Mixing
Can a surjective function have an element in the domain not mapped to the codomain?
$begingroup$
I have seen a lot of definitions for surjectivity stating that every element in the codomain must be mapped to something in the domain. But does the opposite also have to hold true for a function to maintain its surjectivity?
functions elementary-set-theory
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
$endgroup$
add a comment |
$begingroup$
I have seen a lot of definitions for surjectivity stating that every element in the codomain must be mapped to something in the domain. But does the opposite also have to hold true for a function to maintain its surjectivity?
functions elementary-set-theory
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
$endgroup$
1
$begingroup$
It's more correct to say a function is surjective has every element in the codomain mapped from something in the domain, ie. every value in the codomain is attained by a preimage in the domain.
$endgroup$
– hardmath
Dec 17 '14 at 22:40
$begingroup$
Where have you seen all these strange (and non-standard) definitions of the standard concept of surjectivity? Some context is needed to clarify what is meant by "domain" and "codomain".
$endgroup$
– Rob Arthan
Dec 17 '14 at 22:51
add a comment |
$begingroup$
I have seen a lot of definitions for surjectivity stating that every element in the codomain must be mapped to something in the domain. But does the opposite also have to hold true for a function to maintain its surjectivity?
functions elementary-set-theory
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
$endgroup$
I have seen a lot of definitions for surjectivity stating that every element in the codomain must be mapped to something in the domain. But does the opposite also have to hold true for a function to maintain its surjectivity?
functions elementary-set-theory
functions elementary-set-theory
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
edited Dec 17 '14 at 22:48
Asaf Karagila♦
306k33438769
306k33438769
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
asked Dec 17 '14 at 22:38
Mitali PMitali P
112
112
1
$begingroup$
It's more correct to say a function is surjective has every element in the codomain mapped from something in the domain, ie. every value in the codomain is attained by a preimage in the domain.
$endgroup$
– hardmath
Dec 17 '14 at 22:40
$begingroup$
Where have you seen all these strange (and non-standard) definitions of the standard concept of surjectivity? Some context is needed to clarify what is meant by "domain" and "codomain".
$endgroup$
– Rob Arthan
Dec 17 '14 at 22:51
add a comment |
1
$begingroup$
It's more correct to say a function is surjective has every element in the codomain mapped from something in the domain, ie. every value in the codomain is attained by a preimage in the domain.
$endgroup$
– hardmath
Dec 17 '14 at 22:40
$begingroup$
Where have you seen all these strange (and non-standard) definitions of the standard concept of surjectivity? Some context is needed to clarify what is meant by "domain" and "codomain".
$endgroup$
– Rob Arthan
Dec 17 '14 at 22:51
1
1
$begingroup$
It's more correct to say a function is surjective has every element in the codomain mapped from something in the domain, ie. every value in the codomain is attained by a preimage in the domain.
$endgroup$
– hardmath
Dec 17 '14 at 22:40
$begingroup$
It's more correct to say a function is surjective has every element in the codomain mapped from something in the domain, ie. every value in the codomain is attained by a preimage in the domain.
$endgroup$
– hardmath
Dec 17 '14 at 22:40
$begingroup$
Where have you seen all these strange (and non-standard) definitions of the standard concept of surjectivity? Some context is needed to clarify what is meant by "domain" and "codomain".
$endgroup$
– Rob Arthan
Dec 17 '14 at 22:51
$begingroup$
Where have you seen all these strange (and non-standard) definitions of the standard concept of surjectivity? Some context is needed to clarify what is meant by "domain" and "codomain".
$endgroup$
– Rob Arthan
Dec 17 '14 at 22:51
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The definition of a function $f$ from $A$ to $B$, regardless to injectivity or surjectivity, is that the domain of $f$ is $A$, in its entirety.
This means that if $fcolon Ato B$, then for every $ain A$, there is a unique $bin B$ such that the pair $(a,b)in f$.
So the converse holds just for it to be a function from $A$ to $B$.
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
add a comment |
$begingroup$
No (to your title question), a function $f : A to B$, by definition, gives each $x in A$ a value $f(x) in B$.
