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1. • CommentRowNumber1.
   • CommentAuthorEric
   • CommentTimeAug 16th 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexI'm beginning to regret my foray into MO, but tried asking [another question](http://mathoverflow.net/questions/35731/conditions-for-natural-transformations-to-exist). If you have any feedback, feel free to keep it here if you prefer. Here's the question: Given a category $C$ with two objects and one non-identity morphism $$a\to b$$ and another similar category $D$ $$x\to y$$ we can define two functors $F,G:C\to D$ with $$F:a\mapsto x, b\mapsto y$$ and $$G:a\mapsto x, b\mapsto x$$ with morphisms doing the only thing they possibly can. A natural transformation $\alpha:F\Rightarrow G$ would require a component $\alpha_b:F(b)\to G(b)$, but there is no morphism $y\to x$, so if I understand this correctly, there is no natural transformation from $F$ to $G$. Is that correct? Is there a clear set of criteria required for there to exist a natural transformation?

   I’m beginning to regret my foray into MO, but tried asking another question.

   If you have any feedback, feel free to keep it here if you prefer. Here’s the question:

   Given a category $C$ with two objects and one non-identity morphism

   $$a\to b$$

   and another similar category $D$

   $$x\to y$$

   we can define two functors $F,G:C\to D$ with

   $$F:a\mapsto x, b\mapsto y$$

   and

   $$G:a\mapsto x, b\mapsto x$$

   with morphisms doing the only thing they possibly can.

   A natural transformation $\alpha:F\Rightarrow G$ would require a component $\alpha_b:F(b)\to G(b)$, but there is no morphism $y\to x$, so if I understand this correctly, there is no natural transformation from $F$ to $G$.

   Is that correct? Is there a clear set of criteria required for there to exist a natural transformation?

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexMy understanding is that MO is a site essentially for professional mathematicians (including graduate students or undergraduates working at a graduate level) to ask research-level questions. I'm a little surprised that Bruce Westbury reacted in quite the way he did -- wouldn't he recognize your name from the Café? -- but that sort of brusqueness is of itself not surprising to me. Anyway, of course you are correct that there is no natural transformation for your example. Continuing the brusqueness perhaps, this is obvious. Beyond the definition of natural transformation, there is no general set of criteria for existence -- it "is what it is". But there are more specific situations where you can work out what the notion really means. For example, you should try to understand what it means if the domain and codomain categories are discrete categories, or more generally categories coming from posets, or coming from groups. It's perhaps worth noting that the definition of natural transformation between functors $F, G: C \to D$ doesn't involve the category structure of $C$ at all! It only involves the category structure of $D$ (where the commuting naturality squares live). So for all intents and purposes, you could take $C$ to be merely a directed graph, and $F, G$ morphisms of directed graphs to the underlying graph of a category -- the notion of natural transformation still makes sense there.

   My understanding is that MO is a site essentially for professional mathematicians (including graduate students or undergraduates working at a graduate level) to ask research-level questions. I’m a little surprised that Bruce Westbury reacted in quite the way he did – wouldn’t he recognize your name from the Café? – but that sort of brusqueness is of itself not surprising to me.

   Anyway, of course you are correct that there is no natural transformation for your example. Continuing the brusqueness perhaps, this is obvious.

   Beyond the definition of natural transformation, there is no general set of criteria for existence – it “is what it is”. But there are more specific situations where you can work out what the notion really means. For example, you should try to understand what it means if the domain and codomain categories are discrete categories, or more generally categories coming from posets, or coming from groups.

   It’s perhaps worth noting that the definition of natural transformation between functors $F, G: C \to D$ doesn’t involve the category structure of $C$ at all! It only involves the category structure of $D$ (where the commuting naturality squares live). So for all intents and purposes, you could take $C$ to be merely a directed graph, and $F, G$ morphisms of directed graphs to the underlying graph of a category – the notion of natural transformation still makes sense there.

3. • CommentRowNumber3.
   • CommentAuthorEric
   • CommentTimeAug 16th 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItex>Continuing the brusqueness perhaps, this is obvious. No worries about brusqueness there. It is obvious to me too. I just never thought about it. It is mildly interesting (enough to ask about it). >For example, you should try to understand what it means if the domain and codomain categories are discrete categories, or more generally categories coming from posets, or coming from groups. Ok. I will try. Thanks :) PS: Welcome back. I hope you had a great vacation :)

   Continuing the brusqueness perhaps, this is obvious.

