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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 30th 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexRecall that there is a canonical ordinal sum $\boxplus: \Delta\times\Delta\to \Delta$. This is not exactly a monoidal product, but it is a bifunctor. Composing with the Yoneda embedding $h_{(\cdot)}:\Delta\hookrightarrow \operatorname{Set}^{\Delta^{op}}$, we get a bifunctor $\Delta\times\Delta \to \operatorname{Set}^{\Delta^{op}}$. Since the target is cocomplete, this factors through the Yoneda embedding on the source, which gives us a functor $ \operatorname{Set}^{(\Delta\times\Delta)^{op}}\to \operatorname{Set}^{\Delta^{op}}$. Is there any way to continue this to get the actual join? I was thinking that maybe instead of the last step, use the cartesian-closed structure of $Cat$ to exponentiate $ \operatorname{Set}^{\Delta^{op}}$ by $\Delta$ under that adjunction, then apply the second Yoneda embedding, then apply the cartesian adjunction again to end up with a functor $\Delta\times \operatorname{Set}^{\Delta^{op}}\to \operatorname{Set}^{\Delta^{op}}$, but I'm not sure that will work.

   Recall that there is a canonical ordinal sum $\boxplus: \Delta\times\Delta\to \Delta$. This is not exactly a monoidal product, but it is a bifunctor. Composing with the Yoneda embedding $h_{(\cdot)}:\Delta\hookrightarrow \operatorname{Set}^{\Delta^{op}}$, we get a bifunctor $\Delta\times\Delta \to \operatorname{Set}^{\Delta^{op}}$. Since the target is cocomplete, this factors through the Yoneda embedding on the source, which gives us a functor $\operatorname{Set}^{(\Delta\times\Delta)^{op}}\to \operatorname{Set}^{\Delta^{op}}$.

   Is there any way to continue this to get the actual join? I was thinking that maybe instead of the last step, use the cartesian-closed structure of $Cat$ to exponentiate $\operatorname{Set}^{\Delta^{op}}$ by $\Delta$ under that adjunction, then apply the second Yoneda embedding, then apply the cartesian adjunction again to end up with a functor $\Delta\times \operatorname{Set}^{\Delta^{op}}\to \operatorname{Set}^{\Delta^{op}}$, but I’m not sure that will work.

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
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   Author: Todd_Trimble
   Format: MarkdownItex> This is not exactly a monoidal product It's not? > Is there any way to continue this to get the actual join? Do you mean to take the evident composite $$Set^{\Delta^{op}} \times Set^{\Delta^{op}} \stackrel{\Lambda}{\to} Set^{\Delta^{op} \times \Delta^{op}} \stackrel{Lan_{y(\Delta \times \Delta)} y(\Delta) \circ \boxplus}{\to} Set^{\Delta^{op}}$$ The first arrow $\Lambda$ is the adjunct of $$Set^{\Delta^{op}} \times Set^{\Delta^{op}} \times \Delta^{op} \times \Delta^{op} \simeq Set^{\Delta^{op}} \times \Delta^{op} \times Set^{\Delta^{op}} \times \Delta^{op} \stackrel{ev \times ev}{\to} Set \times Set \stackrel{\prod}{\to} Set$$ But the real way to think about this is that for any small categories $X$, $Y$, that $Set^{X^{op} \times Y^{op}}$ is the correct tensor product of $Set^{X^{op}}$ and $Set^{Y^{op}}$ with respect to cocontinuous maps, so that given any separately cocontinuous functor to a cocomplete category, $$F: Set^{X^{op}} \times Set^{Y^{op}} \to D,$$ there exists up to uniquely determined isomorphism a cocontinuous functor $F': Set^{X^{op} \times Y^{op}} \to D$ and an isomorphism $$F \cong F' \circ \Lambda$$ where $\Lambda: Set^{X^{op}} \times Set^{Y^{op}} \to Set^{X^{op} \times Y^{op}}$ is the universal separately cocontinuous functor.

   This is not exactly a monoidal product

   It’s not?

