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1. • CommentRowNumber1.
   • CommentAuthorEric
   • CommentTimeAug 29th 2010
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   Author: Eric
   Format: MarkdownItexHi, I was going to add some details to [[Godement product]], but I can't reproduce what is there and suspect a typo. >For categories $A,B,C$, if $\alpha: F_1\to G_1 : A\to B$ and $\beta: F_2\to G_2 : B\to C$ are natural transformations of functors, the components $(\alpha * \beta)_M$ of the Godement product $\alpha * \beta: F_2\circ F_1\to G_2\circ G_1$ are defined by any of the two equivalent formulas: $$ (\beta * \alpha)_M = \beta_{F_2 M}\circ G_1(\alpha_M) $$ $$ (\beta * \alpha)_M = G_2(\alpha_M)\circ\beta_{F_1 M} $$ Following MacLane (page 42), the natural transformation $\alpha:F_1\Rightarrow G_1$ implies the existence of a morphism $\alpha_M:F_1(M)\to G_1(M)$. This, together with the natural transformation $\beta:F_2\Rightarrow G_2$, implies $$ \array{ F_2\circ F_1(M) & \stackrel{F_2(\alpha_M)}{\to} & F_2\circ G_1(M) \\ \beta_{F_1(M)}\downarrow && \downarrow \beta_{G_1(M)} \\ G_2\circ F_1(M) & \stackrel{G_2(\alpha_M)}{\to} & G_2\circ G_1(M) } \,. $$ I thought the component of the Godement product, i.e. horizontal composition in Cat, should be the diagonal of this diagram so that $$(\beta\circ\alpha)_M = \beta_{G_1(M)}\circ F_2(\alpha_M) = G_2(\alpha_M)\circ \beta_{F_1(M)}.$$ Is there a typo on the page or am I completely missing the mark? Note: Interchanging $F_2\leftrightarrow G_1$ in the two formulas on the page would give my two formulas.

   Hi,

   I was going to add some details to Godement product, but I can’t reproduce what is there and suspect a typo.

   For categories $A,B,C$, if $\alpha: F_1\to G_1 : A\to B$ and $\beta: F_2\to G_2 : B\to C$ are natural transformations of functors, the components $(\alpha * \beta)_M$ of the Godement product $\alpha * \beta: F_2\circ F_1\to G_2\circ G_1$ are defined by any of the two equivalent formulas:

   $$(\beta * \alpha)_M = \beta_{F_2 M}\circ G_1(\alpha_M)$$ $$(\beta * \alpha)_M = G_2(\alpha_M)\circ\beta_{F_1 M}$$

   Following MacLane (page 42), the natural transformation $\alpha:F_1\Rightarrow G_1$ implies the existence of a morphism $\alpha_M:F_1(M)\to G_1(M)$. This, together with the natural transformation $\beta:F_2\Rightarrow G_2$, implies

   $$\array{ F_2\circ F_1(M) & \stackrel{F_2(\alpha_M)}{\to} & F_2\circ G_1(M) \\ \beta_{F_1(M)}\downarrow && \downarrow \beta_{G_1(M)} \\ G_2\circ F_1(M) & \stackrel{G_2(\alpha_M)}{\to} & G_2\circ G_1(M) } \,.$$

   I thought the component of the Godement product, i.e. horizontal composition in Cat, should be the diagonal of this diagram so that

   $$(\beta\circ\alpha)_M = \beta_{G_1(M)}\circ F_2(\alpha_M) = G_2(\alpha_M)\circ \beta_{F_1(M)}.$$

   Is there a typo on the page or am I completely missing the mark?

   Note: Interchanging $F_2\leftrightarrow G_1$ in the two formulas on the page would give my two formulas.

2. • CommentRowNumber2.
   • CommentAuthorFinnLawler
   • CommentTimeAug 29th 2010
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   Author: FinnLawler
   Format: MarkdownItexYou are right and the page is wrong. Whoever wrote that bit must have mixed up the types of $\alpha$ and $\beta$, as you indicate.

   You are right and the page is wrong. Whoever wrote that bit must have mixed up the types of $\alpha$ and $\beta$, as you indicate.

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeAug 29th 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexI've fixed the page, using Eric's diagram, and made the notation consistent too.

   I’ve fixed the page, using Eric’s diagram, and made the notation consistent too.

4. • CommentRowNumber4.
   • CommentAuthorEric
   • CommentTimeAug 30th 2010
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   Author: Eric
   Format: MarkdownItexThanks. This is kind of neat. Each object $M$ of $A$ gets sent to a morphism $\alpha_M$ in $B$ via $\alpha:F_1\Rightarrow G_1$. Then this morphism gets sent to a commuting square in $C$ via $\beta:F_2\Rightarrow G_2$. In effect, horizontal composition sends each object of $A$ to a commuting square of components in $C$.

