Commutative diagram question
Clash Royale CLAN TAG#URR8PPP
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?
$endgroup$
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?
$endgroup$
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?
$endgroup$
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?
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 lefthand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the lefthand 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 bottomright, commutativity of the outer and right squares implies commutativity of the lefthand 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 righthand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the lefthand square commutes, and so $f=g$. So $s$ is monic.
$endgroup$

$begingroup$
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58 
$begingroup$
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00

$begingroup$
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50 
1$begingroup$
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52

$begingroup$
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59

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, [[“$”, “$”], [“\\(“,”\\)”]]);
});
});
}, “mathjaxediting”);
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=”iconimgurwhite” href=”https://imgur.com/”u003eu003c/au003e”,
contentPolicyHtml: “User contributions licensed under u003ca href=”https://creativecommons.org/licenses/bysa/3.0/”u003ecc bysa 3.0 with attribution requiredu003c/au003e u003ca href=”https://stackoverflow.com/legal/contentpolicy”u003e(content policy)u003c/au003e”,
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: “.discardanswer”
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin(‘.newpostlogin’, ‘https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062839%2fcommutativediagramquestion%23newanswer’, ‘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 lefthand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the lefthand 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 bottomright, commutativity of the outer and right squares implies commutativity of the lefthand 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 righthand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the lefthand square commutes, and so $f=g$. So $s$ is monic.
$endgroup$

$begingroup$
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58 
$begingroup$
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00

$begingroup$
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50 
1$begingroup$
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52

$begingroup$
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59

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 lefthand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the lefthand 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 bottomright, commutativity of the outer and right squares implies commutativity of the lefthand 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 righthand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the lefthand square commutes, and so $f=g$. So $s$ is monic.
$endgroup$

$begingroup$
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58 
$begingroup$
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00

$begingroup$
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50 
1$begingroup$
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52

$begingroup$
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59

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 lefthand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the lefthand 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 bottomright, commutativity of the outer and right squares implies commutativity of the lefthand 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 righthand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the lefthand square commutes, and so $f=g$. So $s$ is monic.
$endgroup$
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 lefthand square can be anything at all, commutative or otherwise.
If $s$ is monic, then the lefthand 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 bottomright, commutativity of the outer and right squares implies commutativity of the lefthand 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 righthand square commutes trivially and the outer square commutes since $s circ f = s circ g$. Hence the lefthand square commutes, and so $f=g$. So $s$ is monic.

$begingroup$
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58 
$begingroup$
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00

$begingroup$
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50 
1$begingroup$
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52

$begingroup$
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59

show 4 more comments

$begingroup$
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58 
$begingroup$
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00

$begingroup$
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50 
1$begingroup$
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52

$begingroup$
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58
🙂 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 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 15:58
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00
@PraphullaKoushik: The condition that $s$ is monic does not imply that $s$ is invertible (or even that $s$ has a leftinverse—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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50
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?
$endgroup$
– Praphulla Koushik
Jan 5 at 16:50
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52
@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’.
$endgroup$
– Clive Newstead
Jan 5 at 16:52
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59
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.. 🙂
$endgroup$
– Praphulla Koushik
Jan 5 at 16:59

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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin(‘.newpostlogin’, ‘https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3062839%2fcommutativediagramquestion%23newanswer’, ‘question_page’);
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave(‘#loginlink’);
});
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