What is the one-one and onto mapping? [closed]
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I'm reading the following paper, and here I can't understand the first step of the algorithm.
Please explain what is the one-one and onto mapping?
And how he gets these tables ?
Commutative Associative Binary Operations on a Set with Three Elements
Yogesh Kumar Sarita
abstract-algebra group-theory semigroups associativity
asked Apr 25 '16 at 18:19
David Tsaturyan
529
closed as off-topic by user1729, Jendrik Stelzner, user21820, Did, user99914 Sep 12 at 1:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
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up vote
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I'm reading the following paper, and here I can't understand the first step of the algorithm.
Please explain what is the one-one and onto mapping?
And how he gets these tables ?
Commutative Associative Binary Operations on a Set with Three Elements
Yogesh Kumar Sarita
abstract-algebra group-theory semigroups associativity
asked Apr 25 '16 at 18:19
David Tsaturyan
529
closed as off-topic by user1729, Jendrik Stelzner, user21820, Did, user99914 Sep 12 at 1:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
1
A mapping $f$ from a set $A$ to another set $B$ is one-one (also called injective) if no two elements of $A$ are mapped to the same element of $B$. ie if $f(a_1)=f(a_2)$ only holds if $a_1=a_2$ already holds. A mapping is called onto (I hate that word, use surjective instead) if every element of $B$ is the image of some element of $A$ via $f$, ie if for any $b in B$ there is a $a in A$ so that $f(a)=b$.
– s.harp
Apr 25 '16 at 18:21
Thank you. And I have another question. It is from my paper Here, if we define an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a. Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c). How he gets them?
– David Tsaturyan
Apr 25 '16 at 18:42
The "paper" is published in a predatory journal, so perhaps we should chose a different noun to describe it?
– user1729
Sep 3 at 13:45
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm reading the following paper, and here I can't understand the first step of the algorithm.
Please explain what is the one-one and onto mapping?
And how he gets these tables ?
Commutative Associative Binary Operations on a Set with Three Elements
Yogesh Kumar Sarita
abstract-algebra group-theory semigroups associativity
asked Apr 25 '16 at 18:19
David Tsaturyan
529
I'm reading the following paper, and here I can't understand the first step of the algorithm.
Please explain what is the one-one and onto mapping?
And how he gets these tables ?
Commutative Associative Binary Operations on a Set with Three Elements
Yogesh Kumar Sarita
abstract-algebra group-theory semigroups associativity
abstract-algebra group-theory semigroups associativity
asked Apr 25 '16 at 18:19
David Tsaturyan
529
asked Apr 25 '16 at 18:19
David Tsaturyan
529
asked Apr 25 '16 at 18:19
David Tsaturyan
529
asked Apr 25 '16 at 18:19
David Tsaturyan
529
asked Apr 25 '16 at 18:19
David Tsaturyan
529
529
closed as off-topic by user1729, Jendrik Stelzner, user21820, Did, user99914 Sep 12 at 1:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
closed as off-topic by user1729, Jendrik Stelzner, user21820, Did, user99914 Sep 12 at 1:48
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jendrik Stelzner, user21820, Did, Community
1
A mapping $f$ from a set $A$ to another set $B$ is one-one (also called injective) if no two elements of $A$ are mapped to the same element of $B$. ie if $f(a_1)=f(a_2)$ only holds if $a_1=a_2$ already holds. A mapping is called onto (I hate that word, use surjective instead) if every element of $B$ is the image of some element of $A$ via $f$, ie if for any $b in B$ there is a $a in A$ so that $f(a)=b$.
– s.harp
Apr 25 '16 at 18:21
Thank you. And I have another question. It is from my paper Here, if we define an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a. Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c). How he gets them?
– David Tsaturyan
Apr 25 '16 at 18:42
The "paper" is published in a predatory journal, so perhaps we should chose a different noun to describe it?
– user1729
Sep 3 at 13:45
add a comment |Â
1
A mapping $f$ from a set $A$ to another set $B$ is one-one (also called injective) if no two elements of $A$ are mapped to the same element of $B$. ie if $f(a_1)=f(a_2)$ only holds if $a_1=a_2$ already holds. A mapping is called onto (I hate that word, use surjective instead) if every element of $B$ is the image of some element of $A$ via $f$, ie if for any $b in B$ there is a $a in A$ so that $f(a)=b$.
– s.harp
Apr 25 '16 at 18:21
Thank you. And I have another question. It is from my paper Here, if we define an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a. Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c). How he gets them?