   No worries about brusqueness there. It is obvious to me too. I just never thought about it. It is mildly interesting (enough to ask about it).

   For example, you should try to understand what it means if the domain and codomain categories are discrete categories, or more generally categories coming from posets, or coming from groups.

   Ok. I will try. Thanks :)

   PS: Welcome back. I hope you had a great vacation :)

4. • CommentRowNumber4.
   • CommentAuthorEric
   • CommentTimeAug 16th 2010
   • PermaLink
   Author: Eric
   Format: MarkdownItexBy the way, I jotted this toy example down trying to find a natural transformation that doesn't have a [[ericforgy:Boundary of a 2-Morphism and the Interchange Law|boundary]]. There actually is an endofunctor $\partial\alpha:D\to D$ such that $$\partial\alpha\circ F = G$$ but the opposite direction does not have a boundary. But even this example does not admit an actual natural transformation. So... there are conditions for a boundary to exist and there are also conditions (maybe not obvious or systematic ones) for a natural transformation to exist. I'm trying to get a sense for when they overlap. Still haven't had a chance to take your advice and focus on discrete categories and posets, but wanted to quickly note the motivation before heading to bed.

   By the way, I jotted this toy example down trying to find a natural transformation that doesn’t have a boundary. There actually is an endofunctor $\partial\alpha:D\to D$ such that

   $$\partial\alpha\circ F = G$$

   but the opposite direction does not have a boundary. But even this example does not admit an actual natural transformation.

   So… there are conditions for a boundary to exist and there are also conditions (maybe not obvious or systematic ones) for a natural transformation to exist. I’m trying to get a sense for when they overlap. Still haven’t had a chance to take your advice and focus on discrete categories and posets, but wanted to quickly note the motivation before heading to bed.

5. • CommentRowNumber5.
   • CommentAuthorEric
   • CommentTimeAug 16th 2010
   • (edited Aug 16th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexNote: I must have forgotten to hit submit the first time I wrote this. Argh. >For example, you should try to understand what it means if the domain and codomain categories are discrete categories... Since discrete categories, i.e. sets, have only identity 1-morphisms, the components of any natural transformation must necessarily be identity morphisms. Therefore, with sets $X$ and $Y$ and functions $f,g:X\to Y$, we must have (from the definition of natural transformation) $f(x) = g(x)$ for every $x\in X$, i.e. the functions must be equal. The only natural transformations are the identity 2-morphisms $1_f:f\Rightarrow f$. Note that in this case, whenever the natural transformation exists, so does its [[ericforgy:Boundary of a 2-Morphism and the Interchange Law|boundary]] since $$\partial 1_f\circ f = f$$ implies $$\partial 1_f = 1_Y.$$ ${}$ >... or more generally categories coming from posets I'll need to think about this one some more, but by now I've drawn enough doodles so that I can make a reasonable conjecture that whenever a natural transformation between functors between posets is defined, so is its boundary. Edit: By the way, why are you so certain that no criteria exists? Couldn't it just be that no one has ever thought of it yet? Aleks Kissinger provides his answer on MO, but the strongest statement he makes is "there are probably no conditions".

   Note: I must have forgotten to hit submit the first time I wrote this. Argh.

   For example, you should try to understand what it means if the domain and codomain categories are discrete categories…

   Since discrete categories, i.e. sets, have only identity 1-morphisms, the components of any natural transformation must necessarily be identity morphisms. Therefore, with sets $X$ and $Y$ and functions $f,g:X\to Y$, we must have (from the definition of natural transformation) $f(x) = g(x)$ for every $x\in X$, i.e. the functions must be equal. The only natural transformations are the identity 2-morphisms $1_f:f\Rightarrow f$.

   Note that in this case, whenever the natural transformation exists, so does its boundary since

   $$\partial 1_f\circ f = f$$

   implies

   $$\partial 1_f = 1_Y.$$

   ${}$

   … or more generally categories coming from posets

   I’ll need to think about this one some more, but by now I’ve drawn enough doodles so that I can make a reasonable conjecture that whenever a natural transformation between functors between posets is defined, so is its boundary.

