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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeJan 15th 2014
   • (edited Jan 15th 2014)
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   Author: Mirco Richter
   Format: MarkdownItexEither I miss the point here, or reflexive globular sets are not describeable as presheaves on the reflexive globular category. To see what I mean, let $\mathbf{G}$ be the reflexive globular category, that is objects are the $[n]$th and morphisms are generated by maps $s_{n}:[n]\to[n+1]$, $t_{n}:[n]\to[n+1]$ and $i_n:[n+1]\to [n]$ devoted to the structure equations $s\circ s=t\circ s$, $t\circ t=s\circ t$ and $i\circ s=id$ as well as $i\circ t=id$. Now from the last two equations we see, that both $s$ and $t$ have to be injective, but then the first two equations say that they have to be equal. ?? That's not what the globular shape is all about! (presheaves from here to set are not what one has in mind with globular sets, since source and targets can be different in many situations) Maybe its better to not just copy the experience from simplicial sets, but to think of them covariant?

   Either I miss the point here, or reflexive globular sets are not describeable as presheaves on the reflexive globular category.

   To see what I mean, let $\mathbf{G}$ be the reflexive globular category, that is objects are the $[n]$th and morphisms are generated by maps $s_{n}:[n]\to[n+1]$, $t_{n}:[n]\to[n+1]$ and $i_n:[n+1]\to [n]$ devoted to the structure equations

   $s\circ s=t\circ s$, $t\circ t=s\circ t$ and $i\circ s=id$ as well as $i\circ t=id$.

   Now from the last two equations we see, that both $s$ and $t$ have to be injective, but then the first two equations say that they have to be equal. ?? That’s not what the globular shape is all about! (presheaves from here to set are not what one has in mind with globular sets, since source and targets can be different in many situations)

   Maybe its better to not just copy the experience from simplicial sets, but to think of them covariant?

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeJan 15th 2014
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   Author: DavidRoberts
   Format: MarkdownItex>but then the first two equations say that they have to be equal. I don't understand what you mean. Think of the cellular structure on a circle with two 0-cells, two 1-cells and a 2-cell, as a model for a 2-cell in $G$. (I think something may be wrong with the composition order in what you wrote)

   but then the first two equations say that they have to be equal.

   I don’t understand what you mean. Think of the cellular structure on a circle with two 0-cells, two 1-cells and a 2-cell, as a model for a 2-cell in $G$. (I think something may be wrong with the composition order in what you wrote)

3. • CommentRowNumber3.
   • CommentAuthorMirco Richter
   • CommentTimeJan 15th 2014
   • (edited Jan 15th 2014)
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   Author: Mirco Richter
   Format: MarkdownItexThe composition order has to be reversed, if a globular set is seen as a presheaf $\mathbf{G}\to Set$. I mean that IF this is true, then a globular set is not the same as the globular sets we know, because in $\mathbf{G}$ there is necessarily $s=t$, since otherwise the structure equations can't be satisfied. Can you write down the presheaf that gives your example? Or give me $s_0:[0]\to[1]$ and $t_0:[0]\to[1]$ and $s_1:[1]\to[2]$ and $t_1:[1]\to[2]$ and some $i$'s such that the structure equations are satisfied.

   The composition order has to be reversed, if a globular set is seen as a presheaf $\mathbf{G}\to Set$. I mean that IF this is true, then a globular set is not the same as the globular sets we know, because in $\mathbf{G}$ there is necessarily $s=t$, since otherwise the structure equations can’t be satisfied.

   Can you write down the presheaf that gives your example? Or give me $s_0:[0]\to[1]$ and $t_0:[0]\to[1]$ and $s_1:[1]\to[2]$ and $t_1:[1]\to[2]$ and some $i$’s such that the structure equations are satisfied.

4. • CommentRowNumber4.
   • CommentAuthorMirco Richter
   • CommentTimeJan 15th 2014
   • (edited Jan 15th 2014)
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   Author: Mirco Richter
   Format: MarkdownItexI think David, in fact you supported my argument, since you see globular sets as covariant functors $\mathbf{G}'\to Set$, too? Am I right? Where $\mathbf{G}'$ still has to be defined...

