$lambda$-pure morphisms in $lambda$-accessible categories are monos, unclear proof
This is Proposition 2.29 from the book Locally Presentable and Accessible Categories by Jiří Adámek and Jiří Rosický.
Above is a proof that $lambda$-pure morphisms in $lambda$-accessible categories are monos. It is unclear to me why do we create new $h$ instead of taking $bar{v}$ in the line -4 and why such a $h$ exists in the canonical diagram of $B$ by the displayed line -5th?
category-theory morphism monomorphisms
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
asked Nov 26 at 18:22
user122424
1,0771616
add a comment |
This is Proposition 2.29 from the book Locally Presentable and Accessible Categories by Jiří Adámek and Jiří Rosický.
Above is a proof that $lambda$-pure morphisms in $lambda$-accessible categories are monos. It is unclear to me why do we create new $h$ instead of taking $bar{v}$ in the line -4 and why such a $h$ exists in the canonical diagram of $B$ by the displayed line -5th?
category-theory morphism monomorphisms
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
asked Nov 26 at 18:22
user122424
1,0771616
1
I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post.
– Kevin Carlson
Nov 26 at 19:44
Can you provide me with a link to an upload site for math.stackexchange? Thank you.
– user122424
Nov 26 at 20:19
By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet.
– Kevin Carlson
Nov 26 at 20:31
I've corrected the above link to point to the page with jpg.
– user122424
Nov 26 at 20:39
I inserted the image in your question.
– Kevin Carlson
Nov 26 at 21:52
add a comment |
This is Proposition 2.29 from the book Locally Presentable and Accessible Categories by Jiří Adámek and Jiří Rosický.
Above is a proof that $lambda$-pure morphisms in $lambda$-accessible categories are monos. It is unclear to me why do we create new $h$ instead of taking $bar{v}$ in the line -4 and why such a $h$ exists in the canonical diagram of $B$ by the displayed line -5th?
category-theory morphism monomorphisms
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
asked Nov 26 at 18:22
user122424
1,0771616
This is Proposition 2.29 from the book Locally Presentable and Accessible Categories by Jiří Adámek and Jiří Rosický.
Above is a proof that $lambda$-pure morphisms in $lambda$-accessible categories are monos. It is unclear to me why do we create new $h$ instead of taking $bar{v}$ in the line -4 and why such a $h$ exists in the canonical diagram of $B$ by the displayed line -5th?
category-theory morphism monomorphisms
category-theory morphism monomorphisms
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
asked Nov 26 at 18:22
user122424
1,0771616
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
asked Nov 26 at 18:22
user122424
1,0771616
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
edited Dec 18 at 7:20
Asaf Karagila♦
301k32423755
301k32423755
asked Nov 26 at 18:22
user122424
1,0771616
asked Nov 26 at 18:22
user122424
1,0771616
asked Nov 26 at 18:22
user122424
1,0771616
1,0771616
1
I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post.
– Kevin Carlson
Nov 26 at 19:44
Can you provide me with a link to an upload site for math.stackexchange? Thank you.
– user122424
Nov 26 at 20:19
By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet.
– Kevin Carlson
Nov 26 at 20:31
I've corrected the above link to point to the page with jpg.
– user122424
Nov 26 at 20:39
I inserted the image in your question.
– Kevin Carlson
Nov 26 at 21:52
add a comment |
1
I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post.
– Kevin Carlson
Nov 26 at 19:44
Can you provide me with a link to an upload site for math.stackexchange? Thank you.
– user122424
Nov 26 at 20:19
By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet.
– Kevin Carlson
Nov 26 at 20:31
I've corrected the above link to point to the page with jpg.
– user122424
Nov 26 at 20:39
I inserted the image in your question.
– Kevin Carlson
Nov 26 at 21:52
1
1
I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post.
– Kevin Carlson
Nov 26 at 19:44
I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post.
– Kevin Carlson
Nov 26 at 19:44
Can you provide me with a link to an upload site for math.stackexchange? Thank you.
– user122424
Nov 26 at 20:19
Can you provide me with a link to an upload site for math.stackexchange? Thank you.
– user122424
Nov 26 at 20:19
By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet.
– Kevin Carlson
Nov 26 at 20:31
By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet.
– Kevin Carlson
Nov 26 at 20:31
I've corrected the above link to point to the page with jpg.
– user122424
Nov 26 at 20:39
I've corrected the above link to point to the page with jpg.
– user122424
Nov 26 at 20:39
I inserted the image in your question.
– Kevin Carlson
Nov 26 at 21:52
I inserted the image in your question.