   Edit: By the way, why are you so certain that no criteria exists? Couldn’t it just be that no one has ever thought of it yet? Aleks Kissinger provides his answer on MO, but the strongest statement he makes is “there are probably no conditions”.

6. • CommentRowNumber6.
   • CommentAuthorTobyBartels
   • CommentTimeAug 16th 2010
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   Author: TobyBartels
   Format: MarkdownItex>I can make a reasonable conjecture that whenever a natural transformation between functors between posets is defined, so is its boundary I'm pretty sure that this is false, so I encourage you to work out what a natural transformation is in this case.

   I can make a reasonable conjecture that whenever a natural transformation between functors between posets is defined, so is its boundary

   I’m pretty sure that this is false, so I encourage you to work out what a natural transformation is in this case.

7. • CommentRowNumber7.
   • CommentAuthorEric
   • CommentTimeAug 17th 2010
   • (edited Aug 17th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItex_Note: Some corrections have been made after initially posting this comment based on feedback from Toby in his subsequent comment below._ >I'm pretty sure that this is false, so I encourage you to work out what a natural transformation is in this case. Note: I started writing this comment hoping to provide the (admittedly weak) evidence, i.e. Cases 1-3 below, I used for the conjecture, but when I got down to Case 4, it provided a counter example, so you are correct. I'll leave the earlier cases here because: 1.) I already typed them and 2.) I think they are somewhat interesting :) As in the original comment, we can start with category $C$ given by $a\to b$ and $D$ given by $x\to y$. Based on my understanding, we have only 3 possible functors in this scenario: $$F:a\mapsto x, b\mapsto y, (a\to b)\mapsto (x\to y)$$ $$G:a\mapsto x, b\mapsto x, (a\to b)\mapsto 1_x$$ $$H:a\mapsto y, b\mapsto y, (a\to b)\mapsto 1_y.$$ Let me denote "would be" natural transformations $$\alpha:F\Rightarrow G, \beta:F\Rightarrow H, \gamma:G\Rightarrow H$$ with "would be" natural transformations in the opposite direction by primes $\alpha'$, $\beta'$, and $\gamma'$. **Case 1:** We've already seen that $\alpha$ is not a natural transformation. Nonetheless, we can still define $\partial\alpha$ by $$\partial\alpha: x\mapsto x, y\mapsto x, (x\to y)\mapsto 1_x$$ so that $$\partial\alpha\circ F = G.$$ **Case 2:** There is a natural transformation $\beta:F\Rightarrow H$ and I believe it is unique with components $$\beta_a = (x\to y)$$ $$\beta_b = 1_y.$$ The boundary is given by $$\partial\beta: x\mapsto y, y\mapsto y, (x\to y)\mapsto 1_y$$ so that $$\partial\beta\circ F = H.$$ **Case 3:** There is a natural transformation $\gamma:G\Rightarrow H$ and I believe it is unique with components $$\gamma_a = (x\to y)$$ $$\gamma_b = (x\to y).$$ The boundary is given by $$\partial\gamma: x \mapsto y, y \mapsto y, (x \to y) \mapsto 1_y$$ so that $$\partial\gamma\circ G = H.$$ **Case 4:** There is a natural transformation $\alpha':G\Rightarrow F$ with components $$\alpha'_a = 1_x$$ $$\alpha'_b = (x\to y).$$ The boundary is given by $$\partial\alpha': $$ Oops! There is no boundary. This is actually the first example I drew up that does not have a boundary, but then I got distracted by the fact that $\alpha$ was not even a natural transformation (yet had a boundary, i.e. Case 1 above) and didn't yet observe that $\alpha'$ IS a natural transformation. This provides an example of a natural transformation that does not have a boundary. Good. This marks some progress. Edit: Since there are only two cases remaining, I might as well complete them. **Case 5:** $\beta'$ is not a natural transformation and it does not have a boundary. **Case 6:** $\gamma'$ is not a natural transformation, but it has a boundary given by $$\partial\gamma': x \mapsto x, y \mapsto x, (x \to y) \mapsto 1_x.$$ _Thanks again Toby._

   Note: Some corrections have been made after initially posting this comment based on feedback from Toby in his subsequent comment below.