   I think David, in fact you supported my argument, since you see globular sets as covariant functors $\mathbf{G}'\to Set$, too? Am I right? Where $\mathbf{G}'$ still has to be defined…

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeJan 15th 2014
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   Author: DavidRoberts
   Format: MarkdownItexWhere did.you get your definition of G from? And do you mean G^op -> Set?

   Where did.you get your definition of G from? And do you mean G^op -> Set?

6. • CommentRowNumber6.
   • CommentAuthorMirco Richter
   • CommentTimeJan 15th 2014
   • (edited Jan 15th 2014)
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   Author: Mirco Richter
   Format: MarkdownItexOk maybe I was a little to sloppy,posting this. Sry. Here is the thing in more detail: When one looks at the nLab entry http://ncatlab.org/nlab/show/globular+set, it states that globular sets can be seen as presheaves. I think this is wrong. To be more precise: Let $G$ be the category, which elements are the ordinal numbers $[n]:=(0,....,n)$ for any $n \in \{0,1,2,...\}$ and morphism are generated by the following maps $s: [n] \to [n+1]$ $t: [n] \to [n+1]$ $i: [n+1] \to [n]$ subject to the conditions: 1.) $ss = ts$ 2.) $tt = st$ 3.) $is = Id$ 4.) $it = Id$ Then it is said, that a globular set, is a functor $F:G^{op} \to Set$ =================== On a first guess, one would say, ok this is similar to the situation of simplicial sets and since the functor is contravariant we get the 'globular structure equations' for $S:=F(s)$, $T:=F(t)$ and $I:=F(i)$ 1.) $SS = ST$ 2.) $TT = TS$ 3.) $SI = Id$ 4.) $TI = Id$ and hence get a globular set. BUT the thing is, that for general globular sets, we obviously can choose $S \neq T$ in most cases, while in the category $G$ as given above, this is not true! In particular from $is=Id$ and $it=Id$, we see that the costructure maps $s$ and $t$ have left inverses and hence are injective. But then follows $s=t$ from $ss=ts$ and $tt=st$ at least for $[n] \leq [1]$. So the contradiction is, while there are globular sets with $S\neq T$ for $n\leq 1$, there are no corresponding maps $s$ and $t$ in $G$! Hence such a globular set, can't be a functor $F;G^{op}\to Set$. Remark: (Here I wrote just $s$, $t$ as well as $S$ and $T$ for what is really $s_n$, $t_n$, $S_n$ and $T_n$. Just to not discourage the reader by indices.)

   Ok maybe I was a little to sloppy,posting this. Sry. Here is the thing in more detail:

   When one looks at the nLab entry http://ncatlab.org/nlab/show/globular+set, it states that globular sets can be seen as presheaves. I think this is wrong. To be more precise:

   Let $G$ be the category, which elements are the ordinal numbers $[n]:=(0,....,n)$ for any $n \in \{0,1,2,...\}$ and morphism are generated by the following maps

   $s: [n] \to [n+1]$

   $t: [n] \to [n+1]$

   $i: [n+1] \to [n]$

   subject to the conditions:

   1.) $ss = ts$

   2.) $tt = st$

   3.) $is = Id$

   4.) $it = Id$

   Then it is said, that a globular set, is a functor

   $F:G^{op} \to Set$

   ===================

   On a first guess, one would say, ok this is similar to the situation of simplicial sets and since the functor is contravariant we get the ’globular structure equations’ for $S:=F(s)$, $T:=F(t)$ and $I:=F(i)$

   1.) $SS = ST$

   2.) $TT = TS$

   3.) $SI = Id$

   4.) $TI = Id$

   and hence get a globular set. BUT the thing is, that for general globular sets, we obviously can choose $S \neq T$ in most cases, while in the category $G$ as given above, this is not true!

   In particular from $is=Id$ and $it=Id$, we see that the costructure maps $s$ and $t$ have left inverses and hence are injective. But then follows $s=t$ from $ss=ts$ and $tt=st$ at least for $[n] \leq [1]$.

   So the contradiction is, while there are globular sets with $S\neq T$ for $n\leq 1$, there are no corresponding maps $s$ and $t$ in $G$! Hence such a globular set, can’t be a functor $F;G^{op}\to Set$.