   Is there any way to continue this to get the actual join?

   Do you mean to take the evident composite

   $$Set^{\Delta^{op}} \times Set^{\Delta^{op}} \stackrel{\Lambda}{\to} Set^{\Delta^{op} \times \Delta^{op}} \stackrel{Lan_{y(\Delta \times \Delta)} y(\Delta) \circ \boxplus}{\to} Set^{\Delta^{op}}$$

   The first arrow $\Lambda$ is the adjunct of

   $$Set^{\Delta^{op}} \times Set^{\Delta^{op}} \times \Delta^{op} \times \Delta^{op} \simeq Set^{\Delta^{op}} \times \Delta^{op} \times Set^{\Delta^{op}} \times \Delta^{op} \stackrel{ev \times ev}{\to} Set \times Set \stackrel{\prod}{\to} Set$$

   But the real way to think about this is that for any small categories $X$, $Y$, that $Set^{X^{op} \times Y^{op}}$ is the correct tensor product of $Set^{X^{op}}$ and $Set^{Y^{op}}$ with respect to cocontinuous maps, so that given any separately cocontinuous functor to a cocomplete category,

   $$F: Set^{X^{op}} \times Set^{Y^{op}} \to D,$$

   there exists up to uniquely determined isomorphism a cocontinuous functor $F': Set^{X^{op} \times Y^{op}} \to D$ and an isomorphism

   $$F \cong F' \circ \Lambda$$

   where $\Lambda: Set^{X^{op}} \times Set^{Y^{op}} \to Set^{X^{op} \times Y^{op}}$ is the universal separately cocontinuous functor.

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
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   Author: Harry Gindi
   Format: MarkdownItex> It's not? Nope. It's monoidal on the augmented simplex category. It lifts to a monoidal functor on the presheaf category of $\Delta$ (with monoidal unit given by the empty presheaf) because of the separate cocontinuity thing. Anyway, thanks! Very nice! The join has some more structure that I left out. In particular, if we fix a simplicial set $Y$, it is a cocontinuous functor $Psh(\Delta)\to Y\downarrow Psh(\Delta)$, which is necessitated by the empty presheaf being the identity. This seems like it might make things a bit more difficult do deal with.

   It’s not?

   Nope. It’s monoidal on the augmented simplex category. It lifts to a monoidal functor on the presheaf category of $\Delta$ (with monoidal unit given by the empty presheaf) because of the separate cocontinuity thing.

   Anyway, thanks! Very nice!

   The join has some more structure that I left out.

   In particular, if we fix a simplicial set $Y$, it is a cocontinuous functor $Psh(\Delta)\to Y\downarrow Psh(\Delta)$, which is necessitated by the empty presheaf being the identity. This seems like it might make things a bit more difficult do deal with.

4. • CommentRowNumber4.
   • CommentAuthorTim_Porter
   • CommentTimeNov 30th 2010
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   Author: Tim_Porter
   Format: MarkdownItex@Harry What do you mean by 'the actual join'? As any simplicial set can be augmented in at least one way (if it is connected) and at least two ways if not, there are potentially several joins of a pair of arbitrary simplicial sets, but what are to be the ways of saying which is the 'actual' one?

   @Harry What do you mean by ’the actual join’? As any simplicial set can be augmented in at least one way (if it is connected) and at least two ways if not, there are potentially several joins of a pair of arbitrary simplicial sets, but what are to be the ways of saying which is the ’actual’ one?

5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexI mean the empty augmentation.

   I mean the empty augmentation.

