Do “embedding” and “injective homomorphism” mean the same thing?
$begingroup$
On page 6 of A Shorter Model Theory, it says
For example if $G$ and $H$ are groups and $f : G to H$ is a
homomorphism, then (2.1) says that $f(1^G)=1^H$ and $f(a^{(-1)^{G}}) =
> f(a)^{(-1)^{H}}$. This is exactly the usual definition of homomorphism
between groups. Clause (2.2) adds nothing in this case since there are
no relation symbols in the signature. For the same reason (2.4) is
vacuous for groups. So a homomorphism between groups is an embedding
if and only if it is an injective homomorphism.
Emphasis added.
Hodges always bolds terms when providing a definition like this, which I think means that embedding and injective homomorphism refer to separately-defined notions that happen to line up in this specific case.
Intuitively I think of an embedding and an injective homomorphism as the same thing, not even differing intensionally, but I think this intuition is wrong.
How do embedding and injective homomorphism differ in meaning? Is there a good example that demonstrates this?
definition
$endgroup$
add a comment |
$begingroup$
On page 6 of A Shorter Model Theory, it says
For example if $G$ and $H$ are groups and $f : G to H$ is a
homomorphism, then (2.1) says that $f(1^G)=1^H$ and $f(a^{(-1)^{G}}) =
> f(a)^{(-1)^{H}}$. This is exactly the usual definition of homomorphism
between groups. Clause (2.2) adds nothing in this case since there are
no relation symbols in the signature. For the same reason (2.4) is
vacuous for groups. So a homomorphism between groups is an embedding
if and only if it is an injective homomorphism.
Emphasis added.
Hodges always bolds terms when providing a definition like this, which I think means that embedding and injective homomorphism refer to separately-defined notions that happen to line up in this specific case.
Intuitively I think of an embedding and an injective homomorphism as the same thing, not even differing intensionally, but I think this intuition is wrong.
How do embedding and injective homomorphism differ in meaning? Is there a good example that demonstrates this?
definition
$endgroup$
$begingroup$
The question is what an embedding is, since injective group homomorphism is perfectly clear. The answer depends on the context, e.g., see this question (and now also Matt's answer).
$endgroup$
– Dietrich Burde
Dec 26 '18 at 20:05
$begingroup$
As, it turns out embedding is defined on the previous page as an injective homomorphism where $vec{a} in R^A iff f(vec{a}) in R^B$ instead of the weaker condition $vec{a} in R^A implies f(vec{a}) in R^B$ where $R$ is an arbitrary predicate symbol in our signature and $R^A$ refers to the particular predicate $R$ in the structure $A$.
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:19
add a comment |
$begingroup$
On page 6 of A Shorter Model Theory, it says
For example if $G$ and $H$ are groups and $f : G to H$ is a
homomorphism, then (2.1) says that $f(1^G)=1^H$ and $f(a^{(-1)^{G}}) =
> f(a)^{(-1)^{H}}$. This is exactly the usual definition of homomorphism
between groups. Clause (2.2) adds nothing in this case since there are
no relation symbols in the signature. For the same reason (2.4) is
vacuous for groups. So a homomorphism between groups is an embedding
if and only if it is an injective homomorphism.
Emphasis added.
Hodges always bolds terms when providing a definition like this, which I think means that embedding and injective homomorphism refer to separately-defined notions that happen to line up in this specific case.
Intuitively I think of an embedding and an injective homomorphism as the same thing, not even differing intensionally, but I think this intuition is wrong.
How do embedding and injective homomorphism differ in meaning? Is there a good example that demonstrates this?
definition
$endgroup$
On page 6 of A Shorter Model Theory, it says
For example if $G$ and $H$ are groups and $f : G to H$ is a
homomorphism, then (2.1) says that $f(1^G)=1^H$ and $f(a^{(-1)^{G}}) =
> f(a)^{(-1)^{H}}$. This is exactly the usual definition of homomorphism
between groups. Clause (2.2) adds nothing in this case since there are
no relation symbols in the signature. For the same reason (2.4) is
vacuous for groups. So a homomorphism between groups is an embedding
if and only if it is an injective homomorphism.
Emphasis added.
Hodges always bolds terms when providing a definition like this, which I think means that embedding and injective homomorphism refer to separately-defined notions that happen to line up in this specific case.
Intuitively I think of an embedding and an injective homomorphism as the same thing, not even differing intensionally, but I think this intuition is wrong.
