injective order preserving homomorphism ϕ of two ordered abelian groups satisfies a<b⇔ϕ(a)<ϕ(b).
0
$begingroup$
Exercise:
Prove that every injective order preserving homomorphism $ϕ$ of two ordered abelian groups satisfies $a < b ⇔ ϕ(a) < ϕ(b)$.
Definition:
A homomorphism $ϕ : G → H$ of ordered abelian groups is called order
preserving if $a ≤ b$ implies that $ ϕ(a) ≤ ϕ(b)$.
Proof of exercise:
$implies:$ from Definition we have $a ≤ b implies ϕ(a) ≤ ϕ(b)$. Homomorphism $ϕ$ is injective so we have $a < b implies ϕ(a) < ϕ(b)$ (?)
$Longleftarrow: ϕ(a) < ϕ(b)$ $and$ $ϕ$ is order preserving injection. I am not sure what to do know. Should I use the fact that kernel of injective function has only one element?
group-theory order-theory
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
$endgroup$
|
show 5 more comments
0
$begingroup$
Exercise:
Prove that every injective order preserving homomorphism $ϕ$ of two ordered abelian groups satisfies $a < b ⇔ ϕ(a) < ϕ(b)$.
Definition:
A homomorphism $ϕ : G → H$ of ordered abelian groups is called order
preserving if $a ≤ b$ implies that $ ϕ(a) ≤ ϕ(b)$.
Proof of exercise:
$implies:$ from Definition we have $a ≤ b implies ϕ(a) ≤ ϕ(b)$. Homomorphism $ϕ$ is injective so we have $a < b implies ϕ(a) < ϕ(b)$ (?)
$Longleftarrow: ϕ(a) < ϕ(b)$ $and$ $ϕ$ is order preserving injection. I am not sure what to do know. Should I use the fact that kernel of injective function has only one element?
group-theory order-theory
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
$endgroup$
4
$begingroup$
If $alt b$, then $aleq b$ and $aneq b$, so $phi(a)leqphi(b)$ and $phi(a)neqphi(b)$, hence $phi(a)lt phi(b)$. So that part is correct, but needs expansion. For the converse, well, we must either have $alt b$, $a=b$, or $agt b$. Why is only the first one possible, given your assumption and what you already proved in the forward direction?
$endgroup$
– Arturo Magidin
Jan 9 at 18:43
$begingroup$
If $ϕ(a)=ϕ(b)$ then $a=b$. Why do we need consider $ϕ(a)=ϕ(b)$, not $ϕ(a) neq ϕ(b)$?
$endgroup$
– Hikicianka
Jan 9 at 19:15
$begingroup$
Huh? I don't understand the question.
$endgroup$
– Arturo Magidin
Jan 9 at 19:19
1
$begingroup$
Your "forward" direction (proving that if $alt b$ then $phi(a)ltphi(b)$) is essentially correct; you just needed a bit more words: $alt b$ is equivalent to $aleq b$ and $aneq b$. So to prove that $phi(a)ltphi(b)$, it is enough to show that $phi(a)leqphi(b)$ and that $phi(a)neqphi(b)$; you get $phi(a)leqphi(b)$ from he fact that $phi$ is order-preserving, and you get that $phi(a)neqphi(b)$ from the fact that $phi$ is one-to-one . You don't need to consider any cases.
$endgroup$
– Arturo Magidin
Jan 9 at 19:59
1
$begingroup$
I am suggesting that to prove the converse, "backward" direction (that if $phi(a)lt phi(b)$, then $alt b$), you could consider cases. You assume that $phi(a)ltphi(b)$. Now, because the group is ordered, you must have either $agt b$, $a=b$, or $alt b$. But if $ageq b$, then since $phi$ is order preserving, then... and therefore, the only possible option is that...
$endgroup$
– Arturo Magidin
Jan 9 at 20:00
|
show 5 more comments
0
0
0
$begingroup$
Exercise:
Prove that every injective order preserving homomorphism $ϕ$ of two ordered abelian groups satisfies $a < b ⇔ ϕ(a) < ϕ(b)$.