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexHmm.. I think I know how it should be done now: You do it like this: We have the ordinal sum $\boxplus: \Delta_a\times\Delta_a\to \Delta_a$ which is honestly monoidal. Then you upgrade this to a cocontinuous functor $\operatorname{Set}^{(\Delta_a\times\Delta_a)^{op}}\to \operatorname{Set}^{\Delta_a^{op}}$. Then you apply the pullback of the inclusion $\Delta\hookrightarrow \Delta_a$ to everything in sight, which gives an induced functor $\operatorname{Set}^{(\Delta\times\Delta)^{op}}\to \operatorname{Set}^{\Delta^{op}}$, which is not cocontinuous. Then you apply what you did above to this functor to get Joyal's join. The important point I realized is that when the unit is the initial object, the join cannot actually be in either argument. It is however cocontinuous in each argument as a functor $Y\star(\cdot):Psh(\Delta)\to Y\downarrow Psh(\Delta)$. I may still have it screwed up, but there is something subtle you have to do here to pull the join back to honest simplicial sets and have the unit be the empty presheaf. To clarify, the thing I'm trying to induce is the join of Joyal and Lurie. The join of two simplicial sets in this sense is defined here: http://arxiv.org/pdf/math/0608040v4#page=42

   Hmm.. I think I know how it should be done now:

   You do it like this: We have the ordinal sum $\boxplus: \Delta_a\times\Delta_a\to \Delta_a$ which is honestly monoidal. Then you upgrade this to a cocontinuous functor $\operatorname{Set}^{(\Delta_a\times\Delta_a)^{op}}\to \operatorname{Set}^{\Delta_a^{op}}$. Then you apply the pullback of the inclusion $\Delta\hookrightarrow \Delta_a$ to everything in sight, which gives an induced functor $\operatorname{Set}^{(\Delta\times\Delta)^{op}}\to \operatorname{Set}^{\Delta^{op}}$, which is not cocontinuous. Then you apply what you did above to this functor to get Joyal’s join.

   The important point I realized is that when the unit is the initial object, the join cannot actually be in either argument. It is however cocontinuous in each argument as a functor $Y\star(\cdot):Psh(\Delta)\to Y\downarrow Psh(\Delta)$.

   I may still have it screwed up, but there is something subtle you have to do here to pull the join back to honest simplicial sets and have the unit be the empty presheaf.

   To clarify, the thing I’m trying to induce is the join of Joyal and Lurie. The join of two simplicial sets in this sense is defined here: http://arxiv.org/pdf/math/0608040v4#page=42

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
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   Author: Todd_Trimble
   Format: MarkdownItex> Nope. It's monoidal on the augmented simplex category. Oh sure, right. > the empty augmentation Eh? The component in dimension -1 can't be empty if there is to be an augmentation map $X_0 \to X_{-1}$ (unless $X$ itself is empty). The two canonical choices are left and right adjoint to the restriction functor $Set^{\Delta_{a}^{op}} \to Set^{\Delta^{op}}$. The left adjoint sets $X_{-1}$ equal to the set of connected components, which has the canonical projection $X_0 \to \pi_0(X)$ as the augmentation, and the right adjoint puts $X_{-1} = 1$. Did you mean either of those? If you choose the left Kan extension $L: Set^{\Delta^{op}} \to Set^{\Delta_{a}^{op}}$, which is cocontinuous, then you could apply $$Set^{\Delta^{op}} \times Set^{\Delta^{op}} \stackrel{L \times L}{\to} Set^{\Delta_{a}^{op}} \times Set^{\Delta_{a}^{op}} \to Set^{\Delta_{a}^{op} \times \Delta_{a}^{op}} \to Set^{\Delta_{a}^{op}} \stackrel{res}{\to} Set^{\Delta^{op}}$$ which is separately cocontinuous.

   Nope. It’s monoidal on the augmented simplex category.

   Oh sure, right.

   the empty augmentation

   Eh? The component in dimension -1 can’t be empty if there is to be an augmentation map $X_0 \to X_{-1}$ (unless $X$ itself is empty).

   The two canonical choices are left and right adjoint to the restriction functor $Set^{\Delta_{a}^{op}} \to Set^{\Delta^{op}}$. The left adjoint sets $X_{-1}$ equal to the set of connected components, which has the canonical projection $X_0 \to \pi_0(X)$ as the augmentation, and the right adjoint puts $X_{-1} = 1$. Did you mean either of those?