– Kevin Carlson
Nov 26 at 21:52
add a comment |
1 Answer
1
active
oldest
votes
The goal of the first part of the proof is to find a morphism of $lambda$-presentable objects which coequalizes $p'$ and $q'$ to which to apply the assumption of purity on $f$. There is no reason why $bar fp'$ should equal $bar f q'$, so it would be useless to apply purity before constructing $h$. That said, the reason $h$ exists is that the canonical diagram of $B$ is filtered. We have parallel maps $bar f p'$ and $bar f q'$ in that diagram, and so there must exist a map $h$ in the diagram coequalizing them.
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
1
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
1
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
|
show 3 more comments
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The goal of the first part of the proof is to find a morphism of $lambda$-presentable objects which coequalizes $p'$ and $q'$ to which to apply the assumption of purity on $f$. There is no reason why $bar fp'$ should equal $bar f q'$, so it would be useless to apply purity before constructing $h$. That said, the reason $h$ exists is that the canonical diagram of $B$ is filtered. We have parallel maps $bar f p'$ and $bar f q'$ in that diagram, and so there must exist a map $h$ in the diagram coequalizing them.
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
1
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
1
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
|
show 3 more comments
The goal of the first part of the proof is to find a morphism of $lambda$-presentable objects which coequalizes $p'$ and $q'$ to which to apply the assumption of purity on $f$. There is no reason why $bar fp'$ should equal $bar f q'$, so it would be useless to apply purity before constructing $h$. That said, the reason $h$ exists is that the canonical diagram of $B$ is filtered. We have parallel maps $bar f p'$ and $bar f q'$ in that diagram, and so there must exist a map $h$ in the diagram coequalizing them.
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
1
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
1
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
|
show 3 more comments
The goal of the first part of the proof is to find a morphism of $lambda$-presentable objects which coequalizes $p'$ and $q'$ to which to apply the assumption of purity on $f$. There is no reason why $bar fp'$ should equal $bar f q'$, so it would be useless to apply purity before constructing $h$. That said, the reason $h$ exists is that the canonical diagram of $B$ is filtered. We have parallel maps $bar f p'$ and $bar f q'$ in that diagram, and so there must exist a map $h$ in the diagram coequalizing them.
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
The goal of the first part of the proof is to find a morphism of $lambda$-presentable objects which coequalizes $p'$ and $q'$ to which to apply the assumption of purity on $f$. There is no reason why $bar fp'$ should equal $bar f q'$, so it would be useless to apply purity before constructing $h$. That said, the reason $h$ exists is that the canonical diagram of $B$ is filtered. We have parallel maps $bar f p'$ and $bar f q'$ in that diagram, and so there must exist a map $h$ in the diagram coequalizing them.
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
answered Nov 27 at 16:31
Kevin Carlson
32.4k23271
32.4k23271
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
1
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
1
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
|
show 3 more comments
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
1
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
1
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
OK. Why do we need coequalizing $bar f p'$ and $bar f q'$ rather than just precomposing them with $h$ and making them equal by this precomposition?
– user122424
Nov 27 at 17:28
1
1
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
The map $hbar{f}$ coequalizes $p'$ and $q'$, i.e. the equation $(hbar{f})p' = (hbar{f})q'$holds.
– Marc Olschok
Nov 27 at 19:22
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
Why such an coequalizer exits ? I know, however, that the diagram is filtetred.
– user122424
Nov 28 at 14:53
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
@user122424 What definition of filtered do you favor? This follows either instantly or quickly from any of them. To be clear, this isn't a coequalizer in the sense of a colimit, it's just an arbitrary cocone over the diagram given by two parallel arrows.
– Kevin Carlson
Nov 28 at 18:06
1
1
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
@user122424 OK, sorry if my wording confused you. It's common to say that a morphism "coequalizes" two other morphisms without meaning that it's actually a coequalizer. But that language doesn't appear in the Adamek-Rosicky proof, and I was just explaining the argument they had already made. Is everything clear now?
– Kevin Carlson
Nov 28 at 18:53
|
show 3 more comments
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1
I get a link to a Czech page containing a link to an .xps file, whatever that is. It's best to embed your images directly in the post.
– Kevin Carlson
Nov 26 at 19:44
Can you provide me with a link to an upload site for math.stackexchange? Thank you.
– user122424
Nov 26 at 20:19
By clicking on the landscape image in the edit mode, you can upload, drag and drop, or copy and paste an image from your computer or from elsewhere on the Internet.
– Kevin Carlson
Nov 26 at 20:31
I've corrected the above link to point to the page with jpg.
– user122424
Nov 26 at 20:39
I inserted the image in your question.
– Kevin Carlson
Nov 26 at 21:52