Definition:
A homomorphism $ϕ : G → H$ of ordered abelian groups is called order
preserving if $a ≤ b$ implies that $ ϕ(a) ≤ ϕ(b)$.
Proof of exercise:
$implies:$ from Definition we have $a ≤ b implies ϕ(a) ≤ ϕ(b)$. Homomorphism $ϕ$ is injective so we have $a < b implies ϕ(a) < ϕ(b)$ (?)
$Longleftarrow: ϕ(a) < ϕ(b)$ $and$ $ϕ$ is order preserving injection. I am not sure what to do know. Should I use the fact that kernel of injective function has only one element?
group-theory order-theory
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
$endgroup$
Exercise:
Prove that every injective order preserving homomorphism $ϕ$ of two ordered abelian groups satisfies $a < b ⇔ ϕ(a) < ϕ(b)$.
Definition:
A homomorphism $ϕ : G → H$ of ordered abelian groups is called order
preserving if $a ≤ b$ implies that $ ϕ(a) ≤ ϕ(b)$.
Proof of exercise:
$implies:$ from Definition we have $a ≤ b implies ϕ(a) ≤ ϕ(b)$. Homomorphism $ϕ$ is injective so we have $a < b implies ϕ(a) < ϕ(b)$ (?)
$Longleftarrow: ϕ(a) < ϕ(b)$ $and$ $ϕ$ is order preserving injection. I am not sure what to do know. Should I use the fact that kernel of injective function has only one element?
group-theory order-theory
group-theory order-theory
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
asked Jan 9 at 18:38
HikiciankaHikicianka
1448
1448
4
$begingroup$
If $alt b$, then $aleq b$ and $aneq b$, so $phi(a)leqphi(b)$ and $phi(a)neqphi(b)$, hence $phi(a)lt phi(b)$. So that part is correct, but needs expansion. For the converse, well, we must either have $alt b$, $a=b$, or $agt b$. Why is only the first one possible, given your assumption and what you already proved in the forward direction?
$endgroup$
– Arturo Magidin
Jan 9 at 18:43
$begingroup$
If $ϕ(a)=ϕ(b)$ then $a=b$. Why do we need consider $ϕ(a)=ϕ(b)$, not $ϕ(a) neq ϕ(b)$?
$endgroup$
– Hikicianka
Jan 9 at 19:15
$begingroup$
Huh? I don't understand the question.
$endgroup$
– Arturo Magidin
Jan 9 at 19:19
1
$begingroup$
Your "forward" direction (proving that if $alt b$ then $phi(a)ltphi(b)$) is essentially correct; you just needed a bit more words: $alt b$ is equivalent to $aleq b$ and $aneq b$. So to prove that $phi(a)ltphi(b)$, it is enough to show that $phi(a)leqphi(b)$ and that $phi(a)neqphi(b)$; you get $phi(a)leqphi(b)$ from he fact that $phi$ is order-preserving, and you get that $phi(a)neqphi(b)$ from the fact that $phi$ is one-to-one . You don't need to consider any cases.
$endgroup$
– Arturo Magidin
Jan 9 at 19:59
1
$begingroup$
I am suggesting that to prove the converse, "backward" direction (that if $phi(a)lt phi(b)$, then $alt b$), you could consider cases. You assume that $phi(a)ltphi(b)$. Now, because the group is ordered, you must have either $agt b$, $a=b$, or $alt b$. But if $ageq b$, then since $phi$ is order preserving, then... and therefore, the only possible option is that...
$endgroup$
– Arturo Magidin
Jan 9 at 20:00
|
show 5 more comments
4
$begingroup$
If $alt b$, then $aleq b$ and $aneq b$, so $phi(a)leqphi(b)$ and $phi(a)neqphi(b)$, hence $phi(a)lt phi(b)$. So that part is correct, but needs expansion. For the converse, well, we must either have $alt b$, $a=b$, or $agt b$. Why is only the first one possible, given your assumption and what you already proved in the forward direction?