To be precise, you should have said that "for every element in the codomain, there is (at least) one element which is mapped to it by $f$" for your definition of surjectivity.
answered Dec 17 '14 at 22:42
bknbkn
1612
$endgroup$
add a comment |
$begingroup$
I know exactly why that is confusing, and I will try to explain this step by step. The definition of surjectivity does not mention the domain being overfull. That throws some people off when learning this the first time. The reason for your confusion is likely that you skipped over the Cartesian product and function definition topics without understanding their significance (like I did). In short, the fact that every element of a domain maps to a codomain is implied by the definition of a function (it must be well-defined...read more if you're interested).
Long-winded explanation
So what is a function $f$? Well in set theory, it is a set whose elements are from a relation set, which is called a Cartesian product of sets.
What's that? So given sets $A := {1,2,3}$ and $B := {a,b }$, we could define a relation set between them, but first we have to understand that a relation set is. It is every possible ordered combination (tuples) of elements from each set where the first element corresponds to the first set and the second corresponds to the second set (and so on with $>2$ sets). For example, $C := {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}$ (but not including ${(a,1), (a,2), (b,1), (b,2), (a,3), (b,3)}$, because $anotin A, bnotin A$). Books explain this with a conditional statement like “$A times B := {(x,y)}$, where the $x$'s are from A and the $y$'s are from B”.
So now the critical thing to understand is that a function called $f$ that maps sets $A$ to $B$ ($f : A to B$) takes its elements not directly from $A$, but rather from $C$ above. Now the question is, which elements of $C$ may I use in $f$ such that we can still call $f$ a function?
The answer you've been waiting for is here. Your function only exists out of elements of set $C$ (a subset of $C$).
So we could make our $f: {(1,a), (2,a), (3,a)}$, which is well-defined, because all elements of the domain are accounted for (i.e. our functions knows how to handle any input)—not injective, because $a$ is mapped to by multiple elements of the domain.
The real question
Your question is really, what if I define $f: {(1,a), (2,b)}$ and leave out $(3,a), (3,b)$. Your function can not handle all domain inputs, therefore is not well defined, and is also not a valid function as most authors would advocate. While you could program a "function" like this, the mathematical definition assumes "well-definedness"—not to mention that your program would be prone to error!
It is possible that your function does not map to all of the codomain, but impossible that your function could "ignore" an element from the domain side of the tuple.
Furthermore, there is a rule of correspondence that says the for any one element on the domain side of the tuple, it may only map it to one element of the codomain side of the tuple; in other words, I cannot use both of these tuples: $(1,a),(1,b)$. My domain element $1$ may not map to both $a$ and $b$ (different elements) of the codomain.
Conclusion
If a function is surjective, then you know two things for certain:
All of the elements of the codomain correspond to elements of the domain.
Because of the definition of a function being well defined, the function must know how to handle all elements of the domain.
See also
Injective? Surjective? Bijective? None?
Can surjective functions map an element from the domain...
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
The definition of a function $f$ from $A$ to $B$, regardless to injectivity or surjectivity, is that the domain of $f$ is $A$, in its entirety.
This means that if $fcolon Ato B$, then for every $ain A$, there is a unique $bin B$ such that the pair $(a,b)in f$.
So the converse holds just for it to be a function from $A$ to $B$.
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
add a comment |
$begingroup$
The definition of a function $f$ from $A$ to $B$, regardless to injectivity or surjectivity, is that the domain of $f$ is $A$, in its entirety.
This means that if $fcolon Ato B$, then for every $ain A$, there is a unique $bin B$ such that the pair $(a,b)in f$.
So the converse holds just for it to be a function from $A$ to $B$.
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
add a comment |
$begingroup$
The definition of a function $f$ from $A$ to $B$, regardless to injectivity or surjectivity, is that the domain of $f$ is $A$, in its entirety.
This means that if $fcolon Ato B$, then for every $ain A$, there is a unique $bin B$ such that the pair $(a,b)in f$.
So the converse holds just for it to be a function from $A$ to $B$.
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
The definition of a function $f$ from $A$ to $B$, regardless to injectivity or surjectivity, is that the domain of $f$ is $A$, in its entirety.
This means that if $fcolon Ato B$, then for every $ain A$, there is a unique $bin B$ such that the pair $(a,b)in f$.
So the converse holds just for it to be a function from $A$ to $B$.
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
answered Dec 17 '14 at 22:42
Asaf Karagila♦Asaf Karagila
306k33438769
306k33438769
add a comment |
add a comment |
$begingroup$
No (to your title question), a function $f : A to B$, by definition, gives each $x in A$ a value $f(x) in B$.