   Thanks. This is kind of neat.

   Each object $M$ of $A$ gets sent to a morphism $\alpha_M$ in $B$ via $\alpha:F_1\Rightarrow G_1$. Then this morphism gets sent to a commuting square in $C$ via $\beta:F_2\Rightarrow G_2$.

   In effect, horizontal composition sends each object of $A$ to a commuting square of components in $C$.

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeSep 1st 2010
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItex>In effect, horizontal composition sends each object of $A$ to a commuting square of components in $C$. Right. Then we ignore some of the information; we have a morphism in $C$ which has been decomposed in two different ways (giving the two sides of the square) but we forget about this and only remember the morphism in $C$. (As an exercise in negative thinking: the horizontal composition of *no* transformations is the identity natural transformation on the identity functor on $A$. So each object $M$ of $A$ gives us simply an object of $A$, which is $M$ itself. But we want a morphism of $A$ so we forget that we have an object and use the identity morphism on $M$ instead.)

   In effect, horizontal composition sends each object of $A$ to a commuting square of components in $C$.

   Right. Then we ignore some of the information; we have a morphism in $C$ which has been decomposed in two different ways (giving the two sides of the square) but we forget about this and only remember the morphism in $C$.

   (As an exercise in negative thinking: the horizontal composition of no transformations is the identity natural transformation on the identity functor on $A$. So each object $M$ of $A$ gives us simply an object of $A$, which is $M$ itself. But we want a morphism of $A$ so we forget that we have an object and use the identity morphism on $M$ instead.)

6. • CommentRowNumber6.
   • CommentAuthorEric
   • CommentTimeSep 1st 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexI'm _really_ missing my train rides... >(As an exercise in negative thinking: the horizontal composition of *no* transformations is the identity natural transformation on the identity functor on $A$. So each object $M$ of $A$ gives us simply an object of $A$, which is $M$ itself. But we want a morphism of $A$ so we forget that we have an object and use the identity morphism on $M$ instead.) Is this another way of saying the identity natural transformation on the identity functor on $A$ is the identity with respect to horizontal composition? Negative thinking is always fun, but I'm not very good at it :) I have some notes on this subject that are probably too elementary for you guys to bother with, but might be worth putting on the nLab some time. I'll put it on my personal web first and work out any bugs before transferring it. Time allowing of course. Basically at [[internalization]] (oops, I mean [[internal category]]), I asked if we could walk through a specific example. Now I think a good example to walk through is [[Cat]], i.e. a [[double category]] is a category internal to Cat. Before trying to tackle that, I should write down the Set example in detail. PS: Does a double category (category internal to Cat) count as "categorification" of a category (category internal to Set), i.e. replacing sets with categories?

   I’m really missing my train rides…

   (As an exercise in negative thinking: the horizontal composition of no transformations is the identity natural transformation on the identity functor on $A$. So each object $M$ of $A$ gives us simply an object of $A$, which is $M$ itself. But we want a morphism of $A$ so we forget that we have an object and use the identity morphism on $M$ instead.)

   Is this another way of saying the identity natural transformation on the identity functor on $A$ is the identity with respect to horizontal composition? Negative thinking is always fun, but I’m not very good at it :)

   I have some notes on this subject that are probably too elementary for you guys to bother with, but might be worth putting on the nLab some time. I’ll put it on my personal web first and work out any bugs before transferring it. Time allowing of course.

   Basically at internalization (oops, I mean internal category), I asked if we could walk through a specific example. Now I think a good example to walk through is Cat, i.e. a double category is a category internal to Cat. Before trying to tackle that, I should write down the Set example in detail.

   PS: Does a double category (category internal to Cat) count as “categorification” of a category (category internal to Set), i.e. replacing sets with categories?

7. • CommentRowNumber7.
   • CommentAuthorDavidRoberts
   • CommentTimeSep 1st 2010
   • (edited Sep 2nd 2010)
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItex>PS: Does a double category (category internal to Cat) count as "categorification" of a category (category internal to Set), i.e. replacing sets with categories? sort of. Double groupoids are considered higher dimensional algebra, and this is pretty much along the categorification theme. An amazing result that Ronnie Brown told Grothendieck in their 1980's correspondence (due to a third party - I can't quite remember who at present) is that n-fold groupoids model all homotopy n-types. In particular, double groupoids model all 2-types. This is perhaps not too impressive, as 2-groupoids model all 2-types and these are not too dissimilar, but 3-fold groupoids (groupoids internal to double groupoids) model all 3-types, and one needs genuinely weak 3-groupoids to do this ordinarily. I believe Julie Bergner (Edit: I meant Simona Paoli and got confused) has worked on this more recently (see her article in [[Towards Higher Categories]]).

