Mixing
How can one do mapping from the Cartesian product S x S to S?
$begingroup$
If there is a finite set S, can one map S x S to S?
My guess is that each element of S has an image in S x S. Am I correct? or is there a better explanation?
functions
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
$endgroup$
add a comment |
$begingroup$
If there is a finite set S, can one map S x S to S?
My guess is that each element of S has an image in S x S. Am I correct? or is there a better explanation?
functions
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
$endgroup$
$begingroup$
What about the canonical projections?
$endgroup$
– Hayden
Feb 8 '15 at 1:44
$begingroup$
Oh I don't know what that means ! I'm reading one book and it says Given a set S, a composition law x of S into itself is a mapping from the Cartesian product S x S to S. I don't understand how that can be.
$endgroup$
– Sebastian Santana
Feb 8 '15 at 1:45
$begingroup$
Ah, you're talking about binary operations on a set (Artin, for example, uses the term "composition law"). There are plenty of functions from $Stimes S$ into $S$ for a finite set $S$, but in general we care more about particular sets and particular such functions. Just find any example of a finite group and the group product.
$endgroup$
– Hayden
Feb 8 '15 at 1:49
$begingroup$
For "each element of $S$ [to have] an image in $Stimes S$" you would need to be talking about a map $Sto Stimes S$.
$endgroup$
– whacka
Feb 8 '15 at 2:37
$begingroup$
Ya, its a map / function
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:51
add a comment |
$begingroup$
If there is a finite set S, can one map S x S to S?
My guess is that each element of S has an image in S x S. Am I correct? or is there a better explanation?
functions
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
$endgroup$
If there is a finite set S, can one map S x S to S?
My guess is that each element of S has an image in S x S. Am I correct? or is there a better explanation?
functions
functions
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
edited Feb 8 '15 at 2:29
Andrés E. Caicedo
65.5k8159250
65.5k8159250
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
asked Feb 8 '15 at 1:41
Sebastian SantanaSebastian Santana
1047
1047
$begingroup$
What about the canonical projections?
$endgroup$
– Hayden
Feb 8 '15 at 1:44
$begingroup$
Oh I don't know what that means ! I'm reading one book and it says Given a set S, a composition law x of S into itself is a mapping from the Cartesian product S x S to S. I don't understand how that can be.
$endgroup$
– Sebastian Santana
Feb 8 '15 at 1:45
$begingroup$
Ah, you're talking about binary operations on a set (Artin, for example, uses the term "composition law"). There are plenty of functions from $Stimes S$ into $S$ for a finite set $S$, but in general we care more about particular sets and particular such functions. Just find any example of a finite group and the group product.
$endgroup$
– Hayden
Feb 8 '15 at 1:49
$begingroup$
For "each element of $S$ [to have] an image in $Stimes S$" you would need to be talking about a map $Sto Stimes S$.
$endgroup$
– whacka
Feb 8 '15 at 2:37
$begingroup$
Ya, its a map / function
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:51
add a comment |
$begingroup$
What about the canonical projections?
$endgroup$
– Hayden
Feb 8 '15 at 1:44
$begingroup$
Oh I don't know what that means ! I'm reading one book and it says Given a set S, a composition law x of S into itself is a mapping from the Cartesian product S x S to S. I don't understand how that can be.
$endgroup$
– Sebastian Santana
Feb 8 '15 at 1:45
$begingroup$
Ah, you're talking about binary operations on a set (Artin, for example, uses the term "composition law"). There are plenty of functions from $Stimes S$ into $S$ for a finite set $S$, but in general we care more about particular sets and particular such functions. Just find any example of a finite group and the group product.
$endgroup$
– Hayden
Feb 8 '15 at 1:49
$begingroup$
For "each element of $S$ [to have] an image in $Stimes S$" you would need to be talking about a map $Sto Stimes S$.
$endgroup$
– whacka
Feb 8 '15 at 2:37
$begingroup$
Ya, its a map / function
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:51
$begingroup$
What about the canonical projections?
$endgroup$
– Hayden
Feb 8 '15 at 1:44
$begingroup$
What about the canonical projections?
$endgroup$
– Hayden
Feb 8 '15 at 1:44
$begingroup$
Oh I don't know what that means ! I'm reading one book and it says Given a set S, a composition law x of S into itself is a mapping from the Cartesian product S x S to S. I don't understand how that can be.
