Mixing
Commutative diagram question
$begingroup$
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 Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
$endgroup$
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?
category-theory
category-theory
asked Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
asked Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
edited Jan 5 at 18:40
Praphulla Koushik
asked Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
asked Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
asked Jan 5 at 15:37
Praphulla KoushikPraphulla Koushik
27317
27317
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
$endgroup$
$begingroup$
:) 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 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.
$endgroup$
– 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
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1 Answer
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1 Answer
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$begingroup$
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 Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
$endgroup$
$begingroup$
:) 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 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.
$endgroup$
– 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
$begingroup$
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 Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
$endgroup$
$begingroup$
:) 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 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.
$endgroup$
– 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
$begingroup$
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 Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
$endgroup$
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 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 Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
edited Jan 5 at 16:00
answered Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
answered Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
answered Jan 5 at 15:53
Clive NewsteadClive Newstead
51k474134
51k474134
$begingroup$
:) 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 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.
$endgroup$
– 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 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.
$endgroup$
– 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
$begingroup$
:) 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
$begingroup$
:) 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 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.
$endgroup$
– Clive Newstead
Jan 5 at 16:00
$begingroup$
@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.
$endgroup$
– 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
$begingroup$
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
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
$begingroup$
@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
$begingroup$
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
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