   I’m pretty sure that this is false, so I encourage you to work out what a natural transformation is in this case.

   Note: I started writing this comment hoping to provide the (admittedly weak) evidence, i.e. Cases 1-3 below, I used for the conjecture, but when I got down to Case 4, it provided a counter example, so you are correct. I’ll leave the earlier cases here because: 1.) I already typed them and 2.) I think they are somewhat interesting :)

   As in the original comment, we can start with category $C$ given by $a\to b$ and $D$ given by $x\to y$. Based on my understanding, we have only 3 possible functors in this scenario:

   $$F:a\mapsto x, b\mapsto y, (a\to b)\mapsto (x\to y)$$ $$G:a\mapsto x, b\mapsto x, (a\to b)\mapsto 1_x$$ $$H:a\mapsto y, b\mapsto y, (a\to b)\mapsto 1_y.$$

   Let me denote “would be” natural transformations

   $$\alpha:F\Rightarrow G, \beta:F\Rightarrow H, \gamma:G\Rightarrow H$$

   with “would be” natural transformations in the opposite direction by primes $\alpha'$, $\beta'$, and $\gamma'$.

   Case 1:

   We’ve already seen that $\alpha$ is not a natural transformation. Nonetheless, we can still define $\partial\alpha$ by

   $$\partial\alpha: x\mapsto x, y\mapsto x, (x\to y)\mapsto 1_x$$

   so that

   $$\partial\alpha\circ F = G.$$

   Case 2:

   There is a natural transformation $\beta:F\Rightarrow H$ and I believe it is unique with components

   $$\beta_a = (x\to y)$$ $$\beta_b = 1_y.$$

   The boundary is given by

   $$\partial\beta: x\mapsto y, y\mapsto y, (x\to y)\mapsto 1_y$$

   so that

   $$\partial\beta\circ F = H.$$

   Case 3:

   There is a natural transformation $\gamma:G\Rightarrow H$ and I believe it is unique with components

   $$\gamma_a = (x\to y)$$ $$\gamma_b = (x\to y).$$

   The boundary is given by

   $$\partial\gamma: x \mapsto y, y \mapsto y, (x \to y) \mapsto 1_y$$

   so that

   $$\partial\gamma\circ G = H.$$

   Case 4:

   There is a natural transformation $\alpha':G\Rightarrow F$ with components

   $$\alpha'_a = 1_x$$ $$\alpha'_b = (x\to y).$$

   The boundary is given by

   $$\partial\alpha':$$

   Oops! There is no boundary. This is actually the first example I drew up that does not have a boundary, but then I got distracted by the fact that $\alpha$ was not even a natural transformation (yet had a boundary, i.e. Case 1 above) and didn’t yet observe that $\alpha'$ IS a natural transformation.

   This provides an example of a natural transformation that does not have a boundary.

   Good. This marks some progress.

   Edit: Since there are only two cases remaining, I might as well complete them.

   Case 5:

   $\beta'$ is not a natural transformation and it does not have a boundary.

   Case 6:

   $\gamma'$ is not a natural transformation, but it has a boundary given by

   $$\partial\gamma': x \mapsto x, y \mapsto x, (x \to y) \mapsto 1_x.$$

   Thanks again Toby.

8. • CommentRowNumber8.
   • CommentAuthorTobyBartels
   • CommentTimeAug 17th 2010
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   Author: TobyBartels
   Format: MarkdownItexI agree with everything you say except for the typo in the first line of case 2 and the boundaries of $\gamma$ and $\gamma'$. I think that $\partial\gamma$ should be $(x \mapsto y, y \mapsto y, (x \to y) \mapsto 1_y)$. I also think that $\partial\gamma'$ exists and should be $(x \mapsto x, y \mapsto x, (x \to y) \mapsto 1_x)$.

   I agree with everything you say except for the typo in the first line of case 2 and the boundaries of $\gamma$ and $\gamma'$.

   I think that $\partial\gamma$ should be $(x \mapsto y, y \mapsto y, (x \to y) \mapsto 1_y)$.