$endgroup$
– Arturo Magidin
Jan 9 at 18:43
$begingroup$
If $ϕ(a)=ϕ(b)$ then $a=b$. Why do we need consider $ϕ(a)=ϕ(b)$, not $ϕ(a) neq ϕ(b)$?
$endgroup$
– Hikicianka
Jan 9 at 19:15
$begingroup$
Huh? I don't understand the question.
$endgroup$
– Arturo Magidin
Jan 9 at 19:19
1
$begingroup$
Your "forward" direction (proving that if $alt b$ then $phi(a)ltphi(b)$) is essentially correct; you just needed a bit more words: $alt b$ is equivalent to $aleq b$ and $aneq b$. So to prove that $phi(a)ltphi(b)$, it is enough to show that $phi(a)leqphi(b)$ and that $phi(a)neqphi(b)$; you get $phi(a)leqphi(b)$ from he fact that $phi$ is order-preserving, and you get that $phi(a)neqphi(b)$ from the fact that $phi$ is one-to-one . You don't need to consider any cases.
$endgroup$
– Arturo Magidin
Jan 9 at 19:59
1
$begingroup$
I am suggesting that to prove the converse, "backward" direction (that if $phi(a)lt phi(b)$, then $alt b$), you could consider cases. You assume that $phi(a)ltphi(b)$. Now, because the group is ordered, you must have either $agt b$, $a=b$, or $alt b$. But if $ageq b$, then since $phi$ is order preserving, then... and therefore, the only possible option is that...
$endgroup$
– Arturo Magidin
Jan 9 at 20:00
4
4
$begingroup$
If $alt b$, then $aleq b$ and $aneq b$, so $phi(a)leqphi(b)$ and $phi(a)neqphi(b)$, hence $phi(a)lt phi(b)$. So that part is correct, but needs expansion. For the converse, well, we must either have $alt b$, $a=b$, or $agt b$. Why is only the first one possible, given your assumption and what you already proved in the forward direction?
$endgroup$
– Arturo Magidin
Jan 9 at 18:43
$begingroup$
If $alt b$, then $aleq b$ and $aneq b$, so $phi(a)leqphi(b)$ and $phi(a)neqphi(b)$, hence $phi(a)lt phi(b)$. So that part is correct, but needs expansion. For the converse, well, we must either have $alt b$, $a=b$, or $agt b$. Why is only the first one possible, given your assumption and what you already proved in the forward direction?
$endgroup$
– Arturo Magidin
Jan 9 at 18:43
$begingroup$
If $ϕ(a)=ϕ(b)$ then $a=b$. Why do we need consider $ϕ(a)=ϕ(b)$, not $ϕ(a) neq ϕ(b)$?
$endgroup$
– Hikicianka
Jan 9 at 19:15
$begingroup$
If $ϕ(a)=ϕ(b)$ then $a=b$. Why do we need consider $ϕ(a)=ϕ(b)$, not $ϕ(a) neq ϕ(b)$?
$endgroup$
– Hikicianka
Jan 9 at 19:15
$begingroup$
Huh? I don't understand the question.
$endgroup$
– Arturo Magidin
Jan 9 at 19:19
$begingroup$
Huh? I don't understand the question.
$endgroup$
– Arturo Magidin
Jan 9 at 19:19
1
1
$begingroup$
Your "forward" direction (proving that if $alt b$ then $phi(a)ltphi(b)$) is essentially correct; you just needed a bit more words: $alt b$ is equivalent to $aleq b$ and $aneq b$. So to prove that $phi(a)ltphi(b)$, it is enough to show that $phi(a)leqphi(b)$ and that $phi(a)neqphi(b)$; you get $phi(a)leqphi(b)$ from he fact that $phi$ is order-preserving, and you get that $phi(a)neqphi(b)$ from the fact that $phi$ is one-to-one . You don't need to consider any cases.