   Remark: (Here I wrote just $s$, $t$ as well as $S$ and $T$ for what is really $s_n$, $t_n$, $S_n$ and $T_n$. Just to not discourage the reader by indices.)

7. • CommentRowNumber7.
   • CommentAuthorMirco Richter
   • CommentTimeJan 15th 2014
   • (edited Jan 15th 2014)
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   Author: Mirco Richter
   Format: MarkdownItexMaybe an example is a good thing here: Consider a two cell $\alpha: f\Rightarrow g: A \to B$. (Unfortunately I can't write such a diagram in my browser). Then the globular structure is clear: For $n=2$ we have $S(\alpha)=f$, $T(\alpha)=g$ For $n=1$ we have $S(f)=A$, $S(g)=A$ as well as $T(f)=B$, $T(g)=B$ and $I(f)=id^2_f$, $I(g)=id^2_g$. For $n=0$ we have $I(A)=id_A$ and $I(B)=id_B$ Obviously the globular structure equations are satisfied. ======================== Now consider $A\neq B$ and try to express this $2$-cell as a functor $F: G^{op}\to Set$, where you explicitly give the maps $s_1:[1]\to [2]$ and $t_1:[1]\to[2]$. I would say that it is not possible!

   Maybe an example is a good thing here:

   Consider a two cell $\alpha: f\Rightarrow g: A \to B$. (Unfortunately I can’t write such a diagram in my browser). Then the globular structure is clear:

   For $n=2$ we have $S(\alpha)=f$, $T(\alpha)=g$

   For $n=1$ we have $S(f)=A$, $S(g)=A$ as well as $T(f)=B$, $T(g)=B$ and $I(f)=id^2_f$, $I(g)=id^2_g$.

   For $n=0$ we have $I(A)=id_A$ and $I(B)=id_B$

   Obviously the globular structure equations are satisfied.

   ========================

   Now consider $A\neq B$ and try to express this $2$-cell as a functor $F: G^{op}\to Set$, where you explicitly give the maps $s_1:[1]\to [2]$ and $t_1:[1]\to[2]$. I would say that it is not possible!

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 15th 2014
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   Author: Todd_Trimble
   Format: MarkdownItexMirco, I'm sorry, but you seem to be pretty confused here. > But then follows $s=t$ from $ss=ts$ and $tt=st$ at least for $[n] \leq [1]$. You could conclude that if $s$ were an *epimorphism*. But not from the fact (that you correctly observed) that $s$ is a *monomorphism*. To understand better the globular category, it might help to picture $n$ as representing an $n$-disk, $s: n \to n+1$ as the inclusion of an $n$-disk onto the "northern hemisphere" of the boundary of an $(n+1)$-disk, and $t: n \to n+1$ as the inclusion onto the southern hemisphere. The retraction $n+1 \to n$ could be pictured as the projection of an $(n+1)$-disk or ball onto an equatorial cross-section. (And indeed, there is a covariant functor $\mathbf{G} \to Top$ which carries exactly this interpretation.)

   Mirco, I’m sorry, but you seem to be pretty confused here.

   But then follows $s=t$ from $ss=ts$ and $tt=st$ at least for $[n] \leq [1]$.

   You could conclude that if $s$ were an epimorphism. But not from the fact (that you correctly observed) that $s$ is a monomorphism.

   To understand better the globular category, it might help to picture $n$ as representing an $n$-disk, $s: n \to n+1$ as the inclusion of an $n$-disk onto the “northern hemisphere” of the boundary of an $(n+1)$-disk, and $t: n \to n+1$ as the inclusion onto the southern hemisphere. The retraction $n+1 \to n$ could be pictured as the projection of an $(n+1)$-disk or ball onto an equatorial cross-section. (And indeed, there is a covariant functor $\mathbf{G} \to Top$ which carries exactly this interpretation.)