   PS: Does a double category (category internal to Cat) count as “categorification” of a category (category internal to Set), i.e. replacing sets with categories?

   sort of. Double groupoids are considered higher dimensional algebra, and this is pretty much along the categorification theme. An amazing result that Ronnie Brown told Grothendieck in their 1980’s correspondence (due to a third party - I can’t quite remember who at present) is that n-fold groupoids model all homotopy n-types. In particular, double groupoids model all 2-types. This is perhaps not too impressive, as 2-groupoids model all 2-types and these are not too dissimilar, but 3-fold groupoids (groupoids internal to double groupoids) model all 3-types, and one needs genuinely weak 3-groupoids to do this ordinarily.

   I believe Julie Bergner (Edit: I meant Simona Paoli and got confused) has worked on this more recently (see her article in Towards Higher Categories).

8. • CommentRowNumber8.
   • CommentAuthorEric
   • CommentTimeSep 1st 2010
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   Author: Eric
   Format: MarkdownItexThanks! >I believe Julie Bergner has worked on this more recently (see her article in [[Towards Higher Categories]]). Here is John's version of that page: [[johnbaez:Towards Higher Categories]]. Here is a direct link to [Julie's paper](http://arxiv.org/abs/math/0610239).

   Thanks!

   I believe Julie Bergner has worked on this more recently (see her article in Towards Higher Categories).

   Here is John’s version of that page: Towards Higher Categories (johnbaez). Here is a direct link to Julie’s paper.

9. • CommentRowNumber9.
   • CommentAuthorDavidRoberts
   • CommentTimeSep 1st 2010
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   Author: DavidRoberts
   Format: MarkdownItexThat being said, internal category theory is not quite like categorification in that it is internal to a category, rather than replacing the objects of interesting with categories. But then again Toby was using '2-spaces' in his thesis, which were categories internal to a category of spaces, and 2-groups are categories internal to Grp. And categorification is an art, so what I'm saying is really just my own (ill-formed) opinion :)

   That being said, internal category theory is not quite like categorification in that it is internal to a category, rather than replacing the objects of interesting with categories.

   But then again Toby was using ’2-spaces’ in his thesis, which were categories internal to a category of spaces, and 2-groups are categories internal to Grp.

   And categorification is an art, so what I’m saying is really just my own (ill-formed) opinion :)

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeSep 1st 2010
    • (edited Sep 1st 2010)
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItexThe lesson is that sometimes categorification is internalisation, and sometimes it is not … Really, double categories and $2$-categories are *both* categorifications of categories, just in different ways. One of them got to use the standard number prefix, so the other one had to settle for the tuple adjective. >Is this another way of saying the identity natural transformation on the identity functor on $A$ is the identity with respect to horizontal composition? Yes.

    The lesson is that sometimes categorification is internalisation, and sometimes it is not …

    Really, double categories and $2$-categories are both categorifications of categories, just in different ways. One of them got to use the standard number prefix, so the other one had to settle for the tuple adjective.

    Is this another way of saying the identity natural transformation on the identity functor on $A$ is the identity with respect to horizontal composition?

    Yes.

11. • CommentRowNumber11.
    • CommentAuthorEric
    • CommentTimeSep 1st 2010
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    Author: Eric
    Format: MarkdownItexThank you! By the way, at this point I need to pause and smile. In this thread, I've received two "rights", one "sort of" and one "yes". I'm on a roll! :)

    Thank you! By the way, at this point I need to pause and smile. In this thread, I’ve received two “rights”, one “sort of” and one “yes”. I’m on a roll! :)

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeSep 1st 2010
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    Author: Urs
    Format: MarkdownItex> I believe Julie Bergner has worked on this more recently On n-fold categories? I am not aware of that. Not in the survey areticle on $(\infty,1)$-categories that you linked to, at least. I am wondering which one you mean?

    I believe Julie Bergner has worked on this more recently

    On n-fold categories? I am not aware of that. Not in the survey areticle on $(\infty,1)$-categories that you linked to, at least. I am wondering which one you mean?