   If you choose the left Kan extension $L: Set^{\Delta^{op}} \to Set^{\Delta_{a}^{op}}$, which is cocontinuous, then you could apply

   $$Set^{\Delta^{op}} \times Set^{\Delta^{op}} \stackrel{L \times L}{\to} Set^{\Delta_{a}^{op}} \times Set^{\Delta_{a}^{op}} \to Set^{\Delta_{a}^{op} \times \Delta_{a}^{op}} \to Set^{\Delta_{a}^{op}} \stackrel{res}{\to} Set^{\Delta^{op}}$$

   which is separately cocontinuous.

8. • CommentRowNumber8.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexWhen you said that the right adjoint sets $X_{-1}$ to $1$, I think that's right, if $1$ means that the augmentation is somehow trivial. The join as described on [page 42](http://arxiv.org/pdf/math/0608040v4#page=42) of HTT is not separately cocontinuous as a bifunctor on simplicial sets. It is only cocontinuous in the following sense: If $Y$ is a simplicial set, $Y\star (\cdot):Psh(\Delta)\to Y\downarrow Psh(\Delta)$ is cocontinuous. This is actually fairly important, since we want the unit of the join to be the empty presheaf, but this means that the join cannot preserve the empty colimit when the target is the category of simplicial sets(since this would force the join of the empty presheaf with anything to be zero). It's similar to how taking the product with an object brings you into the overcategory.

   When you said that the right adjoint sets $X_{-1}$ to $1$, I think that’s right, if $1$ means that the augmentation is somehow trivial. The join as described on page 42 of HTT is not separately cocontinuous as a bifunctor on simplicial sets. It is only cocontinuous in the following sense: If $Y$ is a simplicial set, $Y\star (\cdot):Psh(\Delta)\to Y\downarrow Psh(\Delta)$ is cocontinuous.

   This is actually fairly important, since we want the unit of the join to be the empty presheaf, but this means that the join cannot preserve the empty colimit when the target is the category of simplicial sets(since this would force the join of the empty presheaf with anything to be zero).

   It’s similar to how taking the product with an object brings you into the overcategory.

9. • CommentRowNumber9.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 30th 2010
   • (edited Nov 30th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexI think it's the right adjoint for the following reason: It sends the empty presheaf to the monoidal unit in $Psh(\Delta_a)$. Then evaluating $\emptyset \star X$, we have: $$(\emptyset,X)\mapsto (\Delta_{-1}, Aug_1(X))\mapsto ? \mapsto \Delta_{-1}\star_a Aug(X)=Aug(X)\mapsto X,$$ so that's good. If we use the left adjoint and evaluate $\emptyset \star X$, we have: $$(\emptyset,X)\mapsto (\emptyset,Aug_{\pi_0}(X))\mapsto ? \mapsto \emptyset \star_a Aug_{\pi_0}(X)=\emptyset \mapsto \emptyset.$$ That is, we have no unit, since no simplicial set will map to $\Delta_{-1}$. I'll add this formulation to the nLab later if it turns out to be right.

   I think it’s the right adjoint for the following reason: It sends the empty presheaf to the monoidal unit in $Psh(\Delta_a)$. Then evaluating $\emptyset \star X$, we have:

   $$(\emptyset,X)\mapsto (\Delta_{-1}, Aug_1(X))\mapsto ? \mapsto \Delta_{-1}\star_a Aug(X)=Aug(X)\mapsto X,$$

   so that’s good. If we use the left adjoint and evaluate $\emptyset \star X$, we have:

   $$(\emptyset,X)\mapsto (\emptyset,Aug_{\pi_0}(X))\mapsto ? \mapsto \emptyset \star_a Aug_{\pi_0}(X)=\emptyset \mapsto \emptyset.$$

   That is, we have no unit, since no simplicial set will map to $\Delta_{-1}$.

   I’ll add this formulation to the nLab later if it turns out to be right.