$endgroup$
– Sebastian Santana
Feb 8 '15 at 1:45
$begingroup$
Oh I don't know what that means ! I'm reading one book and it says Given a set S, a composition law x of S into itself is a mapping from the Cartesian product S x S to S. I don't understand how that can be.
$endgroup$
– Sebastian Santana
Feb 8 '15 at 1:45
$begingroup$
Ah, you're talking about binary operations on a set (Artin, for example, uses the term "composition law"). There are plenty of functions from $Stimes S$ into $S$ for a finite set $S$, but in general we care more about particular sets and particular such functions. Just find any example of a finite group and the group product.
$endgroup$
– Hayden
Feb 8 '15 at 1:49
$begingroup$
Ah, you're talking about binary operations on a set (Artin, for example, uses the term "composition law"). There are plenty of functions from $Stimes S$ into $S$ for a finite set $S$, but in general we care more about particular sets and particular such functions. Just find any example of a finite group and the group product.
$endgroup$
– Hayden
Feb 8 '15 at 1:49
$begingroup$
For "each element of $S$ [to have] an image in $Stimes S$" you would need to be talking about a map $Sto Stimes S$.
$endgroup$
– whacka
Feb 8 '15 at 2:37
$begingroup$
For "each element of $S$ [to have] an image in $Stimes S$" you would need to be talking about a map $Sto Stimes S$.
$endgroup$
– whacka
Feb 8 '15 at 2:37
$begingroup$
Ya, its a map / function
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:51
$begingroup$
Ya, its a map / function
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not sure exactly what your question is.
I'm not sure what you mean by an element of $S$ having an image in $S times S$.
Yes, one can map $S times S$ to $S$, and I think that you would find this self-evident if you understood what "a map from $S times S$ to $S$" really is.
In that vein, let's construct an example: First, let's define $S$ as $S = {1,2,3}$.
We can define the function (that is, "a map") $f: S times S to S$ as follows:
$$
f(x,y) = max{x,y}
$$
So, in particular, we would say $f(1,2) = f(2,1) = 2$. For any $(x,y) in S times S$, $f$ gives us a single corresponding element of $S$; hence, $f$ maps $S times S$ to $S$. We would say that the image of $(1,2)$, an element of $S times S$ is $2$, an element of $S$.
A more algebraically interesting map is one which is also associative. For example, we can define addition in the context of modular arithmetic as follows:
Take $S = {0,1,2}$. We define the operation "$+$" of addition, which takes two elements in $S$ and gives us another element in $S$. So, we could say that $+$ is a map from $S times S$ to $S$. The rule for this map would be given by the table
$$
begin{array}{c|ccc}
+&0&1&2\
hline
0 & 0 & 1 & 2\
1 & 1 & 2 & 0\
2 & 2 & 0 & 1\
end{array}
$$
So, for example, $+(2,2) = 2+2 = 1$.
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:49
$begingroup$
If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:51
$begingroup$
But why are they ordered that way? Why would we want ordered pairs of a set ?
$endgroup$
– Sebastian Santana
Feb 9 '15 at 9:45
$begingroup$
I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
$endgroup$
– Omnomnomnom
Feb 9 '15 at 12:21
|
show 2 more comments
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm not sure exactly what your question is.
I'm not sure what you mean by an element of $S$ having an image in $S times S$.
Yes, one can map $S times S$ to $S$, and I think that you would find this self-evident if you understood what "a map from $S times S$ to $S$" really is.
In that vein, let's construct an example: First, let's define $S$ as $S = {1,2,3}$.
We can define the function (that is, "a map") $f: S times S to S$ as follows:
$$
f(x,y) = max{x,y}
$$
So, in particular, we would say $f(1,2) = f(2,1) = 2$. For any $(x,y) in S times S$, $f$ gives us a single corresponding element of $S$; hence, $f$ maps $S times S$ to $S$. We would say that the image of $(1,2)$, an element of $S times S$ is $2$, an element of $S$.