13. • CommentRowNumber13.
    • CommentAuthorDavidRoberts
    • CommentTimeSep 2nd 2010
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    Author: DavidRoberts
    Format: MarkdownItexWhoops, I meant Simona Paoli: >The third paper, by Simona Paoli, focuses on a number of algebraic ways of modelling n-types of topological spaces in terms of strict categorical structures. Her focus is on the role of "internal" structures in higher category theory, that is, structures that live in categories other than the category of sets. She surveys the known comparisons among such algebraic models for n-types. It is a part of the comparison project to relate various notions of weak n-groupoid to these strict algebraic models for topological n-types.

    Whoops, I meant Simona Paoli:

    The third paper, by Simona Paoli, focuses on a number of algebraic ways of modelling n-types of topological spaces in terms of strict categorical structures. Her focus is on the role of “internal” structures in higher category theory, that is, structures that live in categories other than the category of sets. She surveys the known comparisons among such algebraic models for n-types. It is a part of the comparison project to relate various notions of weak n-groupoid to these strict algebraic models for topological n-types.

14. • CommentRowNumber14.
    • CommentAuthorzskoda
    • CommentTimeSep 3rd 2010
    • (edited Sep 3rd 2010)
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    Author: zskoda
    Format: MarkdownItexThere were some typoi due me. Poeple now say horizontal product, traditional people like me who say Godement use also the older traditional notation $\beta\star\alpha$ and not $\beta\circ\alpha$ (the latter for vertical composition, because the components are literally usual compositions); the newer notation with circle interchanged is better for higher categories. So I would say $$ (\beta * \alpha)_M := \beta_{G_1 M}\circ G_1(\alpha_M) $$ $$ (\beta * \alpha)_M := G_2(\alpha_M)\circ\beta_{F_1 M} $$ and my typo was that I wrote $\beta_{F_2 M}$ instead of $\beta_{G_1 M}$ in the first formula and that in the preamble I wrote a funny $\alpha\star\beta$ what does not parse to me.

    There were some typoi due me. Poeple now say horizontal product, traditional people like me who say Godement use also the older traditional notation $\beta\star\alpha$ and not $\beta\circ\alpha$ (the latter for vertical composition, because the components are literally usual compositions); the newer notation with circle interchanged is better for higher categories. So I would say

    $$(\beta * \alpha)_M := \beta_{G_1 M}\circ G_1(\alpha_M)$$ $$(\beta * \alpha)_M := G_2(\alpha_M)\circ\beta_{F_1 M}$$

    and my typo was that I wrote $\beta_{F_2 M}$ instead of $\beta_{G_1 M}$ in the first formula and that in the preamble I wrote a funny $\alpha\star\beta$ what does not parse to me.

15. • CommentRowNumber15.
    • CommentAuthorluidnel.maignan
    • CommentTimeApr 22nd 2020
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    Author: luidnel.maignan
    Format: MarkdownItexHi, I have two propositions for this page. The first one is to use the circle symbol for composition from the start and put the star symbol as a comment (currently this is the other way around). The second one is to discuss the relation with the action of natural transformation on arrows. I mean that $\alpha$ sends an object of A to an arrow in B, so using then $\beta$ to send this arrow in B to an arrow in C seems natural and gives an equivalent definition. What do you think ?

    Hi,

    I have two propositions for this page.
    The first one is to use the circle symbol for composition from the start and put the star symbol as a comment (currently this is the other way around).
    The second one is to discuss the relation with the action of natural transformation on arrows. I mean that $\alpha$ sends an object of A to an arrow in B, so using then $\beta$ to send this arrow in B to an arrow in C seems natural and gives an equivalent definition.

    What do you think ?

16. • CommentRowNumber16.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 22nd 2020
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexThe notation on the page is certainly inconsistent and confusing. I see $\circ$ being used for both the vertical and horizontal composition, and both $\circ$ and $;$ and even $:$ being used, and $\ast$ being used in the presence of $;$ but not otherwise. It's strange how so much waffling appears on a short page. Another option would be to use $\circ_n$ or $\ast_n$ for composition across an $n$-cell, so that the Godement composition discussed on the page would be $\circ_0$ or $\ast_0$. Typographically, $\circ_0$ might annoy some people though. (I find it doesn't annoy me: I just looked at [Wikipedia](https://en.wikipedia.org/wiki/Strict_2-category) to see how they do it, and indeed that was their decision.) Another option would be: simply omit symbols when horizontally composing natural transformations, which is what a lot of people do anyway.

    The notation on the page is certainly inconsistent and confusing. I see $\circ$ being used for both the vertical and horizontal composition, and both $\circ$ and $;$ and even $:$ being used, and $\ast$ being used in the presence of $;$ but not otherwise. It’s strange how so much waffling appears on a short page.