To be precise, you should have said that "for every element in the codomain, there is (at least) one element which is mapped to it by $f$" for your definition of surjectivity.
answered Dec 17 '14 at 22:42
bknbkn
1612
$endgroup$
add a comment |
$begingroup$
No (to your title question), a function $f : A to B$, by definition, gives each $x in A$ a value $f(x) in B$.
To be precise, you should have said that "for every element in the codomain, there is (at least) one element which is mapped to it by $f$" for your definition of surjectivity.
answered Dec 17 '14 at 22:42
bknbkn
1612
$endgroup$
add a comment |
$begingroup$
No (to your title question), a function $f : A to B$, by definition, gives each $x in A$ a value $f(x) in B$.
To be precise, you should have said that "for every element in the codomain, there is (at least) one element which is mapped to it by $f$" for your definition of surjectivity.
answered Dec 17 '14 at 22:42
bknbkn
1612
$endgroup$
No (to your title question), a function $f : A to B$, by definition, gives each $x in A$ a value $f(x) in B$.
To be precise, you should have said that "for every element in the codomain, there is (at least) one element which is mapped to it by $f$" for your definition of surjectivity.
answered Dec 17 '14 at 22:42
bknbkn
1612
answered Dec 17 '14 at 22:42
bknbkn
1612
answered Dec 17 '14 at 22:42
bknbkn
1612
answered Dec 17 '14 at 22:42
bknbkn
1612
1612
add a comment |
add a comment |
$begingroup$
I know exactly why that is confusing, and I will try to explain this step by step. The definition of surjectivity does not mention the domain being overfull. That throws some people off when learning this the first time. The reason for your confusion is likely that you skipped over the Cartesian product and function definition topics without understanding their significance (like I did). In short, the fact that every element of a domain maps to a codomain is implied by the definition of a function (it must be well-defined...read more if you're interested).
Long-winded explanation
So what is a function $f$? Well in set theory, it is a set whose elements are from a relation set, which is called a Cartesian product of sets.
What's that? So given sets $A := {1,2,3}$ and $B := {a,b }$, we could define a relation set between them, but first we have to understand that a relation set is. It is every possible ordered combination (tuples) of elements from each set where the first element corresponds to the first set and the second corresponds to the second set (and so on with $>2$ sets). For example, $C := {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}$ (but not including ${(a,1), (a,2), (b,1), (b,2), (a,3), (b,3)}$, because $anotin A, bnotin A$). Books explain this with a conditional statement like “$A times B := {(x,y)}$, where the $x$'s are from A and the $y$'s are from B”.
So now the critical thing to understand is that a function called $f$ that maps sets $A$ to $B$ ($f : A to B$) takes its elements not directly from $A$, but rather from $C$ above. Now the question is, which elements of $C$ may I use in $f$ such that we can still call $f$ a function?
The answer you've been waiting for is here. Your function only exists out of elements of set $C$ (a subset of $C$).
So we could make our $f: {(1,a), (2,a), (3,a)}$, which is well-defined, because all elements of the domain are accounted for (i.e. our functions knows how to handle any input)—not injective, because $a$ is mapped to by multiple elements of the domain.
The real question
Your question is really, what if I define $f: {(1,a), (2,b)}$ and leave out $(3,a), (3,b)$. Your function can not handle all domain inputs, therefore is not well defined, and is also not a valid function as most authors would advocate. While you could program a "function" like this, the mathematical definition assumes "well-definedness"—not to mention that your program would be prone to error!
It is possible that your function does not map to all of the codomain, but impossible that your function could "ignore" an element from the domain side of the tuple.
Furthermore, there is a rule of correspondence that says the for any one element on the domain side of the tuple, it may only map it to one element of the codomain side of the tuple; in other words, I cannot use both of these tuples: $(1,a),(1,b)$. My domain element $1$ may not map to both $a$ and $b$ (different elements) of the codomain.
Conclusion
If a function is surjective, then you know two things for certain:
All of the elements of the codomain correspond to elements of the domain.
Because of the definition of a function being well defined, the function must know how to handle all elements of the domain.
See also
Injective? Surjective? Bijective? None?
Can surjective functions map an element from the domain...