9. • CommentRowNumber9.
   • CommentAuthorMirco Richter
   • CommentTimeJan 16th 2014
   • (edited Jan 16th 2014)
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   Author: Mirco Richter
   Format: MarkdownItexTodd, thanks for your comment. Intuitively the hemisphere POV is what I have in mind about the globular category. However just to make sure, that my imagination really works,I tried, to explicitly find the appropriate maps and now I'm not entirely sure if it works anymore. To see why, here is the brute force method, giving first all possible injective maps for $n\leq1$: $f_{0}^{0}:[0]\to[1];0\to0$, $f_{1}^{0}:[0]\to[1];0\to1$, $f_{0}^{1}:[1]\to[2];(0,1)\to(0,1)$, $f_{1}^{1}:[1]\to[2];(0,1)\to(0,2)$, $f_{2}^{1}:[1]\to[2];(0,1)\to(1,0)$, $f_{3}^{1}:[1]\to[2];(0,1)\to(1,2)$, $f_{4}^{1}:[1]\to[2];(0,1)\to(2,0)$, $f_{5}^{1}:[1]\to[2];(0,1)\to(2,1)$. There are no other, right? Now I 'brute force' the combinatorics, checking all compositions, if there are some satisfying $f_{j}\circ f_{i}=f_{k}\circ f_{i}$ as well as $f_{k}\circ f_{l}=f_{j}\circ f_{l}$, since then $t^{1}:=f_{k}$, $t^{0}:=f_{l}$, $s^{1}:=f_{j}$ and $s^{0}:=f_{i}$. That is $f_{0}\circ f_{1}=1=f_{5}\circ f_{1}$ but $f_{5}\circ f_{0}=2\neq0=f_{0}\circ f_{0}$ $f_{1}\circ f_{1}=2=f_{3}\circ f_{1}$but $f_{3}\circ f_{0}=1\neq0=f_{1}\circ f_{0}$ $f_{2}\circ f_{1}=0=f_{4}\circ f_{1}$ but $f_{4}\circ f_{0}=2\neq1=f_{2}\circ f_{0}$ $f_{3}\circ f_{1}=2=f_{1}\circ f_{1}$ but $f_{1}\circ f_{0}=0\neq1=f_{3}\circ f_{0}$ $f_{4}\circ f_{1}=0=f_{2}\circ f_{1}$ but $f_{2}\circ f_{0}=1\neq2=f_{4}\circ f_{0}$ $f_{5}\circ f_{1}=1=f_{0}\circ f_{1}$ but $f_{0}\circ f_{0}=0\neq 2=f_{5}\circ f_{0}$ ============ $f_{0}\circ f_{0}=0=f_{1}\circ f_{0}$ but $f_{1}\circ f_{1}=2\neq1=f_{0}\circ f_{1}$ $f_{1}\circ f_{0}=0=f_{0}\circ f_{0}$but $f_{0}\circ f_{1}=1\neq2=f_{1}\circ f_{1}$ $f_{2}\circ f_{0}=1=f_{3}\circ f_{0}$ but $f_{3}\circ f_{1}=2\neq0=f_{2}\circ f_{1}$ $f_{3}\circ f_{0}=1=f_{2}\circ f_{0}$ but $f_{2}\circ f_{1}=0\neq2=f_{3}\circ f_{1}$ $f_{4}\circ f_{0}=2=f_{5}\circ f_{0}$ but $f_{5}\circ f_{1}=1\neq0=f_{4}\circ f_{1}$ $f_{5}\circ f_{0}=2=f_{4}\circ f_{0}$ but $f_{4}\circ f_{1}=0\neq1=f_{5}\circ f_{1}$ ===================== This means that the pair of equations $f_{j}\circ f_{i}=f_{k}\circ f_{i}$ and $f_{k}\circ f_{l}=f_{j}\circ f_{l}$ has no solution for $f_i\neq f_l$ here. Please tell me you know what I'm up to! (Such that this tedious Tex-exercise was not for nothing :-) ) There are no injective maps $t^{1};[1]\to[2]$, $t^{0}:[0]\to[1]$, $s^{1}:[1]\to[2]$ and $s^{0}:[0]\to[1]$, satisfying the structure equations. With $t^0\neq s^0$...

   Todd, thanks for your comment. Intuitively the hemisphere POV is what I have in mind about the globular category. However just to make sure, that my imagination really works,I tried, to explicitly find the appropriate maps and now I’m not entirely sure if it works anymore.