How do embedding and injective homomorphism differ in meaning? Is there a good example that demonstrates this?
definition
definition
asked Dec 26 '18 at 20:02
Gregory NisbetGregory Nisbet
756612
756612
$begingroup$
The question is what an embedding is, since injective group homomorphism is perfectly clear. The answer depends on the context, e.g., see this question (and now also Matt's answer).
$endgroup$
– Dietrich Burde
Dec 26 '18 at 20:05
$begingroup$
As, it turns out embedding is defined on the previous page as an injective homomorphism where $vec{a} in R^A iff f(vec{a}) in R^B$ instead of the weaker condition $vec{a} in R^A implies f(vec{a}) in R^B$ where $R$ is an arbitrary predicate symbol in our signature and $R^A$ refers to the particular predicate $R$ in the structure $A$.
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:19
add a comment |
$begingroup$
The question is what an embedding is, since injective group homomorphism is perfectly clear. The answer depends on the context, e.g., see this question (and now also Matt's answer).
$endgroup$
– Dietrich Burde
Dec 26 '18 at 20:05
$begingroup$
As, it turns out embedding is defined on the previous page as an injective homomorphism where $vec{a} in R^A iff f(vec{a}) in R^B$ instead of the weaker condition $vec{a} in R^A implies f(vec{a}) in R^B$ where $R$ is an arbitrary predicate symbol in our signature and $R^A$ refers to the particular predicate $R$ in the structure $A$.
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:19
$begingroup$
The question is what an embedding is, since injective group homomorphism is perfectly clear. The answer depends on the context, e.g., see this question (and now also Matt's answer).
$endgroup$
– Dietrich Burde
Dec 26 '18 at 20:05
$begingroup$
The question is what an embedding is, since injective group homomorphism is perfectly clear. The answer depends on the context, e.g., see this question (and now also Matt's answer).
$endgroup$
– Dietrich Burde
Dec 26 '18 at 20:05
$begingroup$
As, it turns out embedding is defined on the previous page as an injective homomorphism where $vec{a} in R^A iff f(vec{a}) in R^B$ instead of the weaker condition $vec{a} in R^A implies f(vec{a}) in R^B$ where $R$ is an arbitrary predicate symbol in our signature and $R^A$ refers to the particular predicate $R$ in the structure $A$.
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:19
$begingroup$
As, it turns out embedding is defined on the previous page as an injective homomorphism where $vec{a} in R^A iff f(vec{a}) in R^B$ instead of the weaker condition $vec{a} in R^A implies f(vec{a}) in R^B$ where $R$ is an arbitrary predicate symbol in our signature and $R^A$ refers to the particular predicate $R$ in the structure $A$.
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For groups they are the same thing. For topological spaces, they are not. There can be an injective continuous map $f:Xto Y$ where the image $f(X) $ is not homeomorphic to $X$. A quick example is $X=[0, 2pi)$, $Y=S^1$, $f(x) =e^{ix} $.
$endgroup$
1
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general asan injective homomorphism that additionally satisfies the property [...]
".
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
add a comment |
$begingroup$
Embedding is a term used for the underlying set of space, for instance we may embed the set of integers $mathbb{Z}$ by a list of distinct points ${x_n}inmathbb{R}$. Homomorphism is used for maps that preserve addition and multiplication. For instance we only have two homomorphism $f:mathbb{Z}rightarrowmathbb{R}$, i.e. zero map and identity map. But we can construct many embeddings $g: mathbb{Z}mapstomathbb{R}$ by choosing different lists ${x_n}$
$endgroup$
add a comment |
$begingroup$
Here is a finite example involving partial orders. Intuitively, if we take a partial order that is not a linear order and linearize it, the $le$ relation now has extra pairs of elements in it that were not there originally.
Let $left(*right)^W$ denote the set of pairs inhabiting the $*$ relation in $W$. $*$ itself may be thought of as an abstract predicate or as a set of pairs, $left(*right)^W$ is always a set of pairs.
Let's define a single non-logical binary predicate $le mathop{:} mathcal{D} times mathcal{D} to left{ top, bot right} $ , with $mathcal{D}$ being the domain of the model.
$le$ can also be thought of as a set of pairs $Pleft( mathcal{D} times mathcal{D} right)$ .
The theory of partial orders consists the following laws:
$$ a le a tag{1a}$$
$$ a le b land b le c implies a le c tag{1b} $$
Let's define two partial orders on 3 elements, $A$ and $B$ . Let the three elements be $mathcal{D} = left{ g, h, k right}$ .
A B
g g
/
/ h
h k
k
Let $f : left{ g, h, k right} to left{ g, h, k right} = text{id} $ denote a function from the domain of $A$ to the domain of $B$. I will show that $f$ is an injective homomorphism and that $f$ is not an embedding.