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
$endgroup$
add a comment |
$begingroup$
I know exactly why that is confusing, and I will try to explain this step by step. The definition of surjectivity does not mention the domain being overfull. That throws some people off when learning this the first time. The reason for your confusion is likely that you skipped over the Cartesian product and function definition topics without understanding their significance (like I did). In short, the fact that every element of a domain maps to a codomain is implied by the definition of a function (it must be well-defined...read more if you're interested).
Long-winded explanation
So what is a function $f$? Well in set theory, it is a set whose elements are from a relation set, which is called a Cartesian product of sets.
What's that? So given sets $A := {1,2,3}$ and $B := {a,b }$, we could define a relation set between them, but first we have to understand that a relation set is. It is every possible ordered combination (tuples) of elements from each set where the first element corresponds to the first set and the second corresponds to the second set (and so on with $>2$ sets). For example, $C := {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}$ (but not including ${(a,1), (a,2), (b,1), (b,2), (a,3), (b,3)}$, because $anotin A, bnotin A$). Books explain this with a conditional statement like “$A times B := {(x,y)}$, where the $x$'s are from A and the $y$'s are from B”.
So now the critical thing to understand is that a function called $f$ that maps sets $A$ to $B$ ($f : A to B$) takes its elements not directly from $A$, but rather from $C$ above. Now the question is, which elements of $C$ may I use in $f$ such that we can still call $f$ a function?
The answer you've been waiting for is here. Your function only exists out of elements of set $C$ (a subset of $C$).
So we could make our $f: {(1,a), (2,a), (3,a)}$, which is well-defined, because all elements of the domain are accounted for (i.e. our functions knows how to handle any input)—not injective, because $a$ is mapped to by multiple elements of the domain.
The real question
Your question is really, what if I define $f: {(1,a), (2,b)}$ and leave out $(3,a), (3,b)$. Your function can not handle all domain inputs, therefore is not well defined, and is also not a valid function as most authors would advocate. While you could program a "function" like this, the mathematical definition assumes "well-definedness"—not to mention that your program would be prone to error!
It is possible that your function does not map to all of the codomain, but impossible that your function could "ignore" an element from the domain side of the tuple.
Furthermore, there is a rule of correspondence that says the for any one element on the domain side of the tuple, it may only map it to one element of the codomain side of the tuple; in other words, I cannot use both of these tuples: $(1,a),(1,b)$. My domain element $1$ may not map to both $a$ and $b$ (different elements) of the codomain.
Conclusion
If a function is surjective, then you know two things for certain:
All of the elements of the codomain correspond to elements of the domain.
Because of the definition of a function being well defined, the function must know how to handle all elements of the domain.
See also
Injective? Surjective? Bijective? None?
Can surjective functions map an element from the domain...
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
$endgroup$
add a comment |
$begingroup$
I know exactly why that is confusing, and I will try to explain this step by step. The definition of surjectivity does not mention the domain being overfull. That throws some people off when learning this the first time. The reason for your confusion is likely that you skipped over the Cartesian product and function definition topics without understanding their significance (like I did). In short, the fact that every element of a domain maps to a codomain is implied by the definition of a function (it must be well-defined...read more if you're interested).
Long-winded explanation
So what is a function $f$? Well in set theory, it is a set whose elements are from a relation set, which is called a Cartesian product of sets.
What's that? So given sets $A := {1,2,3}$ and $B := {a,b }$, we could define a relation set between them, but first we have to understand that a relation set is. It is every possible ordered combination (tuples) of elements from each set where the first element corresponds to the first set and the second corresponds to the second set (and so on with $>2$ sets). For example, $C := {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}$ (but not including ${(a,1), (a,2), (b,1), (b,2), (a,3), (b,3)}$, because $anotin A, bnotin A$). Books explain this with a conditional statement like “$A times B := {(x,y)}$, where the $x$'s are from A and the $y$'s are from B”.
So now the critical thing to understand is that a function called $f$ that maps sets $A$ to $B$ ($f : A to B$) takes its elements not directly from $A$, but rather from $C$ above. Now the question is, which elements of $C$ may I use in $f$ such that we can still call $f$ a function?
The answer you've been waiting for is here. Your function only exists out of elements of set $C$ (a subset of $C$).
So we could make our $f: {(1,a), (2,a), (3,a)}$, which is well-defined, because all elements of the domain are accounted for (i.e. our functions knows how to handle any input)—not injective, because $a$ is mapped to by multiple elements of the domain.