   To see why, here is the brute force method, giving first all possible injective maps for $n\leq1$:

   $f_{0}^{0}:[0]\to[1];0\to0$,

   $f_{1}^{0}:[0]\to[1];0\to1$,

   $f_{0}^{1}:[1]\to[2];(0,1)\to(0,1)$,

   $f_{1}^{1}:[1]\to[2];(0,1)\to(0,2)$,

   $f_{2}^{1}:[1]\to[2];(0,1)\to(1,0)$,

   $f_{3}^{1}:[1]\to[2];(0,1)\to(1,2)$,

   $f_{4}^{1}:[1]\to[2];(0,1)\to(2,0)$,

   $f_{5}^{1}:[1]\to[2];(0,1)\to(2,1)$.

   There are no other, right?

   Now I ’brute force’ the combinatorics, checking all compositions, if there are some satisfying $f_{j}\circ f_{i}=f_{k}\circ f_{i}$ as well as $f_{k}\circ f_{l}=f_{j}\circ f_{l}$, since then $t^{1}:=f_{k}$, $t^{0}:=f_{l}$, $s^{1}:=f_{j}$ and $s^{0}:=f_{i}$.

   That is

   $f_{0}\circ f_{1}=1=f_{5}\circ f_{1}$ but $f_{5}\circ f_{0}=2\neq0=f_{0}\circ f_{0}$

   $f_{1}\circ f_{1}=2=f_{3}\circ f_{1}$but $f_{3}\circ f_{0}=1\neq0=f_{1}\circ f_{0}$

   $f_{2}\circ f_{1}=0=f_{4}\circ f_{1}$ but $f_{4}\circ f_{0}=2\neq1=f_{2}\circ f_{0}$

   $f_{3}\circ f_{1}=2=f_{1}\circ f_{1}$ but $f_{1}\circ f_{0}=0\neq1=f_{3}\circ f_{0}$

   $f_{4}\circ f_{1}=0=f_{2}\circ f_{1}$ but $f_{2}\circ f_{0}=1\neq2=f_{4}\circ f_{0}$

   $f_{5}\circ f_{1}=1=f_{0}\circ f_{1}$ but $f_{0}\circ f_{0}=0\neq 2=f_{5}\circ f_{0}$

   ============

   $f_{0}\circ f_{0}=0=f_{1}\circ f_{0}$ but $f_{1}\circ f_{1}=2\neq1=f_{0}\circ f_{1}$

   $f_{1}\circ f_{0}=0=f_{0}\circ f_{0}$but $f_{0}\circ f_{1}=1\neq2=f_{1}\circ f_{1}$

   $f_{2}\circ f_{0}=1=f_{3}\circ f_{0}$ but $f_{3}\circ f_{1}=2\neq0=f_{2}\circ f_{1}$

   $f_{3}\circ f_{0}=1=f_{2}\circ f_{0}$ but $f_{2}\circ f_{1}=0\neq2=f_{3}\circ f_{1}$

   $f_{4}\circ f_{0}=2=f_{5}\circ f_{0}$ but $f_{5}\circ f_{1}=1\neq0=f_{4}\circ f_{1}$

   $f_{5}\circ f_{0}=2=f_{4}\circ f_{0}$ but $f_{4}\circ f_{1}=0\neq1=f_{5}\circ f_{1}$

   =====================

   This means that the pair of equations $f_{j}\circ f_{i}=f_{k}\circ f_{i}$ and $f_{k}\circ f_{l}=f_{j}\circ f_{l}$ has no solution for $f_i\neq f_l$ here.