10. • CommentRowNumber10.
    • CommentAuthorTim_Porter
    • CommentTimeNov 30th 2010
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    Author: Tim_Porter
    Format: MarkdownItex@Harry I think I told you before of Phil Ehlers' thesis (1993) has all this in it at a good level of detail. (He learnt it in correspondence with Jack Duskin and Don van Osdol, who in turn heard it from Lawvere, and probably Street as it goes back to Day). I somehow take exception to it being called the Joyal-Lurie join or similar therefore.It is also a generalisation of the classical join of connected simplicial complexes which is in Spanier. (I wrote up Phil's treatment and it was published in JPAA.(P. J. Ehlers and Tim Porter, Joins for (Augmented) Simplicial Sets, Jour. Pure Applied Algebra, 145 (2000) 37-44) see the nLab page for a discussion of all this. :-)

    @Harry I think I told you before of Phil Ehlers’ thesis (1993) has all this in it at a good level of detail. (He learnt it in correspondence with Jack Duskin and Don van Osdol, who in turn heard it from Lawvere, and probably Street as it goes back to Day). I somehow take exception to it being called the Joyal-Lurie join or similar therefore.It is also a generalisation of the classical join of connected simplicial complexes which is in Spanier. (I wrote up Phil’s treatment and it was published in JPAA.(P. J. Ehlers and Tim Porter, Joins for (Augmented) Simplicial Sets, Jour. Pure Applied Algebra, 145 (2000) 37-44) see the nLab page for a discussion of all this. :-)

11. • CommentRowNumber11.
    • CommentAuthorHarry Gindi
    • CommentTimeNov 30th 2010
    • (edited Nov 30th 2010)
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    Author: Harry Gindi
    Format: MarkdownItex@Tim: I was clarifying by calling it the Joyal-Lurie join, since you seemed to be confused about exactly which augmentation I was talking about! (The only places I've seen this specific join used are Joyal's notes on quasicategories and Lurie's book). I have tried to read Ehlers's thesis, but I don't understand what Dec is, and I still don't understand ends and coends.

    @Tim: I was clarifying by calling it the Joyal-Lurie join, since you seemed to be confused about exactly which augmentation I was talking about! (The only places I’ve seen this specific join used are Joyal’s notes on quasicategories and Lurie’s book).

    I have tried to read Ehlers’s thesis, but I don’t understand what Dec is, and I still don’t understand ends and coends.

12. • CommentRowNumber12.
    • CommentAuthorHarry Gindi
    • CommentTimeDec 1st 2010
    • (edited Dec 1st 2010)
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    Author: Harry Gindi
    Format: MarkdownItexI wonder what's the best way to show that $X\mapsto Y=Y\star\emptyset \to Y\star X$ as a functor $sSet\to Y\downarrow sSet$ preserves colimits. It evidently preserves the initial object by construction, but why should it send colimits to colimits? Is the best way to do it just to construct the adjoint formally, that is, $p:Y\to Z \mapsto (Z_{p/})_n=Hom_{Y\downarrow sSet}(\iota_{Y\star \Delta^n},p)$ and prove it via unit/counit adjunction or by the existence of universal arrows? Is there any obvious way to see it from the definition above (if it is even correct)? Oh, hmmm... I think it has something to do with the idea that if we join $Y$ with a colimit in the augmented simplex category, Y becomes the new augmentation, and the colimit is taken over $Y$, but I'm not sure how to show this precisely.

    I wonder what’s the best way to show that $X\mapsto Y=Y\star\emptyset \to Y\star X$ as a functor $sSet\to Y\downarrow sSet$ preserves colimits. It evidently preserves the initial object by construction, but why should it send colimits to colimits?

    Is the best way to do it just to construct the adjoint formally, that is, $p:Y\to Z \mapsto (Z_{p/})_n=Hom_{Y\downarrow sSet}(\iota_{Y\star \Delta^n},p)$ and prove it via unit/counit adjunction or by the existence of universal arrows?

    Is there any obvious way to see it from the definition above (if it is even correct)?

    Oh, hmmm… I think it has something to do with the idea that if we join $Y$ with a colimit in the augmented simplex category, Y becomes the new augmentation, and the colimit is taken over $Y$, but I’m not sure how to show this precisely.