The real question
Your question is really, what if I define $f: {(1,a), (2,b)}$ and leave out $(3,a), (3,b)$. Your function can not handle all domain inputs, therefore is not well defined, and is also not a valid function as most authors would advocate. While you could program a "function" like this, the mathematical definition assumes "well-definedness"—not to mention that your program would be prone to error!
It is possible that your function does not map to all of the codomain, but impossible that your function could "ignore" an element from the domain side of the tuple.
Furthermore, there is a rule of correspondence that says the for any one element on the domain side of the tuple, it may only map it to one element of the codomain side of the tuple; in other words, I cannot use both of these tuples: $(1,a),(1,b)$. My domain element $1$ may not map to both $a$ and $b$ (different elements) of the codomain.
Conclusion
If a function is surjective, then you know two things for certain:
All of the elements of the codomain correspond to elements of the domain.
Because of the definition of a function being well defined, the function must know how to handle all elements of the domain.
See also
Injective? Surjective? Bijective? None?
Can surjective functions map an element from the domain...
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
$endgroup$
I know exactly why that is confusing, and I will try to explain this step by step. The definition of surjectivity does not mention the domain being overfull. That throws some people off when learning this the first time. The reason for your confusion is likely that you skipped over the Cartesian product and function definition topics without understanding their significance (like I did). In short, the fact that every element of a domain maps to a codomain is implied by the definition of a function (it must be well-defined...read more if you're interested).
Long-winded explanation
So what is a function $f$? Well in set theory, it is a set whose elements are from a relation set, which is called a Cartesian product of sets.
What's that? So given sets $A := {1,2,3}$ and $B := {a,b }$, we could define a relation set between them, but first we have to understand that a relation set is. It is every possible ordered combination (tuples) of elements from each set where the first element corresponds to the first set and the second corresponds to the second set (and so on with $>2$ sets). For example, $C := {(1,a), (1,b), (2,a), (2,b), (3,a), (3,b)}$ (but not including ${(a,1), (a,2), (b,1), (b,2), (a,3), (b,3)}$, because $anotin A, bnotin A$). Books explain this with a conditional statement like “$A times B := {(x,y)}$, where the $x$'s are from A and the $y$'s are from B”.
So now the critical thing to understand is that a function called $f$ that maps sets $A$ to $B$ ($f : A to B$) takes its elements not directly from $A$, but rather from $C$ above. Now the question is, which elements of $C$ may I use in $f$ such that we can still call $f$ a function?
The answer you've been waiting for is here. Your function only exists out of elements of set $C$ (a subset of $C$).
So we could make our $f: {(1,a), (2,a), (3,a)}$, which is well-defined, because all elements of the domain are accounted for (i.e. our functions knows how to handle any input)—not injective, because $a$ is mapped to by multiple elements of the domain.
The real question
Your question is really, what if I define $f: {(1,a), (2,b)}$ and leave out $(3,a), (3,b)$. Your function can not handle all domain inputs, therefore is not well defined, and is also not a valid function as most authors would advocate. While you could program a "function" like this, the mathematical definition assumes "well-definedness"—not to mention that your program would be prone to error!
It is possible that your function does not map to all of the codomain, but impossible that your function could "ignore" an element from the domain side of the tuple.
Furthermore, there is a rule of correspondence that says the for any one element on the domain side of the tuple, it may only map it to one element of the codomain side of the tuple; in other words, I cannot use both of these tuples: $(1,a),(1,b)$. My domain element $1$ may not map to both $a$ and $b$ (different elements) of the codomain.
Conclusion
If a function is surjective, then you know two things for certain:
All of the elements of the codomain correspond to elements of the domain.
Because of the definition of a function being well defined, the function must know how to handle all elements of the domain.
See also
Injective? Surjective? Bijective? None?
Can surjective functions map an element from the domain...
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
edited Jan 26 at 20:31
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
answered Jan 25 at 22:32
Jonathan KomarJonathan Komar
1268
1268
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$begingroup$
It's more correct to say a function is surjective has every element in the codomain mapped from something in the domain, ie. every value in the codomain is attained by a preimage in the domain.
$endgroup$
– hardmath
Dec 17 '14 at 22:40
$begingroup$
Where have you seen all these strange (and non-standard) definitions of the standard concept of surjectivity? Some context is needed to clarify what is meant by "domain" and "codomain".
$endgroup$
– Rob Arthan
Dec 17 '14 at 22:51