   I also think that $\partial\gamma'$ exists and should be $(x \mapsto x, y \mapsto x, (x \to y) \mapsto 1_x)$.

9. • CommentRowNumber9.
   • CommentAuthorEric
   • CommentTimeAug 17th 2010
   • (edited Aug 17th 2010)
   • PermaLink
   Author: Eric
   Format: MarkdownItexThanks Toby! :) I'll fix the typo in Case 2. I don't know what I was thinking when I wrote $\partial\gamma$. I was probably looking at the natural transformation diagram. I'll correct it. I wrote $\partial\gamma'$ down quickly as an afterthought. I'll include your correction. Edit: So to make some observations... There are cases: * Natural transformation exists and the boundary exists * Natural transformation does not exist, yet the boundary exists * Natural transformation exists, yet the boundary does not exist * Neither natural transformation nor boundary exists In terms of numbers, it seems boundaries exist more frequently than natural transformations. This makes sense I suppose because functors are more flexible than natural transformations, which require components. The fact that some natural transformations exist, but do not have boundaries seems like an interesting maths problem. It is also interesting that 2-morphisms in (strict?) 2-groupoids always have boundaries. Or so I think...

   Thanks Toby! :)

   I’ll fix the typo in Case 2.

   I don’t know what I was thinking when I wrote $\partial\gamma$. I was probably looking at the natural transformation diagram. I’ll correct it.

   I wrote $\partial\gamma'$ down quickly as an afterthought. I’ll include your correction.

   Edit: So to make some observations…

   There are cases:

   • Natural transformation exists and the boundary exists
   • Natural transformation does not exist, yet the boundary exists
   • Natural transformation exists, yet the boundary does not exist
   • Neither natural transformation nor boundary exists

   In terms of numbers, it seems boundaries exist more frequently than natural transformations. This makes sense I suppose because functors are more flexible than natural transformations, which require components.

   The fact that some natural transformations exist, but do not have boundaries seems like an interesting maths problem.

   It is also interesting that 2-morphisms in (strict?) 2-groupoids always have boundaries. Or so I think…

10. • CommentRowNumber10.
    • CommentAuthorTobyBartels
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>It is also interesting that 2-morphisms in (strict?) 2-groupoids always have boundaries. I think that, given any natural isomorphism $\alpha\colon F \tilde{\Rightarrow} G\colon C \to D$, there exists an equivalence $E\colon D \to D'$ such that the whiskering $E \rhd \alpha$ has a boundary. However, I haven't checked in detail.

    It is also interesting that 2-morphisms in (strict?) 2-groupoids always have boundaries.

    I think that, given any natural isomorphism $\alpha\colon F \tilde{\Rightarrow} G\colon C \to D$, there exists an equivalence $E\colon D \to D'$ such that the whiskering $E \rhd \alpha$ has a boundary. However, I haven’t checked in detail.

11. • CommentRowNumber11.
    • CommentAuthorEric
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItex>It is also interesting that 2-morphisms in (strict?) 2-groupoids always have boundaries. Or so I think... Although not conclusive, I just convinced myself that if we take the 2-object 1-non-identity morphism categories $C$ and $D$ above and turn them into 2-object 2-non-identity morphism groupoids by added inverse morphisms, then * All natural transformations $\alpha$, $\beta$, $\gamma$, $\alpha'$, $\beta'$, and $\gamma'$ exist * All corresponding boundaries exist

    It is also interesting that 2-morphisms in (strict?) 2-groupoids always have boundaries. Or so I think…

    Although not conclusive, I just convinced myself that if we take the 2-object 1-non-identity morphism categories $C$ and $D$ above and turn them into 2-object 2-non-identity morphism groupoids by added inverse morphisms, then

    • All natural transformations $\alpha$, $\beta$, $\gamma$, $\alpha'$, $\beta'$, and $\gamma'$ exist
    • All corresponding boundaries exist
12. • CommentRowNumber12.
    • CommentAuthorTobyBartels
    • CommentTimeAug 17th 2010
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    Author: TobyBartels
    Format: MarkdownItexI agree that all natural transformations exist, but I still think that $\partial\alpha'$ and $\partial\beta'$ still don't exist (at least, not on the nose). Perhaps you should write down what you think $\partial\alpha'$ is in this case?