– David Tsaturyan
Apr 25 '16 at 18:42
The "paper" is published in a predatory journal, so perhaps we should chose a different noun to describe it?
– user1729
Sep 3 at 13:45
1
1
A mapping $f$ from a set $A$ to another set $B$ is one-one (also called injective) if no two elements of $A$ are mapped to the same element of $B$. ie if $f(a_1)=f(a_2)$ only holds if $a_1=a_2$ already holds. A mapping is called onto (I hate that word, use surjective instead) if every element of $B$ is the image of some element of $A$ via $f$, ie if for any $b in B$ there is a $a in A$ so that $f(a)=b$.
– s.harp
Apr 25 '16 at 18:21
A mapping $f$ from a set $A$ to another set $B$ is one-one (also called injective) if no two elements of $A$ are mapped to the same element of $B$. ie if $f(a_1)=f(a_2)$ only holds if $a_1=a_2$ already holds. A mapping is called onto (I hate that word, use surjective instead) if every element of $B$ is the image of some element of $A$ via $f$, ie if for any $b in B$ there is a $a in A$ so that $f(a)=b$.
– s.harp
Apr 25 '16 at 18:21
Thank you. And I have another question. It is from my paper Here, if we define an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a. Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c). How he gets them?
– David Tsaturyan
Apr 25 '16 at 18:42
Thank you. And I have another question. It is from my paper Here, if we define an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a. Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c). How he gets them?
– David Tsaturyan
Apr 25 '16 at 18:42
The "paper" is published in a predatory journal, so perhaps we should chose a different noun to describe it?
– user1729
Sep 3 at 13:45
The "paper" is published in a predatory journal, so perhaps we should chose a different noun to describe it?
– user1729
Sep 3 at 13:45
add a comment |Â
3 Answers
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I can at least answer the question regarding onto and one-to-one maps. A one-to-one map is an injective map. This means that for a map $T:Vto W$ and $x,yin V$, whenever $T(x)=T(y)$ we must have $x=y$. An onto map is a surjective map. This means that for all $yin W$ there exists some $xin V$ such that $T(x)=y$. Basically, an injective map is one which has a unique input for every output, and a surjective map is one where the image equals the codomain (every $win W$ belongs to the image of $T$).
Hopefully this clarifies your first question at least.
answered Apr 25 '16 at 18:30
Dave
7,91311032
add a comment |Â
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A one-one mapping is a function that maps an element of some domain $V$ to some element in the codomain $W$. It is called one-to-one map if every mapping is unique, that is there are no two elements in the domain $V$ that map to the same element in $W$. For example, if our mapping maps $mathbbR$ to $mathbbR$ and is defined as:
$$T(x) = x$$
Then this is injective (one-to-one). It is easy to see that given some $x$, there is no $y$ such that $y neq x$ and $T(y) = T(x)$ simultaneously.
An onto mapping is when we have a function such that every element in the codomain, there is an element in the domain that maps to it. For example, if our domain is the set of all possible quadratic functions and our co-domain is the set of all positive integers then:
$$T(p(x)) = floor(|p(1)|)$$
Is surjective (onto) since we can always generate a quadratic function to meet every element in our co-domain(to map to $12$ for example, the quadratic $x^2 + 11$ is sufficient).
answered Apr 25 '16 at 18:40
q.Then
2,2271621
add a comment |Â
up vote
0
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A much more intuitive way to think (IMO) is in terms of "at most" and "at least". A function $f$ is injective iff every element of $codom f$ is the image of at most one element of $dom f$. It is surjective iff every element of $codom f$ is the image of at least one element of $dom f$.
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
 |Â
show 6 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I can at least answer the question regarding onto and one-to-one maps. A one-to-one map is an injective map. This means that for a map $T:Vto W$ and $x,yin V$, whenever $T(x)=T(y)$ we must have $x=y$. An onto map is a surjective map. This means that for all $yin W$ there exists some $xin V$ such that $T(x)=y$. Basically, an injective map is one which has a unique input for every output, and a surjective map is one where the image equals the codomain (every $win W$ belongs to the image of $T$).
Hopefully this clarifies your first question at least.
answered Apr 25 '16 at 18:30
Dave
7,91311032
add a comment |Â
up vote
0
down vote
I can at least answer the question regarding onto and one-to-one maps. A one-to-one map is an injective map. This means that for a map $T:Vto W$ and $x,yin V$, whenever $T(x)=T(y)$ we must have $x=y$. An onto map is a surjective map. This means that for all $yin W$ there exists some $xin V$ such that $T(x)=y$. Basically, an injective map is one which has a unique input for every output, and a surjective map is one where the image equals the codomain (every $win W$ belongs to the image of $T$).
