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 yesterday
Praphulla Koushik
23716
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 yesterday
Praphulla Koushik
23716
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 yesterday
Praphulla Koushik
23716
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 yesterday
Praphulla Koushik
23716
asked yesterday
Praphulla Koushik
23716
edited yesterday
asked yesterday
Praphulla Koushik
23716
asked yesterday
Praphulla Koushik
23716
asked yesterday
Praphulla Koushik
23716
23716
add a comment |
add a comment |
1 Answer
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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 yesterday
Clive Newstead
50.7k474133
:) 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
yesterday
@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
yesterday
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
yesterday
1
@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
yesterday
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
yesterday
|
show 4 more comments
Your Answer
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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 yesterday
Clive Newstead
50.7k474133
:) 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
yesterday
@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
yesterday
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
yesterday
1
@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
yesterday
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
yesterday
|
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 yesterday
Clive Newstead
50.7k474133
:) 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
yesterday
@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
yesterday
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
yesterday
1
@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
yesterday
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
yesterday
|
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 yesterday
Clive Newstead
50.7k474133
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 yesterday
Clive Newstead
50.7k474133
edited yesterday
answered yesterday
Clive Newstead
50.7k474133
answered yesterday
Clive Newstead
50.7k474133
answered yesterday
Clive Newstead
50.7k474133
50.7k474133
:) 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
yesterday
@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
yesterday
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
yesterday
1
@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
yesterday
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
yesterday
|
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
yesterday
@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
yesterday
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
yesterday
1
@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
yesterday
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
yesterday
:) 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
yesterday
:) 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
yesterday
@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
yesterday
@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
yesterday
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
yesterday
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
yesterday
1
1
@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
yesterday
@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
yesterday
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
yesterday
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
yesterday
|
show 4 more comments
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