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2

$begingroup$

Let $f: A rightarrow B$ and $g: B rightarrow C$ be two functions.
Then $g circ f$ is clearly defined $forall a in A$ but what about $f circ g$, do we take it as undefined, given that A and C are disjoint? What if A and C are not disjoint?

Since domain of $g$ is the same as the codomain of $f$ hence $g circ f$ exists $forall a in A$.

What I think:
If A and C are disjoint sets, then $f(c)$ is not defined, since $cin C$ is not in the domain of $f$

Am I missing something?

function-and-relation-composition

share|cite|improve this question

edited Mar 14 at 15:48

Aditya

asked Mar 14 at 15:34

AdityaAditya

496

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  your logic is sound.
  $endgroup$
  – Doug M
  Mar 14 at 15:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think a better way to phrase this is that $f(c)$ is not defined, since $cin C$ is not in the domain of $f$
  $endgroup$
  – pwerth
  Mar 14 at 15:39
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f: A rightarrow B$ and $g: B rightarrow C$ be two functions.
Then $g circ f$ is clearly defined $forall a in A$ but what about $f circ g$, do we take it as undefined, given that A and C are disjoint? What if A and C are not disjoint?

Since domain of $g$ is the same as the codomain of $f$ hence $g circ f$ exists $forall a in A$.

What I think:
If A and C are disjoint sets, then $f(c)$ is not defined, since $cin C$ is not in the domain of $f$

Am I missing something?

function-and-relation-composition

share|cite|improve this question

edited Mar 14 at 15:48

Aditya

asked Mar 14 at 15:34

AdityaAditya

496

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  your logic is sound.
  $endgroup$
  – Doug M
  Mar 14 at 15:38
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think a better way to phrase this is that $f(c)$ is not defined, since $cin C$ is not in the domain of $f$
  $endgroup$
  – pwerth
  Mar 14 at 15:39
  

  
  
  
  

add a comment | 

$begingroup$

Let $f: A rightarrow B$ and $g: B rightarrow C$ be two functions.
Then $g circ f$ is clearly defined $forall a in A$ but what about $f circ g$, do we take it as undefined, given that A and C are disjoint? What if A and C are not disjoint?

Since domain of $g$ is the same as the codomain of $f$ hence $g circ f$ exists $forall a in A$.

What I think:
If A and C are disjoint sets, then $f(c)$ is not defined, since $cin C$ is not in the domain of $f$

Am I missing something?

function-and-relation-composition

share|cite|improve this question

edited Mar 14 at 15:48

Aditya

asked Mar 14 at 15:34

AdityaAditya

496

$endgroup$

share|cite|improve this question

edited Mar 14 at 15:48

Aditya

asked Mar 14 at 15:34

AdityaAditya

496

share|cite|improve this question

edited Mar 14 at 15:48

Aditya

asked Mar 14 at 15:34

AdityaAditya

496

asked Mar 14 at 15:34

AdityaAditya

496

asked Mar 14 at 15:34

AdityaAditya

496

1 Answer
1

active

oldest

votes

1

$begingroup$

There is a strict definition for composition:

If $f: A to B$, $g: C to D$ are functions, the composition $f circ g$ is the function $C to B$ given by $c mapsto g(c) mapsto f(g(c))$, which is well defined if and only if $g(c) in A$.

So indeed your intuition is right.

share|cite|improve this answer

answered Mar 14 at 15:57

MariahMariah

1,8231718

$endgroup$

add a comment | 

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1

$begingroup$

There is a strict definition for composition:

If $f: A to B$, $g: C to D$ are functions, the composition $f circ g$ is the function $C to B$ given by $c mapsto g(c) mapsto f(g(c))$, which is well defined if and only if $g(c) in A$.

So indeed your intuition is right.

share|cite|improve this answer

answered Mar 14 at 15:57

MariahMariah

1,8231718

$endgroup$

add a comment | 

1

$begingroup$

There is a strict definition for composition:

If $f: A to B$, $g: C to D$ are functions, the composition $f circ g$ is the function $C to B$ given by $c mapsto g(c) mapsto f(g(c))$, which is well defined if and only if $g(c) in A$.

So indeed your intuition is right.

share|cite|improve this answer

answered Mar 14 at 15:57

MariahMariah

1,8231718

$endgroup$

add a comment | 

$begingroup$

There is a strict definition for composition:

If $f: A to B$, $g: C to D$ are functions, the composition $f circ g$ is the function $C to B$ given by $c mapsto g(c) mapsto f(g(c))$, which is well defined if and only if $g(c) in A$.

So indeed your intuition is right.

share|cite|improve this answer

answered Mar 14 at 15:57

MariahMariah

1,8231718

$endgroup$

share|cite|improve this answer

answered Mar 14 at 15:57

MariahMariah

1,8231718

answered Mar 14 at 15:57

MariahMariah

1,8231718

answered Mar 14 at 15:57

MariahMariah

1,8231718