$endgroup$
– Arturo Magidin
Jan 9 at 19:59
$begingroup$
Your "forward" direction (proving that if $alt b$ then $phi(a)ltphi(b)$) is essentially correct; you just needed a bit more words: $alt b$ is equivalent to $aleq b$ and $aneq b$. So to prove that $phi(a)ltphi(b)$, it is enough to show that $phi(a)leqphi(b)$ and that $phi(a)neqphi(b)$; you get $phi(a)leqphi(b)$ from he fact that $phi$ is order-preserving, and you get that $phi(a)neqphi(b)$ from the fact that $phi$ is one-to-one . You don't need to consider any cases.
$endgroup$
– Arturo Magidin
Jan 9 at 19:59
1
1
$begingroup$
I am suggesting that to prove the converse, "backward" direction (that if $phi(a)lt phi(b)$, then $alt b$), you could consider cases. You assume that $phi(a)ltphi(b)$. Now, because the group is ordered, you must have either $agt b$, $a=b$, or $alt b$. But if $ageq b$, then since $phi$ is order preserving, then... and therefore, the only possible option is that...
$endgroup$
– Arturo Magidin
Jan 9 at 20:00
$begingroup$
I am suggesting that to prove the converse, "backward" direction (that if $phi(a)lt phi(b)$, then $alt b$), you could consider cases. You assume that $phi(a)ltphi(b)$. Now, because the group is ordered, you must have either $agt b$, $a=b$, or $alt b$. But if $ageq b$, then since $phi$ is order preserving, then... and therefore, the only possible option is that...
$endgroup$
– Arturo Magidin
Jan 9 at 20:00
|
show 5 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067808%2finjective-order-preserving-homomorphism-%25cf%2595-of-two-ordered-abelian-groups-satisfie%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067808%2finjective-order-preserving-homomorphism-%25cf%2595-of-two-ordered-abelian-groups-satisfie%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
4
$begingroup$
If $alt b$, then $aleq b$ and $aneq b$, so $phi(a)leqphi(b)$ and $phi(a)neqphi(b)$, hence $phi(a)lt phi(b)$. So that part is correct, but needs expansion. For the converse, well, we must either have $alt b$, $a=b$, or $agt b$. Why is only the first one possible, given your assumption and what you already proved in the forward direction?
$endgroup$
– Arturo Magidin
Jan 9 at 18:43
$begingroup$
If $ϕ(a)=ϕ(b)$ then $a=b$. Why do we need consider $ϕ(a)=ϕ(b)$, not $ϕ(a) neq ϕ(b)$?
$endgroup$
– Hikicianka
Jan 9 at 19:15
$begingroup$
Huh? I don't understand the question.
$endgroup$
– Arturo Magidin
Jan 9 at 19:19
1
$begingroup$
Your "forward" direction (proving that if $alt b$ then $phi(a)ltphi(b)$) is essentially correct; you just needed a bit more words: $alt b$ is equivalent to $aleq b$ and $aneq b$. So to prove that $phi(a)ltphi(b)$, it is enough to show that $phi(a)leqphi(b)$ and that $phi(a)neqphi(b)$; you get $phi(a)leqphi(b)$ from he fact that $phi$ is order-preserving, and you get that $phi(a)neqphi(b)$ from the fact that $phi$ is one-to-one . You don't need to consider any cases.
$endgroup$
– Arturo Magidin
Jan 9 at 19:59
1
$begingroup$
I am suggesting that to prove the converse, "backward" direction (that if $phi(a)lt phi(b)$, then $alt b$), you could consider cases. You assume that $phi(a)ltphi(b)$. Now, because the group is ordered, you must have either $agt b$, $a=b$, or $alt b$. But if $ageq b$, then since $phi$ is order preserving, then... and therefore, the only possible option is that...
$endgroup$
– Arturo Magidin
Jan 9 at 20:00