A more algebraically interesting map is one which is also associative. For example, we can define addition in the context of modular arithmetic as follows:
Take $S = {0,1,2}$. We define the operation "$+$" of addition, which takes two elements in $S$ and gives us another element in $S$. So, we could say that $+$ is a map from $S times S$ to $S$. The rule for this map would be given by the table
$$
begin{array}{c|ccc}
+&0&1&2\
hline
0 & 0 & 1 & 2\
1 & 1 & 2 & 0\
2 & 2 & 0 & 1\
end{array}
$$
So, for example, $+(2,2) = 2+2 = 1$.
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:49
$begingroup$
If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:51
$begingroup$
But why are they ordered that way? Why would we want ordered pairs of a set ?
$endgroup$
– Sebastian Santana
Feb 9 '15 at 9:45
$begingroup$
I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
$endgroup$
– Omnomnomnom
Feb 9 '15 at 12:21
|
show 2 more comments
$begingroup$
I'm not sure exactly what your question is.
I'm not sure what you mean by an element of $S$ having an image in $S times S$.
Yes, one can map $S times S$ to $S$, and I think that you would find this self-evident if you understood what "a map from $S times S$ to $S$" really is.
In that vein, let's construct an example: First, let's define $S$ as $S = {1,2,3}$.
We can define the function (that is, "a map") $f: S times S to S$ as follows:
$$
f(x,y) = max{x,y}
$$
So, in particular, we would say $f(1,2) = f(2,1) = 2$. For any $(x,y) in S times S$, $f$ gives us a single corresponding element of $S$; hence, $f$ maps $S times S$ to $S$. We would say that the image of $(1,2)$, an element of $S times S$ is $2$, an element of $S$.
A more algebraically interesting map is one which is also associative. For example, we can define addition in the context of modular arithmetic as follows:
Take $S = {0,1,2}$. We define the operation "$+$" of addition, which takes two elements in $S$ and gives us another element in $S$. So, we could say that $+$ is a map from $S times S$ to $S$. The rule for this map would be given by the table
$$
begin{array}{c|ccc}
+&0&1&2\
hline
0 & 0 & 1 & 2\
1 & 1 & 2 & 0\
2 & 2 & 0 & 1\
end{array}
$$
So, for example, $+(2,2) = 2+2 = 1$.
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:49
$begingroup$
If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:51
$begingroup$
But why are they ordered that way? Why would we want ordered pairs of a set ?
$endgroup$
– Sebastian Santana
Feb 9 '15 at 9:45
$begingroup$
I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
$endgroup$
– Omnomnomnom
Feb 9 '15 at 12:21
|
show 2 more comments
$begingroup$
I'm not sure exactly what your question is.
I'm not sure what you mean by an element of $S$ having an image in $S times S$.
Yes, one can map $S times S$ to $S$, and I think that you would find this self-evident if you understood what "a map from $S times S$ to $S$" really is.
In that vein, let's construct an example: First, let's define $S$ as $S = {1,2,3}$.
We can define the function (that is, "a map") $f: S times S to S$ as follows:
$$
f(x,y) = max{x,y}
$$
So, in particular, we would say $f(1,2) = f(2,1) = 2$. For any $(x,y) in S times S$, $f$ gives us a single corresponding element of $S$; hence, $f$ maps $S times S$ to $S$. We would say that the image of $(1,2)$, an element of $S times S$ is $2$, an element of $S$.
A more algebraically interesting map is one which is also associative. For example, we can define addition in the context of modular arithmetic as follows:
Take $S = {0,1,2}$. We define the operation "$+$" of addition, which takes two elements in $S$ and gives us another element in $S$. So, we could say that $+$ is a map from $S times S$ to $S$. The rule for this map would be given by the table
$$
begin{array}{c|ccc}
+&0&1&2\
hline
0 & 0 & 1 & 2\
1 & 1 & 2 & 0\
2 & 2 & 0 & 1\
end{array}
$$
So, for example, $+(2,2) = 2+2 = 1$.
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
I'm not sure exactly what your question is.
I'm not sure what you mean by an element of $S$ having an image in $S times S$.
Yes, one can map $S times S$ to $S$, and I think that you would find this self-evident if you understood what "a map from $S times S$ to $S$" really is.
In that vein, let's construct an example: First, let's define $S$ as $S = {1,2,3}$.
