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What Is The Non-Extra Equivalent Of Coends?

0

$begingroup$

As far as I understand, the coend of a diagram $X:I^text{op}times Ito D$ is an object $xin D_0$ together with a natural isomorphism $alpha:[[I,D]](X,Delta^e_*)cong D(x,*)$ in $[D,text{Set}]$ where I denote with $[[I,D]]$ the category of diagrams $I^text{op}times Ito D$ and extranatural transformations between them as well as with $Delta^e_*:Dto[[I,D]]$ the functor taking an object $y$ to the constant diagram $(i_0,i_1)mapsto y$.

I feel now a bit stupid because I cannot tell what the analogous concept for natural transformations is called. I'm sure I have already seen it but, given $X:Ito D$, what exactly is an object $xin D_0$ together with a natural transformation $alpha:[I,D](X,Delta_*)cong D(x,*)$?

Also, what about the notion $int^i X(i,i)$? The coend does not only depend on the $X(i,i)$ values, right?

category-theory

share|cite|improve this question

edited Mar 17 at 17:58

fweth

asked Mar 17 at 17:00

fwethfweth

1,190713

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure I understand what you've written in the first paragraph. For $[[I,D]](X,Delta^e_*)$ to make sense $Delta^e_*$ should be a functor $I^{op}times Ito D$, but you seem to say it's something else. And I've never heard the phrase "constant extranatural transformation" before and am not sure what you mean by it.
  $endgroup$
  – Malice Vidrine
  Mar 17 at 17:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MaliceVidrine $[[I,D]](X,Delta^e_*)$ should be a functor $Dtotext{Set}$ via $ymapsto[[I,D]](X,Delta^e_y)$, $Delta^e_y$, does that make sense? I wrote 'constant extranatural transformation' which makes no sense, fixed it now.
  $endgroup$
  – fweth
  Mar 17 at 17:43
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also I meant $Dto[[I,D]]$, not $Dto[[Ito D]]$, fixed it now.
  $endgroup$
  – fweth
  Mar 17 at 17:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Got confused with extranatural transformations and diagrams. Fixed some more things, hope it makes sense now.
  $endgroup$
  – fweth
  Mar 17 at 17:51
  

  
  
  
  

add a comment | 

0

$begingroup$

As far as I understand, the coend of a diagram $X:I^text{op}times Ito D$ is an object $xin D_0$ together with a natural isomorphism $alpha:[[I,D]](X,Delta^e_*)cong D(x,*)$ in $[D,text{Set}]$ where I denote with $[[I,D]]$ the category of diagrams $I^text{op}times Ito D$ and extranatural transformations between them as well as with $Delta^e_*:Dto[[I,D]]$ the functor taking an object $y$ to the constant diagram $(i_0,i_1)mapsto y$.

I feel now a bit stupid because I cannot tell what the analogous concept for natural transformations is called. I'm sure I have already seen it but, given $X:Ito D$, what exactly is an object $xin D_0$ together with a natural transformation $alpha:[I,D](X,Delta_*)cong D(x,*)$?

Also, what about the notion $int^i X(i,i)$? The coend does not only depend on the $X(i,i)$ values, right?

category-theory

share|cite|improve this question

edited Mar 17 at 17:58

fweth

asked Mar 17 at 17:00

fwethfweth

1,190713

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure I understand what you've written in the first paragraph. For $[[I,D]](X,Delta^e_*)$ to make sense $Delta^e_*$ should be a functor $I^{op}times Ito D$, but you seem to say it's something else. And I've never heard the phrase "constant extranatural transformation" before and am not sure what you mean by it.
  $endgroup$
  – Malice Vidrine
  Mar 17 at 17:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MaliceVidrine $[[I,D]](X,Delta^e_*)$ should be a functor $Dtotext{Set}$ via $ymapsto[[I,D]](X,Delta^e_y)$, $Delta^e_y$, does that make sense? I wrote 'constant extranatural transformation' which makes no sense, fixed it now.
  $endgroup$
  – fweth
  Mar 17 at 17:43
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also I meant $Dto[[I,D]]$, not $Dto[[Ito D]]$, fixed it now.
  $endgroup$
  – fweth
  Mar 17 at 17:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Got confused with extranatural transformations and diagrams. Fixed some more things, hope it makes sense now.
  $endgroup$
  – fweth
  Mar 17 at 17:51
  

  
  
  
  

add a comment | 

$begingroup$

As far as I understand, the coend of a diagram $X:I^text{op}times Ito D$ is an object $xin D_0$ together with a natural isomorphism $alpha:[[I,D]](X,Delta^e_*)cong D(x,*)$ in $[D,text{Set}]$ where I denote with $[[I,D]]$ the category of diagrams $I^text{op}times Ito D$ and extranatural transformations between them as well as with $Delta^e_*:Dto[[I,D]]$ the functor taking an object $y$ to the constant diagram $(i_0,i_1)mapsto y$.

I feel now a bit stupid because I cannot tell what the analogous concept for natural transformations is called. I'm sure I have already seen it but, given $X:Ito D$, what exactly is an object $xin D_0$ together with a natural transformation $alpha:[I,D](X,Delta_*)cong D(x,*)$?