13. • CommentRowNumber13.
    • CommentAuthorTodd_Trimble
    • CommentTimeDec 1st 2010
    • (edited Dec 1st 2010)
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    Author: Todd_Trimble
    Format: MarkdownItexHarry, your description is correct. The choice of the right adjoint corresponds to the convention Lurie gives: $S(\emptyset) = 1 = S'(\emptyset)$, which uniquely determines the augmentation. The rest just follows by free cocompletion abstract nonsense (together with forgetting the augmentation at the end, which is the restriction functor from augmented simplicial sets to simplicial sets). To see the functor $Y \star -: sSet \to Y \downarrow sSet$ preserves colimits, it's enough to show that the right adjoint $sSet \to sSet_a$ preserves wide pushouts. (See the theorem at [[connected limit]] and its proof, or perhaps see especially [[parametric right adjoint]], for the dual statement.) To do this, you just show this component by component. Now in components in dimension $n \geq 0$, the right adjoint behaves as the identity functor, so we're okay in those dimensions. In dimension $-1$, what we have is the constant functor (at the value $1$). But the constant functor preserves wide pushouts. We are done.

    Harry, your description is correct. The choice of the right adjoint corresponds to the convention Lurie gives: $S(\emptyset) = 1 = S'(\emptyset)$, which uniquely determines the augmentation. The rest just follows by free cocompletion abstract nonsense (together with forgetting the augmentation at the end, which is the restriction functor from augmented simplicial sets to simplicial sets).

    To see the functor $Y \star -: sSet \to Y \downarrow sSet$ preserves colimits, it’s enough to show that the right adjoint $sSet \to sSet_a$ preserves wide pushouts. (See the theorem at connected limit and its proof, or perhaps see especially parametric right adjoint, for the dual statement.) To do this, you just show this component by component. Now in components in dimension $n \geq 0$, the right adjoint behaves as the identity functor, so we’re okay in those dimensions. In dimension $-1$, what we have is the constant functor (at the value $1$). But the constant functor preserves wide pushouts. We are done.

14. • CommentRowNumber14.
    • CommentAuthorHarry Gindi
    • CommentTimeDec 1st 2010
    • (edited Dec 1st 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexAh, I think it's something like this: if $F:C\to sSet$ is a functor, this determines a cone $\{\emptyset \to F(c)\}$. Sending this diagram up to AsSet, this determines a cone $\{\Delta^{-1}\to Aug(F(c))\}$, that is, a functor $Aug(F):[0]\star C\to AsSet$, where $[0]\star C$ is the ordinary join of categories. Further, $colim Aug(F)=Aug(colim F)$. Joining with $Aug(Y)$ and using the commutation of colimits with joins in $AsSet$, we have that $$Aug(Y)\star Aug(colim F)\cong Aug(Y)\star \colim Aug(F) \cong colim Aug(Y)\star Aug(F).$$ Mapping back down to sSet via the forgetful functor $j$, we have that $j(colim Aug(Y)\star Aug(F))\cong colim j(Aug(Y)\star Aug(F))$, but the diagram $Aug(Y)\star Aug(F)$ maps to the cone $\{Y\to Y\star F(c)\}$, and the colimit of this cone is exactly $Y\to Y\star colim F$, which proves commutation of colimits. Is that right? (Oh, I didn't see that you posted there. I'll check out your answer. Meanwhile, did I have the right idea?) Edit: (Looking at the articles that you linked, I think I got it right). Edit 2: (The key idea here is that $Aug$ can be factored as a composition of functors $sSet\to \Delta^{-1}\downarrow AsSet \to AsSet$, where the first functor is cocontinuous or something like that? How do we make things more precise?)