    I agree that all natural transformations exist, but I still think that $\partial\alpha'$ and $\partial\beta'$ still don’t exist (at least, not on the nose).

    Perhaps you should write down what you think $\partial\alpha'$ is in this case?

13. • CommentRowNumber13.
    • CommentAuthorEric
    • CommentTimeAug 17th 2010
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    Author: Eric
    Format: MarkdownItexConvincing myself is easy. I drew one natural transformation diagram, it became obvious that all natural transformations exist, and then I grossly/erroneously extrapolated :) Yeah. You're right of course. Invertibility of morphisms does not fix the boundary. $\partial\alpha'$ still does not exist. However, this pushes things forward a little I think. I'm now tempted to think that a sufficient condition for the boundary of a natural transformation to exist is that the source functor has a right inverse. This is what my gut instinct has been saying from the beginning. Sound reasonable? But then $G$ and $H$ do not have right inverses, yet $\gamma$ and $\gamma'$ have boundaries. It would be nice to come up with a necessary and sufficient condition. Maybe your natural isomorphism is the answer.

    Convincing myself is easy. I drew one natural transformation diagram, it became obvious that all natural transformations exist, and then I grossly/erroneously extrapolated :)

    Yeah. You’re right of course. Invertibility of morphisms does not fix the boundary. $\partial\alpha'$ still does not exist.

    However, this pushes things forward a little I think. I’m now tempted to think that a sufficient condition for the boundary of a natural transformation to exist is that the source functor has a right inverse. This is what my gut instinct has been saying from the beginning. Sound reasonable? But then $G$ and $H$ do not have right inverses, yet $\gamma$ and $\gamma'$ have boundaries. It would be nice to come up with a necessary and sufficient condition. Maybe your natural isomorphism is the answer.

14. • CommentRowNumber14.
    • CommentAuthorTobyBartels
    • CommentTimeAug 17th 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>a sufficient condition for the boundary of a natural transformation to exist is that the source functor has a right inverse Do you mean a left inverse? (for the Leibniz order of composition). If $F$ has a left inverse $F'$, then $G \circ F'$ is a boundary from $F$ to $G$, since $$ (G \circ F') \circ F = G \circ (F' \circ F) = G \circ id_C = G .$$ My talk of natural isomorphisms (and replacing $D$ with an equivalent category) is only another sufficient condition. The bottom line, as far as I can see, is that the boundaries from $F$ to $G$ (whether any exist, and how many there are) are pretty much independent of the natural transformations from $F$ to $G$ (whether any exist, and how many there are), and it doesn't make sense to say that a boundary is a the boundary *of* a natural transformation. Anyway, I must go to bed now.

    a sufficient condition for the boundary of a natural transformation to exist is that the source functor has a right inverse

    Do you mean a left inverse? (for the Leibniz order of composition). If $F$ has a left inverse $F'$, then $G \circ F'$ is a boundary from $F$ to $G$, since

    $$(G \circ F') \circ F = G \circ (F' \circ F) = G \circ id_C = G .$$

    My talk of natural isomorphisms (and replacing $D$ with an equivalent category) is only another sufficient condition.

    The bottom line, as far as I can see, is that the boundaries from $F$ to $G$ (whether any exist, and how many there are) are pretty much independent of the natural transformations from $F$ to $G$ (whether any exist, and how many there are), and it doesn’t make sense to say that a boundary is a the boundary of a natural transformation.

    Anyway, I must go to bed now.

15. • CommentRowNumber15.
    • CommentAuthorEric
    • CommentTimeAug 17th 2010
    • (edited Aug 17th 2010)
    • PermaLink
    Author: Eric
    Format: MarkdownItex>and it doesn't make sense to say that a boundary is a the boundary of a natural transformation. This might be true, but I'm not so sure we should dismiss the connection too easily. After all, when the boundary of a 2-morphism $\alpha:f\Rightarrow g$ exists and the source 1-morphism is invertible, the boundary of the 2-morphism is $$\partial\alpha = g\circ f^{-1}.$$ This has every right to be called "the boundary of $\alpha$" because it IS the boundary of $\alpha$ :) When not dealing with groupoids, the meaning becomes less clear, but I think that in many circumstances more general than groupoids there could be a nice intuitive way to think of boundaries. After all, they provide a nice [[ericforgy:Boundary of a 2-Morphism and the Interchange Law|calculus]] from which the interchange law falls out naturally.

    and it doesn’t make sense to say that a boundary is a the boundary of a natural transformation.