Hopefully this clarifies your first question at least.
answered Apr 25 '16 at 18:30
Dave
7,91311032
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I can at least answer the question regarding onto and one-to-one maps. A one-to-one map is an injective map. This means that for a map $T:Vto W$ and $x,yin V$, whenever $T(x)=T(y)$ we must have $x=y$. An onto map is a surjective map. This means that for all $yin W$ there exists some $xin V$ such that $T(x)=y$. Basically, an injective map is one which has a unique input for every output, and a surjective map is one where the image equals the codomain (every $win W$ belongs to the image of $T$).
Hopefully this clarifies your first question at least.
answered Apr 25 '16 at 18:30
Dave
7,91311032
I can at least answer the question regarding onto and one-to-one maps. A one-to-one map is an injective map. This means that for a map $T:Vto W$ and $x,yin V$, whenever $T(x)=T(y)$ we must have $x=y$. An onto map is a surjective map. This means that for all $yin W$ there exists some $xin V$ such that $T(x)=y$. Basically, an injective map is one which has a unique input for every output, and a surjective map is one where the image equals the codomain (every $win W$ belongs to the image of $T$).
Hopefully this clarifies your first question at least.
answered Apr 25 '16 at 18:30
Dave
7,91311032
answered Apr 25 '16 at 18:30
Dave
7,91311032
answered Apr 25 '16 at 18:30
Dave
7,91311032
answered Apr 25 '16 at 18:30
Dave
7,91311032
7,91311032
add a comment |Â
add a comment |Â
up vote
0
down vote
A one-one mapping is a function that maps an element of some domain $V$ to some element in the codomain $W$. It is called one-to-one map if every mapping is unique, that is there are no two elements in the domain $V$ that map to the same element in $W$. For example, if our mapping maps $mathbbR$ to $mathbbR$ and is defined as:
$$T(x) = x$$
Then this is injective (one-to-one). It is easy to see that given some $x$, there is no $y$ such that $y neq x$ and $T(y) = T(x)$ simultaneously.
An onto mapping is when we have a function such that every element in the codomain, there is an element in the domain that maps to it. For example, if our domain is the set of all possible quadratic functions and our co-domain is the set of all positive integers then:
$$T(p(x)) = floor(|p(1)|)$$
Is surjective (onto) since we can always generate a quadratic function to meet every element in our co-domain(to map to $12$ for example, the quadratic $x^2 + 11$ is sufficient).
answered Apr 25 '16 at 18:40
q.Then
2,2271621
add a comment |Â
up vote
0
down vote
A one-one mapping is a function that maps an element of some domain $V$ to some element in the codomain $W$. It is called one-to-one map if every mapping is unique, that is there are no two elements in the domain $V$ that map to the same element in $W$. For example, if our mapping maps $mathbbR$ to $mathbbR$ and is defined as:
$$T(x) = x$$
Then this is injective (one-to-one). It is easy to see that given some $x$, there is no $y$ such that $y neq x$ and $T(y) = T(x)$ simultaneously.
An onto mapping is when we have a function such that every element in the codomain, there is an element in the domain that maps to it. For example, if our domain is the set of all possible quadratic functions and our co-domain is the set of all positive integers then:
$$T(p(x)) = floor(|p(1)|)$$
Is surjective (onto) since we can always generate a quadratic function to meet every element in our co-domain(to map to $12$ for example, the quadratic $x^2 + 11$ is sufficient).
answered Apr 25 '16 at 18:40
q.Then
2,2271621
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A one-one mapping is a function that maps an element of some domain $V$ to some element in the codomain $W$. It is called one-to-one map if every mapping is unique, that is there are no two elements in the domain $V$ that map to the same element in $W$. For example, if our mapping maps $mathbbR$ to $mathbbR$ and is defined as:
$$T(x) = x$$
Then this is injective (one-to-one). It is easy to see that given some $x$, there is no $y$ such that $y neq x$ and $T(y) = T(x)$ simultaneously.
