $g(Y)subset Y$. If that is given, the equation is true as you can see below.

By definition of the restriction, you get $f(x)=fmid_Y(x)$, $g(x)=gmid_Y(x)$ and $fcirc g(x)=(fcirc g)mid_Y(x)$ for all $xin Y$.

For all $xin Y$ this yields
$$
(fmid_Ycirc gmid_Y)(x)=fmid_Y(gmid_Y(x))=fmid_Y(g(x))=f(g(x))=fcirc g(x)=(fcirc g)mid_Y(x).
$$
(At the third equation you use $g(x)in Y$.) Hence, $fmid_Ycirc gmid_Y=(fcirc g)mid_Y$ holds.

share|cite|improve this answer

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The composition $fmid_Ycirc gmid_Y$ is well defined if and only if $g(Y)subset Y$. If that is given, the equation is true as you can see below.

By definition of the restriction, you get $f(x)=fmid_Y(x)$, $g(x)=gmid_Y(x)$ and $fcirc g(x)=(fcirc g)mid_Y(x)$ for all $xin Y$.

For all $xin Y$ this yields
$$
(fmid_Ycirc gmid_Y)(x)=fmid_Y(gmid_Y(x))=fmid_Y(g(x))=f(g(x))=fcirc g(x)=(fcirc g)mid_Y(x).
$$
(At the third equation you use $g(x)in Y$.) Hence, $fmid_Ycirc gmid_Y=(fcirc g)mid_Y$ holds.

share|cite|improve this answer

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

The composition $fmid_Ycirc gmid_Y$ is well defined if and only if $g(Y)subset Y$. If that is given, the equation is true as you can see below.

By definition of the restriction, you get $f(x)=fmid_Y(x)$, $g(x)=gmid_Y(x)$ and $fcirc g(x)=(fcirc g)mid_Y(x)$ for all $xin Y$.

For all $xin Y$ this yields
$$
(fmid_Ycirc gmid_Y)(x)=fmid_Y(gmid_Y(x))=fmid_Y(g(x))=f(g(x))=fcirc g(x)=(fcirc g)mid_Y(x).
$$
(At the third equation you use $g(x)in Y$.) Hence, $fmid_Ycirc gmid_Y=(fcirc g)mid_Y$ holds.

share|cite|improve this answer

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

answered Jul 21 at 21:57

Mundron Schmidt

7,1162727

7,1162727

add a comment | 

add a comment | 

up vote
0
down vote

It does look trivial: $text lhs=f(g(y))=textrhs forall yin Y$.

That is, we have $bigcup_yin Y(y,f(g(y)))$ for both functions.

Note: $fmid _Y=fcirc i:Yrightarrow Y$, where $i:Yhookrightarrow X$ is the inclusion of $Y$ into $X$. I have written $Y$ for the codomain by invariance...

share|cite|improve this answer

edited Jul 22 at 0:41

answered Jul 21 at 22:41

Chris Custer

5,4082622

add a comment | 

up vote
0
down vote

It does look trivial: $text lhs=f(g(y))=textrhs forall yin Y$.

That is, we have $bigcup_yin Y(y,f(g(y)))$ for both functions.

Note: $fmid _Y=fcirc i:Yrightarrow Y$, where $i:Yhookrightarrow X$ is the inclusion of $Y$ into $X$. I have written $Y$ for the codomain by invariance...

share|cite|improve this answer

edited Jul 22 at 0:41

answered Jul 21 at 22:41

Chris Custer

5,4082622

add a comment | 

up vote
0
down vote

up vote
0
down vote

It does look trivial: $text lhs=f(g(y))=textrhs forall yin Y$.

That is, we have $bigcup_yin Y(y,f(g(y)))$ for both functions.

Note: $fmid _Y=fcirc i:Yrightarrow Y$, where $i:Yhookrightarrow X$ is the inclusion of $Y$ into $X$. I have written $Y$ for the codomain by invariance...

share|cite|improve this answer

edited Jul 22 at 0:41

answered Jul 21 at 22:41

Chris Custer

5,4082622

It does look trivial: $text lhs=f(g(y))=textrhs forall yin Y$.

That is, we have $bigcup_yin Y(y,f(g(y)))$ for both functions.

Note: $fmid _Y=fcirc i:Yrightarrow Y$, where $i:Yhookrightarrow X$ is the inclusion of $Y$ into $X$. I have written $Y$ for the codomain by invariance...

share|cite|improve this answer

edited Jul 22 at 0:41

answered Jul 21 at 22:41

Chris Custer

5,4082622

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 0:41

edited Jul 22 at 0:41

edited Jul 22 at 0:41

answered Jul 21 at 22:41

Chris Custer

5,4082622

answered Jul 21 at 22:41

Chris Custer

5,4082622

answered Jul 21 at 22:41

Chris Custer

5,4082622

5,4082622

add a comment | 

add a comment | 

 

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