    Ah, I think it’s something like this:

    if $F:C\to sSet$ is a functor, this determines a cone $\{\emptyset \to F(c)\}$. Sending this diagram up to AsSet, this determines a cone $\{\Delta^{-1}\to Aug(F(c))\}$, that is, a functor $Aug(F):[0]\star C\to AsSet$, where $[0]\star C$ is the ordinary join of categories. Further, $colim Aug(F)=Aug(colim F)$. Joining with $Aug(Y)$ and using the commutation of colimits with joins in $AsSet$, we have that

    $$Aug(Y)\star Aug(colim F)\cong Aug(Y)\star \colim Aug(F) \cong colim Aug(Y)\star Aug(F).$$

    Mapping back down to sSet via the forgetful functor $j$, we have that $j(colim Aug(Y)\star Aug(F))\cong colim j(Aug(Y)\star Aug(F))$, but the diagram $Aug(Y)\star Aug(F)$ maps to the cone $\{Y\to Y\star F(c)\}$, and the colimit of this cone is exactly $Y\to Y\star colim F$, which proves commutation of colimits.

    Is that right?

    (Oh, I didn’t see that you posted there. I’ll check out your answer. Meanwhile, did I have the right idea?)

    Edit: (Looking at the articles that you linked, I think I got it right).

    Edit 2: (The key idea here is that $Aug$ can be factored as a composition of functors $sSet\to \Delta^{-1}\downarrow AsSet \to AsSet$, where the first functor is cocontinuous or something like that? How do we make things more precise?)

15. • CommentRowNumber15.
    • CommentAuthorTim_Porter
    • CommentTimeDec 1st 2010
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    Author: Tim_Porter
    Format: MarkdownItex@Harry I do think that Dec and the end/coend calculus is so useful in all this that it would make your like infinitely easier!!!! For Dec there is very little to understand on how it works and it may pay to go back to Jack Duskin's Memoir AMS where he talks about Dec quite a lot. It depends what you mean by 'understand' however :-) I don't think I am confused by what augmentation you are using, but as the construction does depend on the augmentation used and is most naturally considered in the augmented context, I feel that to look at it in the context of simplicial sets is like trying to dance the salsa in shoes with the laces tied together... if you see what I mean. The dancer tends to at least stumble if not fall flat on their face! (I am also irritated by papers and books that do not give reasonably correct references the origin of ideas as I have suffered from that too much. This is not aimed at you, however, rather at some of the sources.) Ends and coends are tricky to grasp at first but are well worth the effort (and as usual are easy once they click!). I found that thinking of the examples of geometric realisation, and the set of natural transformation between two functors helped me to grasp what was happening, but like so much you have to find the 'right' way of thinking of them for yourself.

    @Harry I do think that Dec and the end/coend calculus is so useful in all this that it would make your like infinitely easier!!!! For Dec there is very little to understand on how it works and it may pay to go back to Jack Duskin’s Memoir AMS where he talks about Dec quite a lot. It depends what you mean by ’understand’ however :-)

    I don’t think I am confused by what augmentation you are using, but as the construction does depend on the augmentation used and is most naturally considered in the augmented context, I feel that to look at it in the context of simplicial sets is like trying to dance the salsa in shoes with the laces tied together… if you see what I mean. The dancer tends to at least stumble if not fall flat on their face! (I am also irritated by papers and books that do not give reasonably correct references the origin of ideas as I have suffered from that too much. This is not aimed at you, however, rather at some of the sources.)

    Ends and coends are tricky to grasp at first but are well worth the effort (and as usual are easy once they click!). I found that thinking of the examples of geometric realisation, and the set of natural transformation between two functors helped me to grasp what was happening, but like so much you have to find the ’right’ way of thinking of them for yourself.

16. • CommentRowNumber16.
    • CommentAuthorHarry Gindi
    • CommentTimeDec 1st 2010
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItex@Tim: I understood the construction for AsSets just fine. The tricky part was figuring it out for sSets and also figuring out how to understand the right adjoint into the undercategory, a phenomenon not present in the augmented case. Regarding ends and coends, I have trouble following computations involving them, and in particular, I don't understand how the heck we're supposed to prove things like the commutation of ends and coends with the enriched hom, when ends and coends decompose, etc. I tried reading Kelly's book, but none of the computations are really covered, and his usage of the word "$\mathcal{V}$-natural" is extremely confusing. Do you know of any book or paper that discusses things from scratch in a different way? Mac Lane's book cheats by using the subdivision category and doesn't bother with the enriched case, so that's a no-go, and Kashiwara-Schapira ignores the subject entirely. Also, I often have trouble understanding what the notation even means.