We can define the function (that is, "a map") $f: S times S to S$ as follows:
$$
f(x,y) = max{x,y}
$$
So, in particular, we would say $f(1,2) = f(2,1) = 2$. For any $(x,y) in S times S$, $f$ gives us a single corresponding element of $S$; hence, $f$ maps $S times S$ to $S$. We would say that the image of $(1,2)$, an element of $S times S$ is $2$, an element of $S$.
A more algebraically interesting map is one which is also associative. For example, we can define addition in the context of modular arithmetic as follows:
Take $S = {0,1,2}$. We define the operation "$+$" of addition, which takes two elements in $S$ and gives us another element in $S$. So, we could say that $+$ is a map from $S times S$ to $S$. The rule for this map would be given by the table
$$
begin{array}{c|ccc}
+&0&1&2\
hline
0 & 0 & 1 & 2\
1 & 1 & 2 & 0\
2 & 2 & 0 & 1\
end{array}
$$
So, for example, $+(2,2) = 2+2 = 1$.
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
edited Feb 8 '15 at 2:01
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
answered Feb 8 '15 at 1:54
OmnomnomnomOmnomnomnom
128k791184
128k791184
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:49
$begingroup$
If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:51
$begingroup$
But why are they ordered that way? Why would we want ordered pairs of a set ?
$endgroup$
– Sebastian Santana
Feb 9 '15 at 9:45
$begingroup$
I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
$endgroup$
– Omnomnomnom
Feb 9 '15 at 12:21
|
show 2 more comments
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:49
$begingroup$
If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:51
$begingroup$
But why are they ordered that way? Why would we want ordered pairs of a set ?
$endgroup$
– Sebastian Santana
Feb 9 '15 at 9:45
$begingroup$
I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
$endgroup$
– Omnomnomnom
Feb 9 '15 at 12:21
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Oh lord ! why am I so dumb and stupid? So S x S defines a set containing ordered pairs (x,y) where y = f(x). Which means for every value of x there is an image y when rule f is applied. y is the element if S. S x S gives the pairs that are related together. That is what Cartesian product is, that S x S. Did I get that right?
$endgroup$
– Sebastian Santana
Feb 8 '15 at 2:50
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
$endgroup$
– Omnomnomnom
Feb 8 '15 at 13:49
$begingroup$
Let's start here: $S times S$ is the Cartesian product of $S$ with itself, which the set of all ordered pairs of elements in $S$.
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– Omnomnomnom
Feb 8 '15 at 13:49
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If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
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– Omnomnomnom
Feb 8 '15 at 13:51
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If we wanted to think of the resulting relation as a set, it would be a subset of $(Stimes S)times S$.
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– Omnomnomnom
Feb 8 '15 at 13:51
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But why are they ordered that way? Why would we want ordered pairs of a set ?
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– Sebastian Santana
Feb 9 '15 at 9:45
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But why are they ordered that way? Why would we want ordered pairs of a set ?
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– Sebastian Santana
Feb 9 '15 at 9:45
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I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
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– Omnomnomnom
Feb 9 '15 at 12:21
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I don't understand the question. $Stimes S$ is the set or ordered pairs of elements of $S$.What kind of object would we consider instead of a map from $Stimes S$ to $S$? If you want an example in which $f(x,y)neq f(y,x)$, try $f(x,y)= x-y$.
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– Omnomnomnom
Feb 9 '15 at 12:21
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What about the canonical projections?
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– Hayden
Feb 8 '15 at 1:44
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Oh I don't know what that means ! I'm reading one book and it says Given a set S, a composition law x of S into itself is a mapping from the Cartesian product S x S to S. I don't understand how that can be.
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– Sebastian Santana
Feb 8 '15 at 1:45
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Ah, you're talking about binary operations on a set (Artin, for example, uses the term "composition law"). There are plenty of functions from $Stimes S$ into $S$ for a finite set $S$, but in general we care more about particular sets and particular such functions. Just find any example of a finite group and the group product.
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– Hayden
Feb 8 '15 at 1:49
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For "each element of $S$ [to have] an image in $Stimes S$" you would need to be talking about a map $Sto Stimes S$.
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– whacka
Feb 8 '15 at 2:37
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Ya, its a map / function
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– Sebastian Santana
Feb 8 '15 at 2:51