    This might be true, but I’m not so sure we should dismiss the connection too easily. After all, when the boundary of a 2-morphism $\alpha:f\Rightarrow g$ exists and the source 1-morphism is invertible, the boundary of the 2-morphism is

    $$\partial\alpha = g\circ f^{-1}.$$

    This has every right to be called “the boundary of $\alpha$” because it IS the boundary of $\alpha$ :)

    When not dealing with groupoids, the meaning becomes less clear, but I think that in many circumstances more general than groupoids there could be a nice intuitive way to think of boundaries. After all, they provide a nice calculus from which the interchange law falls out naturally.

16. • CommentRowNumber16.
    • CommentAuthorTobyBartels
    • CommentTimeAug 17th 2010
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    Author: TobyBartels
    Format: MarkdownItex>This has every right to be called "the boundary of $\alpha$" because it IS the boundary of $\alpha$ :) What I mean is that it is not *of* $\alpha$ particularly; it is determined completely by $f$ and $g$, even though there may be many different $2$-morphisms, or none at all, from $f$ to $g$. Furthermore, you have to bring up $f$ and $g$ before you can bring up $\alpha$ at all, so there is no point in even mentioning $\alpha$ if all that you want to talk about is $g \circ f^{-1}$. Yes, if you have $\alpha$ in hand, then $f$ is the source of $\alpha$, $g$ is the target of $\alpha$, and $g \circ f^{-1}$ (assuming that $f$ is invertible) is the boundary of $\alpha$, but these things are all more basic than $\alpha$. Another problem is that the boundary of $\alpha$ is not always *the* boundary of $\alpha$. When $f$ is invertible, the boundary from $f$ to $g$ is unique, but if $f$ has several left inverses, then $g \circ f'$ will be a boundary from $f$ to $g$ whenever $f'$ is any left inverse of $f$, and this will generally give several boundaries (and there may be yet others, and there may be several boundaries even when $f$ has no left inverse). And picking a natural transformation from $f$ to $g$ does nothing to help choose among the boundaries from $f$ to $g$. That is why I say that natural transformations and boundaries are independent concepts.

    This has every right to be called “the boundary of $\alpha$” because it IS the boundary of $\alpha$ :)

    What I mean is that it is not of $\alpha$ particularly; it is determined completely by $f$ and $g$, even though there may be many different $2$-morphisms, or none at all, from $f$ to $g$. Furthermore, you have to bring up $f$ and $g$ before you can bring up $\alpha$ at all, so there is no point in even mentioning $\alpha$ if all that you want to talk about is $g \circ f^{-1}$. Yes, if you have $\alpha$ in hand, then $f$ is the source of $\alpha$, $g$ is the target of $\alpha$, and $g \circ f^{-1}$ (assuming that $f$ is invertible) is the boundary of $\alpha$, but these things are all more basic than $\alpha$.

    Another problem is that the boundary of $\alpha$ is not always the boundary of $\alpha$. When $f$ is invertible, the boundary from $f$ to $g$ is unique, but if $f$ has several left inverses, then $g \circ f'$ will be a boundary from $f$ to $g$ whenever $f'$ is any left inverse of $f$, and this will generally give several boundaries (and there may be yet others, and there may be several boundaries even when $f$ has no left inverse). And picking a natural transformation from $f$ to $g$ does nothing to help choose among the boundaries from $f$ to $g$. That is why I say that natural transformations and boundaries are independent concepts.

17. • CommentRowNumber17.
    • CommentAuthorTobyBartels
    • CommentTimeAug 17th 2010
    • PermaLink
    Author: TobyBartels
    Format: MarkdownItex>After all, they provide a nice calculus from which the interchange law falls out naturally. I've written about that on your page now.

    After all, they provide a nice calculus from which the interchange law falls out naturally.

    I’ve written about that on your page now.