Also, what about the notion $int^i X(i,i)$? The coend does not only depend on the $X(i,i)$ values, right?

category-theory

share|cite|improve this question

edited Mar 17 at 17:58

fweth

asked Mar 17 at 17:00

fwethfweth

1,190713

$endgroup$

share|cite|improve this question

edited Mar 17 at 17:58

fweth

asked Mar 17 at 17:00

fwethfweth

1,190713

share|cite|improve this question

edited Mar 17 at 17:58

fweth

asked Mar 17 at 17:00

fwethfweth

1,190713

asked Mar 17 at 17:00

fwethfweth

1,190713

asked Mar 17 at 17:00

fwethfweth

1,190713

1 Answer
1

active

oldest

votes

2

$begingroup$

The case for ordinary natural transformations is just that of a colimit. The natural isomorphism $[I,D](X,Delta_*)simeq D(x,*)$ says that there is a natural correspondence between morphisms $xto y$ for any $y$, and cocones under $X$ with vertex $y$ (cocones are just natural transformations to constant functors); and this is just what it means to be a colimit.

And what's important in the notation $int^iX(i,i)$ is that it binds those variable positions--it is not saying that it is only constructed from the "diagonal" values. It can be important to be clear on what variables you're binding because it might happen that you have a diagram $F:Ctimes C^{op}times Cto D$ and can form the coend $int^cF(c,c,c')$ or $int^cF(c',c,c)$ for a given parameter $c'$, and there's no reason these will be the same things; the notation tells you which two arguments you're taking the coend with respect to.

share|cite|improve this answer

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's also true that the coend is a quotient of $coprod X(i,i)$, so the notation captures the sort of "generating" objects fully.
  $endgroup$
  – Kevin Carlson
  Mar 18 at 2:00
  

  
  
  
  

add a comment | 

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2

$begingroup$

The case for ordinary natural transformations is just that of a colimit. The natural isomorphism $[I,D](X,Delta_*)simeq D(x,*)$ says that there is a natural correspondence between morphisms $xto y$ for any $y$, and cocones under $X$ with vertex $y$ (cocones are just natural transformations to constant functors); and this is just what it means to be a colimit.

And what's important in the notation $int^iX(i,i)$ is that it binds those variable positions--it is not saying that it is only constructed from the "diagonal" values. It can be important to be clear on what variables you're binding because it might happen that you have a diagram $F:Ctimes C^{op}times Cto D$ and can form the coend $int^cF(c,c,c')$ or $int^cF(c',c,c)$ for a given parameter $c'$, and there's no reason these will be the same things; the notation tells you which two arguments you're taking the coend with respect to.

share|cite|improve this answer

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's also true that the coend is a quotient of $coprod X(i,i)$, so the notation captures the sort of "generating" objects fully.
  $endgroup$
  – Kevin Carlson
  Mar 18 at 2:00
  

  
  
  
  

add a comment | 

2

$begingroup$

The case for ordinary natural transformations is just that of a colimit. The natural isomorphism $[I,D](X,Delta_*)simeq D(x,*)$ says that there is a natural correspondence between morphisms $xto y$ for any $y$, and cocones under $X$ with vertex $y$ (cocones are just natural transformations to constant functors); and this is just what it means to be a colimit.

And what's important in the notation $int^iX(i,i)$ is that it binds those variable positions--it is not saying that it is only constructed from the "diagonal" values. It can be important to be clear on what variables you're binding because it might happen that you have a diagram $F:Ctimes C^{op}times Cto D$ and can form the coend $int^cF(c,c,c')$ or $int^cF(c',c,c)$ for a given parameter $c'$, and there's no reason these will be the same things; the notation tells you which two arguments you're taking the coend with respect to.

share|cite|improve this answer

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It's also true that the coend is a quotient of $coprod X(i,i)$, so the notation captures the sort of "generating" objects fully.
  $endgroup$
  – Kevin Carlson
  Mar 18 at 2:00
  

  
  
  
  

add a comment | 

$begingroup$

The case for ordinary natural transformations is just that of a colimit. The natural isomorphism $[I,D](X,Delta_*)simeq D(x,*)$ says that there is a natural correspondence between morphisms $xto y$ for any $y$, and cocones under $X$ with vertex $y$ (cocones are just natural transformations to constant functors); and this is just what it means to be a colimit.

And what's important in the notation $int^iX(i,i)$ is that it binds those variable positions--it is not saying that it is only constructed from the "diagonal" values. It can be important to be clear on what variables you're binding because it might happen that you have a diagram $F:Ctimes C^{op}times Cto D$ and can form the coend $int^cF(c,c,c')$ or $int^cF(c',c,c)$ for a given parameter $c'$, and there's no reason these will be the same things; the notation tells you which two arguments you're taking the coend with respect to.

share|cite|improve this answer

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123

$endgroup$

share|cite|improve this answer

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123

answered Mar 17 at 18:32

Malice VidrineMalice Vidrine

6,22421123