Here (3a-3b) are the elements of $left(leright)^A$ and $left(leright)^B$ .
Note that
$$left(=right)^A = left(=right)^B tag{2} $$
Examining $left(leright)$ specifically.
$$ left{ left(g,hright), left(g,kright) right} cup left(=right)^A = left(leright)^A tag{3a} $$
$$ left{ left(g,hright), left(g,kright), left(h,kright) right} cup left(=right)^B = left(leright)^B tag{3b} $$
From (2) and (3a-3b) it is clear that $left(leright)^A subsetneq left(leright)^B$, which means that $f$ is a homomorphism and a non-embedding. $f$ is injective because the graph of $f$ is the identity function, which is a bijection.
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
For groups they are the same thing. For topological spaces, they are not. There can be an injective continuous map $f:Xto Y$ where the image $f(X) $ is not homeomorphic to $X$. A quick example is $X=[0, 2pi)$, $Y=S^1$, $f(x) =e^{ix} $.
$endgroup$
1
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general asan injective homomorphism that additionally satisfies the property [...]
".
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
add a comment |
$begingroup$
For groups they are the same thing. For topological spaces, they are not. There can be an injective continuous map $f:Xto Y$ where the image $f(X) $ is not homeomorphic to $X$. A quick example is $X=[0, 2pi)$, $Y=S^1$, $f(x) =e^{ix} $.
$endgroup$
1
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general asan injective homomorphism that additionally satisfies the property [...]
".
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
add a comment |
$begingroup$
For groups they are the same thing. For topological spaces, they are not. There can be an injective continuous map $f:Xto Y$ where the image $f(X) $ is not homeomorphic to $X$. A quick example is $X=[0, 2pi)$, $Y=S^1$, $f(x) =e^{ix} $.
$endgroup$
For groups they are the same thing. For topological spaces, they are not. There can be an injective continuous map $f:Xto Y$ where the image $f(X) $ is not homeomorphic to $X$. A quick example is $X=[0, 2pi)$, $Y=S^1$, $f(x) =e^{ix} $.
answered Dec 26 '18 at 20:06
Matt SamuelMatt Samuel
38.8k63769
38.8k63769
1
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general asan injective homomorphism that additionally satisfies the property [...]
".
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
add a comment |
1
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general asan injective homomorphism that additionally satisfies the property [...]
".
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
1
1
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
@Gregory An injective homomorphism has a left inverse. However, the left inverse need not be a homomorphism (that's what happens in my example, it's not continuous). You could require that the left inverse be a homomorphism.
$endgroup$
– Matt Samuel
Dec 26 '18 at 20:15
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general as
an injective homomorphism that additionally satisfies the property [...]
".$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
$begingroup$
I deleted my earlier comment because I saw that I missed the answer on the previous page of the book. It said, roughly, "is there a way to define an embedding in general as
an injective homomorphism that additionally satisfies the property [...]
".$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:23
add a comment |
$begingroup$
Embedding is a term used for the underlying set of space, for instance we may embed the set of integers $mathbb{Z}$ by a list of distinct points ${x_n}inmathbb{R}$. Homomorphism is used for maps that preserve addition and multiplication. For instance we only have two homomorphism $f:mathbb{Z}rightarrowmathbb{R}$, i.e. zero map and identity map. But we can construct many embeddings $g: mathbb{Z}mapstomathbb{R}$ by choosing different lists ${x_n}$
$endgroup$
add a comment |
$begingroup$
Embedding is a term used for the underlying set of space, for instance we may embed the set of integers $mathbb{Z}$ by a list of distinct points ${x_n}inmathbb{R}$. Homomorphism is used for maps that preserve addition and multiplication. For instance we only have two homomorphism $f:mathbb{Z}rightarrowmathbb{R}$, i.e. zero map and identity map. But we can construct many embeddings $g: mathbb{Z}mapstomathbb{R}$ by choosing different lists ${x_n}$
$endgroup$
add a comment |
$begingroup$
Embedding is a term used for the underlying set of space, for instance we may embed the set of integers $mathbb{Z}$ by a list of distinct points ${x_n}inmathbb{R}$. Homomorphism is used for maps that preserve addition and multiplication. For instance we only have two homomorphism $f:mathbb{Z}rightarrowmathbb{R}$, i.e. zero map and identity map. But we can construct many embeddings $g: mathbb{Z}mapstomathbb{R}$ by choosing different lists ${x_n}$
$endgroup$
Embedding is a term used for the underlying set of space, for instance we may embed the set of integers $mathbb{Z}$ by a list of distinct points ${x_n}inmathbb{R}$. Homomorphism is used for maps that preserve addition and multiplication. For instance we only have two homomorphism $f:mathbb{Z}rightarrowmathbb{R}$, i.e. zero map and identity map. But we can construct many embeddings $g: mathbb{Z}mapstomathbb{R}$ by choosing different lists ${x_n}$
answered Dec 26 '18 at 20:24
Dianbin BaoDianbin Bao
346
346
add a comment |
add a comment |
$begingroup$
Here is a finite example involving partial orders. Intuitively, if we take a partial order that is not a linear order and linearize it, the $le$ relation now has extra pairs of elements in it that were not there originally.