   Please tell me you know what I’m up to! (Such that this tedious Tex-exercise was not for nothing :-) ) There are no injective maps $t^{1};[1]\to[2]$, $t^{0}:[0]\to[1]$, $s^{1}:[1]\to[2]$ and $s^{0}:[0]\to[1]$, satisfying the structure equations. With $t^0\neq s^0$…

10. • CommentRowNumber10.
    • CommentAuthorMirco Richter
    • CommentTimeJan 16th 2014
    • (edited Jan 16th 2014)
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    Author: Mirco Richter
    Format: MarkdownItexSo to proof me wrong, it is enough to just give maps $s^0:[0]\to[1]$, $t^0:[0]\to[1]$ and $s^1:[1]\to[2]$ as well as $t^1:[1]\to [2]$ such that they are injective with $s^1s^0=t^1s^0$ and $t^1t^0=s^1t^0$. But you can't, at least under the assumption $s^0\neq t^0$ and that's an implication of the INJECTIVITY, call me confused or not. (At least you should take the time to proof that your insult is satisfied, so I have to take it as a correct observation, not as blinded arrogance)

    So to proof me wrong, it is enough to just give maps $s^0:[0]\to[1]$, $t^0:[0]\to[1]$ and $s^1:[1]\to[2]$ as well as $t^1:[1]\to [2]$ such that they are injective with $s^1s^0=t^1s^0$ and $t^1t^0=s^1t^0$. But you can’t, at least under the assumption $s^0\neq t^0$ and that’s an implication of the INJECTIVITY, call me confused or not.

    (At least you should take the time to proof that your insult is satisfied, so I have to take it as a correct observation, not as blinded arrogance)

11. • CommentRowNumber11.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 16th 2014
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    Author: DavidRoberts
    Format: MarkdownItexRe #6, I don't think s, t and i are maps of sets, rather that G is just the free category generated by this data. Then s and t are monomorphisms, but not injective maps.

    Re #6, I don’t think s, t and i are maps of sets, rather that G is just the free category generated by this data. Then s and t are monomorphisms, but not injective maps.

12. • CommentRowNumber12.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 16th 2014
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    Author: DavidRoberts
    Format: MarkdownItexP.S. You have proved that G is not concrete, as Δ is.

    P.S. You have proved that G is not concrete, as Δ is.

13. • CommentRowNumber13.
    • CommentAuthorMirco Richter
    • CommentTimeJan 16th 2014
    • (edited Jan 16th 2014)
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    Author: Mirco Richter
    Format: MarkdownItexYes of course you can just define a category to have objects $\mathbf{n}$ and those morphisms. But at least to my eyes, people mostly think of the objects $\mathbf{n}$ as 'the ordinals' $[n]=\{0,\ldots, n\}$. (At least that is the POV of the nLab entry) However,.. thanks David to clear that point! These tiny bits on the lowest level, commonly taking for granted, can really screw me, if they don't work out as expected. .... But I think we can still use the ordinals $[n]$ as we know them from the simplex category, if we shift the grading. Maybe this will require some negative thinking. I'll post it, if I know what I mean...

    Yes of course you can just define a category to have objects $\mathbf{n}$ and those morphisms. But at least to my eyes, people mostly think of the objects $\mathbf{n}$ as ’the ordinals’ $[n]=\{0,\ldots, n\}$. (At least that is the POV of the nLab entry)

    However,.. thanks David to clear that point! These tiny bits on the lowest level, commonly taking for granted, can really screw me, if they don’t work out as expected.

    ….

    But I think we can still use the ordinals $[n]$ as we know them from the simplex category, if we shift the grading. Maybe this will require some negative thinking. I’ll post it, if I know what I mean…

14. • CommentRowNumber14.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 16th 2014
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    Author: Todd_Trimble
    Format: MarkdownItexYeah. *Show* me where the objects $n$ are regarded as ordinals in that article. Show me *any* place in that article that said we are supposed to regard the $n$ as finite sets and the morphisms as functions. If you find such a concrete representation (which might be possible; I'm not denying the possibility), then go ahead and add your finding to the entry. I would like to add, Mirco, that I never insulted you. On the other hand, if you claim that many scores of people who have examined this material before you are wrong about something, then obviously the burden of proof is on you. You made repeated claims that the injectivity of $s$ and $t$ forced $s = t$, to which I responded (and where I thought you were confused).

    Yeah. Show me where the objects $n$ are regarded as ordinals in that article. Show me any place in that article that said we are supposed to regard the $n$ as finite sets and the morphisms as functions.

    If you find such a concrete representation (which might be possible; I’m not denying the possibility), then go ahead and add your finding to the entry.