    Another option would be to use $\circ_n$ or $\ast_n$ for composition across an $n$-cell, so that the Godement composition discussed on the page would be $\circ_0$ or $\ast_0$. Typographically, $\circ_0$ might annoy some people though. (I find it doesn’t annoy me: I just looked at Wikipedia to see how they do it, and indeed that was their decision.)

    Another option would be: simply omit symbols when horizontally composing natural transformations, which is what a lot of people do anyway.

17. • CommentRowNumber17.
    • CommentAuthorluidnel.maignan
    • CommentTimeApr 25th 2020
    • PermaLink
    Author: luidnel.maignan
    Format: MarkdownItexI like the idea of $\circ_n$ for many reasons: 0-cells not composable, 1-cells composable with $\circ_0$, 2-cells composable with $\circ_0$ and $\circ_1$, etc. This is easy, avoid graphical thinking (horizontal, vertical, left, right, etc), and allows to note only afterwards that $\circ_0$ is coherently used and is a functor, etc. Since the page is short, maybe a little section on notations can be added.

    I like the idea of $\circ_n$ for many reasons: 0-cells not composable, 1-cells composable with $\circ_0$, 2-cells composable with $\circ_0$ and $\circ_1$, etc.
    This is easy, avoid graphical thinking (horizontal, vertical, left, right, etc), and allows to note only afterwards that $\circ_0$ is coherently used and is a functor, etc.
    Since the page is short, maybe a little section on notations can be added.

18. • CommentRowNumber18.
    • CommentAuthornLab edit announcer
    • CommentTimeApr 28th 2020
    • PermaLink
    Author: nLab edit announcer
    Format: MarkdownItexUse $\circ$ as the primary notation and mention $\ast$ instead of the other way around in order to match with the equations. Luidnel Maignan <a href="https://ncatlab.org/nlab/revision/diff/Godement+product/12">diff</a>, <a href="https://ncatlab.org/nlab/revision/Godement+product/12">v12</a>, <a href="https://ncatlab.org/nlab/show/Godement+product">current</a>

    Use $\circ$ as the primary notation and mention $\ast$ instead of the other way around in order to match with the equations.

    Luidnel Maignan

    diff, v12, current

19. • CommentRowNumber19.
    • CommentAuthorluidnel.maignan
    • CommentTimeApr 28th 2020
    • PermaLink
    Author: luidnel.maignan
    Format: TextI guess it is simpler actually to propose a change, so I tried one. Feel free to revert it if this is not appropriate. For now, this is only about the composition symbol, not about my second proposition.

    I guess it is simpler actually to propose a change, so I tried one. Feel free to revert it if this is not appropriate.
    For now, this is only about the composition symbol, not about my second proposition.
20. • CommentRowNumber20.
    • CommentAuthornLab edit announcer
    • CommentTimeApr 28th 2020
    • PermaLink
    Author: nLab edit announcer
    Format: MarkdownItexAdd the equivalent definition $(\beta\circ\alpha)_M = \beta(\alpha_M)$ Luidnel Maignan <a href="https://ncatlab.org/nlab/revision/diff/Godement+product/13">diff</a>, <a href="https://ncatlab.org/nlab/revision/Godement+product/13">v13</a>, <a href="https://ncatlab.org/nlab/show/Godement+product">current</a>

    Add the equivalent definition $(\beta\circ\alpha)_M = \beta(\alpha_M)$

    Luidnel Maignan

    diff, v13, current

21. • CommentRowNumber21.
    • CommentAuthorluidnel.maignan
    • CommentTimeApr 28th 2020
    • PermaLink
    Author: luidnel.maignan
    Format: TextI also made a try for my second proposition which involved minor changes in [[natural transformations]] to be able to refer to it properly.

    I also made a try for my second proposition which involved minor changes in natural transformations to be able to refer to it properly.
22. • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 28th 2020
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexBut it seems to be the same symbol as used for vertical composition. Let me work on it a little later today. The article could also use a little TikZ love.

    But it seems to be the same symbol as used for vertical composition.

    Let me work on it a little later today. The article could also use a little TikZ love.

23. • CommentRowNumber23.
    • CommentAuthorjademaster
    • CommentTimeOct 28th 2021
    • PermaLink
    Author: jademaster
    Format: MarkdownItexFixed broken diagram <a href="https://ncatlab.org/nlab/revision/diff/Godement+product/14">diff</a>, <a href="https://ncatlab.org/nlab/revision/Godement+product/14">v14</a>, <a href="https://ncatlab.org/nlab/show/Godement+product">current</a>

    Fixed broken diagram

    diff, v14, current