Let $left(*right)^W$ denote the set of pairs inhabiting the $*$ relation in $W$. $*$ itself may be thought of as an abstract predicate or as a set of pairs, $left(*right)^W$ is always a set of pairs.
Let's define a single non-logical binary predicate $le mathop{:} mathcal{D} times mathcal{D} to left{ top, bot right} $ , with $mathcal{D}$ being the domain of the model.
$le$ can also be thought of as a set of pairs $Pleft( mathcal{D} times mathcal{D} right)$ .
The theory of partial orders consists the following laws:
$$ a le a tag{1a}$$
$$ a le b land b le c implies a le c tag{1b} $$
Let's define two partial orders on 3 elements, $A$ and $B$ . Let the three elements be $mathcal{D} = left{ g, h, k right}$ .
A B
g g
/
/ h
h k
k
Let $f : left{ g, h, k right} to left{ g, h, k right} = text{id} $ denote a function from the domain of $A$ to the domain of $B$. I will show that $f$ is an injective homomorphism and that $f$ is not an embedding.
Here (3a-3b) are the elements of $left(leright)^A$ and $left(leright)^B$ .
Note that
$$left(=right)^A = left(=right)^B tag{2} $$
Examining $left(leright)$ specifically.
$$ left{ left(g,hright), left(g,kright) right} cup left(=right)^A = left(leright)^A tag{3a} $$
$$ left{ left(g,hright), left(g,kright), left(h,kright) right} cup left(=right)^B = left(leright)^B tag{3b} $$
From (2) and (3a-3b) it is clear that $left(leright)^A subsetneq left(leright)^B$, which means that $f$ is a homomorphism and a non-embedding. $f$ is injective because the graph of $f$ is the identity function, which is a bijection.
$endgroup$
add a comment |
$begingroup$
Here is a finite example involving partial orders. Intuitively, if we take a partial order that is not a linear order and linearize it, the $le$ relation now has extra pairs of elements in it that were not there originally.
Let $left(*right)^W$ denote the set of pairs inhabiting the $*$ relation in $W$. $*$ itself may be thought of as an abstract predicate or as a set of pairs, $left(*right)^W$ is always a set of pairs.
Let's define a single non-logical binary predicate $le mathop{:} mathcal{D} times mathcal{D} to left{ top, bot right} $ , with $mathcal{D}$ being the domain of the model.
$le$ can also be thought of as a set of pairs $Pleft( mathcal{D} times mathcal{D} right)$ .
The theory of partial orders consists the following laws:
$$ a le a tag{1a}$$
$$ a le b land b le c implies a le c tag{1b} $$
Let's define two partial orders on 3 elements, $A$ and $B$ . Let the three elements be $mathcal{D} = left{ g, h, k right}$ .
A B
g g
/
/ h
h k
k
Let $f : left{ g, h, k right} to left{ g, h, k right} = text{id} $ denote a function from the domain of $A$ to the domain of $B$. I will show that $f$ is an injective homomorphism and that $f$ is not an embedding.
Here (3a-3b) are the elements of $left(leright)^A$ and $left(leright)^B$ .
Note that
$$left(=right)^A = left(=right)^B tag{2} $$
Examining $left(leright)$ specifically.
$$ left{ left(g,hright), left(g,kright) right} cup left(=right)^A = left(leright)^A tag{3a} $$
$$ left{ left(g,hright), left(g,kright), left(h,kright) right} cup left(=right)^B = left(leright)^B tag{3b} $$
From (2) and (3a-3b) it is clear that $left(leright)^A subsetneq left(leright)^B$, which means that $f$ is a homomorphism and a non-embedding. $f$ is injective because the graph of $f$ is the identity function, which is a bijection.
$endgroup$
add a comment |
$begingroup$
Here is a finite example involving partial orders. Intuitively, if we take a partial order that is not a linear order and linearize it, the $le$ relation now has extra pairs of elements in it that were not there originally.
