Commutative diagram question
Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$
Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?
Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??
If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?
category-theory
asked 2 hours ago
Praphulla Koushik
22215
add a comment |
Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$
Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?
Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??
If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?
category-theory
asked 2 hours ago
Praphulla Koushik
22215
add a comment |
Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$
Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?
Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??
If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?
category-theory
asked 2 hours ago
Praphulla Koushik
22215
Suppose outer square in above diagram is comutative i.e., $ccirc qcirc p=scirc rcirc a$
Further, suppose right side square is commutative i.e., $ccirc q=scirc b$?
Does it imply left side square is commutative i.e., $bcirc p=rcirc a$??
If $s$ has an inverse, it follows trivially. Are there other conditions that confirm this?
category-theory
category-theory
asked 2 hours ago
Praphulla Koushik
22215
asked 2 hours ago
Praphulla Koushik
22215
asked 2 hours ago
Praphulla Koushik
22215
asked 2 hours ago
Praphulla Koushik
22215
asked 2 hours ago
Praphulla Koushik
22215
22215
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the left-hand square commutes, since
$$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$
In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.
So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.
answered 1 hour ago
Clive Newstead
50.6k474133
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
|
show 4 more comments
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
});
}
});
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%2f3062839%2fcommutative-diagram-question%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the left-hand square commutes, since
$$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$
In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.
So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.
answered 1 hour ago
Clive Newstead
50.6k474133
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
|
show 4 more comments
In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the left-hand square commutes, since
$$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$
In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.
So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.
answered 1 hour ago
Clive Newstead
50.6k474133
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
|
show 4 more comments
In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the left-hand square commutes, since
$$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$
In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.
So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.
answered 1 hour ago
Clive Newstead
50.6k474133
In general, no. If $F$ is a terminal object, then the whole diagram and the right square commute automatically (since all composites to $F$ are the unique morphism to the terminal object), and then the left-hand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the left-hand square commutes, since
$$s circ r circ a = c circ q circ p = s circ b circ p quad Rightarrow quad r circ a = b circ p$$
In fact, $s$ being monic is, in a sense, equivalent to the condition in your question. Indeed, suppose $s$ is fixed and, for all such diagrams with $s$ in the bottom-right, commutativity of the outer and right squares implies commutativity of the left-hand square. We prove that $s$ is monic.
So let $f,g : B to E$ and suppose $s circ f = s circ g$. Form the diagram with $a=p=q=mathrm{id}_B$, $b=f$, $r=g$ and $c = s circ f$. Then the right-hand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the left-hand square commutes, and so $f=g$. So $s$ is monic.
answered 1 hour ago
Clive Newstead
50.6k474133
edited 1 hour ago
answered 1 hour ago
Clive Newstead
50.6k474133
answered 1 hour ago
Clive Newstead
50.6k474133
answered 1 hour ago
Clive Newstead
50.6k474133
50.6k474133
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
|
show 4 more comments
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
:) If $s$ is monic, it follows trivially just because of the definition.. I should have added this along with "If $s$ is invertible, it follows immediately"...Thanks anyways :)
– Praphulla Koushik
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a left-inverse—that's split monic). I've updated my answer to prove that, relative to fixed $s$, the condition in your question is equivalent to the assertion that $s$ is monic.
– Clive Newstead
1 hour ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
I dont know whats split monic.. I only know monic and that means it has left inverse.. So, I said for monic it follows... what is monic for you if not left invertible arrow?
– Praphulla Koushik
48 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
@PraphullaKoushik: Having a left inverse implies monic, but being monic does not imply having a left inverse. See monic versus split monic on nLab, and also on Wikipedia. In some categories (e.g. the category of sets) all monomorphisms are split, but that is not the case in all categories. I don't know of any credible source on category theory that defines 'monomorphism' to mean 'has a left inverse'.
– Clive Newstead
45 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
Ok Ok... An arrow $g$ is monic if $gcirc f=gcirc h$ implies $f=h$... An arrow $g$ is called split monic if there exists an arrow $t$ such that $tcirc g=1$... Suppose $g$ is split monic and $gcirc f=gcirc h$.. We have $tcirc gcirc f=tcirc gcirc h$.. As $tcirc g=1$, we have $f=h$... YOu are only using monic in your answer... Though this is not what I expected, this is interesting :) :) Thanks, I learned some new terminology.. :) In some sense, having a section of $s$ confirms what I asked.. :)
– Praphulla Koushik
38 mins ago
|
show 4 more comments
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
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%2f3062839%2fcommutative-diagram-question%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