An onto mapping is when we have a function such that every element in the codomain, there is an element in the domain that maps to it. For example, if our domain is the set of all possible quadratic functions and our co-domain is the set of all positive integers then:
$$T(p(x)) = floor(|p(1)|)$$
Is surjective (onto) since we can always generate a quadratic function to meet every element in our co-domain(to map to $12$ for example, the quadratic $x^2 + 11$ is sufficient).
answered Apr 25 '16 at 18:40
q.Then
2,2271621
A one-one mapping is a function that maps an element of some domain $V$ to some element in the codomain $W$. It is called one-to-one map if every mapping is unique, that is there are no two elements in the domain $V$ that map to the same element in $W$. For example, if our mapping maps $mathbbR$ to $mathbbR$ and is defined as:
$$T(x) = x$$
Then this is injective (one-to-one). It is easy to see that given some $x$, there is no $y$ such that $y neq x$ and $T(y) = T(x)$ simultaneously.
An onto mapping is when we have a function such that every element in the codomain, there is an element in the domain that maps to it. For example, if our domain is the set of all possible quadratic functions and our co-domain is the set of all positive integers then:
$$T(p(x)) = floor(|p(1)|)$$
Is surjective (onto) since we can always generate a quadratic function to meet every element in our co-domain(to map to $12$ for example, the quadratic $x^2 + 11$ is sufficient).
answered Apr 25 '16 at 18:40
q.Then
2,2271621
answered Apr 25 '16 at 18:40
q.Then
2,2271621
answered Apr 25 '16 at 18:40
q.Then
2,2271621
answered Apr 25 '16 at 18:40
q.Then
2,2271621
2,2271621
add a comment |Â
add a comment |Â
up vote
0
down vote
A much more intuitive way to think (IMO) is in terms of "at most" and "at least". A function $f$ is injective iff every element of $codom f$ is the image of at most one element of $dom f$. It is surjective iff every element of $codom f$ is the image of at least one element of $dom f$.
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
 |Â
show 6 more comments
up vote
0
down vote
A much more intuitive way to think (IMO) is in terms of "at most" and "at least". A function $f$ is injective iff every element of $codom f$ is the image of at most one element of $dom f$. It is surjective iff every element of $codom f$ is the image of at least one element of $dom f$.
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
 |Â
show 6 more comments
up vote
0
down vote
up vote
0
down vote
A much more intuitive way to think (IMO) is in terms of "at most" and "at least". A function $f$ is injective iff every element of $codom f$ is the image of at most one element of $dom f$. It is surjective iff every element of $codom f$ is the image of at least one element of $dom f$.
A much more intuitive way to think (IMO) is in terms of "at most" and "at least". A function $f$ is injective iff every element of $codom f$ is the image of at most one element of $dom f$. It is surjective iff every element of $codom f$ is the image of at least one element of $dom f$.
answered Apr 25 '16 at 19:40
user325466
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
 |Â
show 6 more comments
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
Please explain this question: We have the set of 3 elements S = a,b,c, and it's defined an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a, after this is written " Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c) " how is it obtained?
– David Tsaturyan
Apr 25 '16 at 19:49
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
There's a missing operator. By definition, $f (a)f (a) = bb $. Maybe you mean $f (f (a)) = c $?
– user325466
Apr 25 '16 at 19:55
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Please look at the picture I added in the question, in the picture all is written.
– David Tsaturyan
Apr 25 '16 at 19:58
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
Maybe you can expand a bit on exactly what's puzzling you. f takes elements to elements, and the way they combine is set by the binary operator, not f.
– user325466
Apr 25 '16 at 20:16
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
I should study the algorithm (the paper attached in my question) for computing all semigroups of order n(here n = 5), and I couldn't understand the first step of the algorithm.
– David Tsaturyan
Apr 25 '16 at 20:28
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1
A mapping $f$ from a set $A$ to another set $B$ is one-one (also called injective) if no two elements of $A$ are mapped to the same element of $B$. ie if $f(a_1)=f(a_2)$ only holds if $a_1=a_2$ already holds. A mapping is called onto (I hate that word, use surjective instead) if every element of $B$ is the image of some element of $A$ via $f$, ie if for any $b in B$ there is a $a in A$ so that $f(a)=b$.
– s.harp
Apr 25 '16 at 18:21
Thank you. And I have another question. It is from my paper Here, if we define an isomorphic mapping from S to S by f(a)=b, f(b)=c, f(c)=a. Then we get f(b)=f(a)f(a), f(a)=f(a)f(b), f(a)=f(a)f(c). How he gets them?
– David Tsaturyan
Apr 25 '16 at 18:42
The "paper" is published in a predatory journal, so perhaps we should chose a different noun to describe it?
– user1729
Sep 3 at 13:45