Let $left(*right)^W$ denote the set of pairs inhabiting the $*$ relation in $W$. $*$ itself may be thought of as an abstract predicate or as a set of pairs, $left(*right)^W$ is always a set of pairs.
Let's define a single non-logical binary predicate $le mathop{:} mathcal{D} times mathcal{D} to left{ top, bot right} $ , with $mathcal{D}$ being the domain of the model.
$le$ can also be thought of as a set of pairs $Pleft( mathcal{D} times mathcal{D} right)$ .
The theory of partial orders consists the following laws:
$$ a le a tag{1a}$$
$$ a le b land b le c implies a le c tag{1b} $$
Let's define two partial orders on 3 elements, $A$ and $B$ . Let the three elements be $mathcal{D} = left{ g, h, k right}$ .
A B
g g
/
/ h
h k
k
Let $f : left{ g, h, k right} to left{ g, h, k right} = text{id} $ denote a function from the domain of $A$ to the domain of $B$. I will show that $f$ is an injective homomorphism and that $f$ is not an embedding.
Here (3a-3b) are the elements of $left(leright)^A$ and $left(leright)^B$ .
Note that
$$left(=right)^A = left(=right)^B tag{2} $$
Examining $left(leright)$ specifically.
$$ left{ left(g,hright), left(g,kright) right} cup left(=right)^A = left(leright)^A tag{3a} $$
$$ left{ left(g,hright), left(g,kright), left(h,kright) right} cup left(=right)^B = left(leright)^B tag{3b} $$
From (2) and (3a-3b) it is clear that $left(leright)^A subsetneq left(leright)^B$, which means that $f$ is a homomorphism and a non-embedding. $f$ is injective because the graph of $f$ is the identity function, which is a bijection.
$endgroup$
Here is a finite example involving partial orders. Intuitively, if we take a partial order that is not a linear order and linearize it, the $le$ relation now has extra pairs of elements in it that were not there originally.
Let $left(*right)^W$ denote the set of pairs inhabiting the $*$ relation in $W$. $*$ itself may be thought of as an abstract predicate or as a set of pairs, $left(*right)^W$ is always a set of pairs.
Let's define a single non-logical binary predicate $le mathop{:} mathcal{D} times mathcal{D} to left{ top, bot right} $ , with $mathcal{D}$ being the domain of the model.
$le$ can also be thought of as a set of pairs $Pleft( mathcal{D} times mathcal{D} right)$ .
The theory of partial orders consists the following laws:
$$ a le a tag{1a}$$
$$ a le b land b le c implies a le c tag{1b} $$
Let's define two partial orders on 3 elements, $A$ and $B$ . Let the three elements be $mathcal{D} = left{ g, h, k right}$ .
A B
g g
/
/ h
h k
k
Let $f : left{ g, h, k right} to left{ g, h, k right} = text{id} $ denote a function from the domain of $A$ to the domain of $B$. I will show that $f$ is an injective homomorphism and that $f$ is not an embedding.
Here (3a-3b) are the elements of $left(leright)^A$ and $left(leright)^B$ .
Note that
$$left(=right)^A = left(=right)^B tag{2} $$
Examining $left(leright)$ specifically.
$$ left{ left(g,hright), left(g,kright) right} cup left(=right)^A = left(leright)^A tag{3a} $$
$$ left{ left(g,hright), left(g,kright), left(h,kright) right} cup left(=right)^B = left(leright)^B tag{3b} $$
From (2) and (3a-3b) it is clear that $left(leright)^A subsetneq left(leright)^B$, which means that $f$ is a homomorphism and a non-embedding. $f$ is injective because the graph of $f$ is the identity function, which is a bijection.
edited Dec 31 '18 at 3:07
answered Dec 30 '18 at 18:56
Gregory NisbetGregory Nisbet
756612
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$begingroup$
The question is what an embedding is, since injective group homomorphism is perfectly clear. The answer depends on the context, e.g., see this question (and now also Matt's answer).
$endgroup$
– Dietrich Burde
Dec 26 '18 at 20:05
$begingroup$
As, it turns out embedding is defined on the previous page as an injective homomorphism where $vec{a} in R^A iff f(vec{a}) in R^B$ instead of the weaker condition $vec{a} in R^A implies f(vec{a}) in R^B$ where $R$ is an arbitrary predicate symbol in our signature and $R^A$ refers to the particular predicate $R$ in the structure $A$.
$endgroup$
– Gregory Nisbet
Dec 26 '18 at 20:19