    I would like to add, Mirco, that I never insulted you. On the other hand, if you claim that many scores of people who have examined this material before you are wrong about something, then obviously the burden of proof is on you. You made repeated claims that the injectivity of $s$ and $t$ forced $s = t$, to which I responded (and where I thought you were confused).

15. • CommentRowNumber15.
    • CommentAuthorMirco Richter
    • CommentTimeJan 16th 2014
    • (edited Jan 16th 2014)
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexFrom http://ncatlab.org/nlab/show/globular+set The globe category $\mathbb{G}$ is the category whose objects are the natural numbers, denoted here $[n] \in \mathbb{N}$ and whose morphisms are generated from $ \sigma_n : [n] \to [n+1] $ $ \tau_n : [n] \to [n+1] $ for all $n \in \mathbb{N}$ subject to the relations. I bet that most mathematicians would read this as $[n]=\{0,\ldots, n\}$ and $\sigma / \tau : [n] \to [n+1]$ as a map (of sets). Especially regarding the article http://ncatlab.org/nlab/show/ordinal+number, where its said, that any natural number is a finite ordinal and ordinals can be seen as ordered sets. I don't say, that there are no other possibilities, but at least it should be mentioned, as David pointed out, that the globe category can't be seen as a concrete category that way and that $\sigma_n$ as well as $\tau_n$ are not set maps.

    From http://ncatlab.org/nlab/show/globular+set

    The globe category $\mathbb{G}$ is the category whose objects are the natural numbers, denoted here $[n] \in \mathbb{N}$ and whose morphisms are generated from

    $\sigma_n : [n] \to [n+1]$

    $\tau_n : [n] \to [n+1]$

    for all $n \in \mathbb{N}$ subject to the relations. I bet that most mathematicians would read this as $[n]=\{0,\ldots, n\}$ and $\sigma / \tau : [n] \to [n+1]$ as a map (of sets). Especially regarding the article http://ncatlab.org/nlab/show/ordinal+number, where its said, that any natural number is a finite ordinal and ordinals can be seen as ordered sets.

    I don’t say, that there are no other possibilities, but at least it should be mentioned, as David pointed out, that the globe category can’t be seen as a concrete category that way and that $\sigma_n$ as well as $\tau_n$ are not set maps.

16. • CommentRowNumber16.
    • CommentAuthorDavidRoberts
    • CommentTimeJan 16th 2014
    • PermaLink
    Author: DavidRoberts
    Format: MarkdownItex@Mirco >it should be mentioned, go ahead :-)

    @Mirco

    it should be mentioned,

    go ahead :-)

17. • CommentRowNumber17.
    • CommentAuthorTodd_Trimble
    • CommentTimeJan 16th 2014
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexSIGH. I bet most mathematicians who read category theory know about defining a category using generators and relations, which is what the article does. It says so right there. They are furthermore used to the idea that not all categories are described as concrete categories in their definitions (and also the fact that there might not be any concretizations, or there may be many). I doubt that *most* mathematicians would make your mistake, and even doubt many of them would. But if you INSIST they would, then one remedy might be to change all those $[n]$'s to something else like $\mathbf{n}$'s that don't carry the same associations.

    SIGH.

    I bet most mathematicians who read category theory know about defining a category using generators and relations, which is what the article does. It says so right there. They are furthermore used to the idea that not all categories are described as concrete categories in their definitions (and also the fact that there might not be any concretizations, or there may be many).

    I doubt that most mathematicians would make your mistake, and even doubt many of them would. But if you INSIST they would, then one remedy might be to change all those $[n]$’s to something else like $\mathbf{n}$’s that don’t carry the same associations.

18. • CommentRowNumber18.
    • CommentAuthorMirco Richter
    • CommentTimeJan 16th 2014
    • PermaLink
    Author: Mirco Richter
    Format: MarkdownItexOk maybe its better to let these speculations about mathematicians aside. I see nothing to gain here anymore and I shouldn't have started into that direction. If this is not a common pitfall but an act of isolated confusion, there is no need to change the entry.

    Ok maybe its better to let these speculations about mathematicians aside. I see nothing to gain here anymore and I shouldn’t have started into that direction.

    If this is not a common pitfall but an act of isolated confusion, there is no need to change the entry.