    @Tim: I understood the construction for AsSets just fine. The tricky part was figuring it out for sSets and also figuring out how to understand the right adjoint into the undercategory, a phenomenon not present in the augmented case.

    Regarding ends and coends, I have trouble following computations involving them, and in particular, I don’t understand how the heck we’re supposed to prove things like the commutation of ends and coends with the enriched hom, when ends and coends decompose, etc. I tried reading Kelly’s book, but none of the computations are really covered, and his usage of the word “$\mathcal{V}$-natural” is extremely confusing. Do you know of any book or paper that discusses things from scratch in a different way? Mac Lane’s book cheats by using the subdivision category and doesn’t bother with the enriched case, so that’s a no-go, and Kashiwara-Schapira ignores the subject entirely.

    Also, I often have trouble understanding what the notation even means.

17. • CommentRowNumber17.
    • CommentAuthorTim_Porter
    • CommentTimeDec 1st 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItexI will think about where you might find an approachable version of ends and coends. For the enriched case, I suspect that you need indexed ends and coends anyway (for simplicially enriched categories Jean-Marc and I had a go at this in out TAMS paper and there are papers by other well known people on indexed coherent ends etc. which no doubt you have looked at.) The use of the subdivision category is not really a cheat but is a set-enriched version of the more general indexing that is needed (I think that is right). I found myself recently needing the subdivision category for something else and found it strange and inadequately justified in most of the literature. I must rush now. I did put ends and coends in the Menagerie but not in the bit that is relatively stable. We clearly need to work on the nLab version with possibly input from yourself as to have something written by someone who is finding something tricky to master is a good way of ensuring that the 'oh that is so simple' syndrome doesn't take over!!!! (I watched one of Tom Lehrer's songs the other day which was exactly on that syndrome!)

    I will think about where you might find an approachable version of ends and coends. For the enriched case, I suspect that you need indexed ends and coends anyway (for simplicially enriched categories Jean-Marc and I had a go at this in out TAMS paper and there are papers by other well known people on indexed coherent ends etc. which no doubt you have looked at.)

    The use of the subdivision category is not really a cheat but is a set-enriched version of the more general indexing that is needed (I think that is right). I found myself recently needing the subdivision category for something else and found it strange and inadequately justified in most of the literature.

    I must rush now. I did put ends and coends in the Menagerie but not in the bit that is relatively stable. We clearly need to work on the nLab version with possibly input from yourself as to have something written by someone who is finding something tricky to master is a good way of ensuring that the ’oh that is so simple’ syndrome doesn’t take over!!!! (I watched one of Tom Lehrer’s songs the other day which was exactly on that syndrome!)

18. • CommentRowNumber18.
    • CommentAuthorTim_Porter
    • CommentTimeDec 1st 2010
    • PermaLink
    Author: Tim_Porter
    Format: MarkdownItex@Harry If you want to look at Chapter 12 of the Menagerie and have not seen it, I can send you a copy. Oh, total Dec is simply composition with ordinal sum, and the two forms of decalage are then the composition along $\oplus [0]$, and $[0]\oplus -$ and everything else about them follows from that. (but I do think that it is worth exploring them as the left and right cones on the augmentation. (There is a lovely paper by Eric Goubault and his brother which uses dec and its geometric meaning in the context of intuitionistic modal logic. I keep on meaning to write up an account for the lab but there are so many things I should do!!!)

    @Harry If you want to look at Chapter 12 of the Menagerie and have not seen it, I can send you a copy.

    Oh, total Dec is simply composition with ordinal sum, and the two forms of decalage are then the composition along $\oplus [0]$, and $[0]\oplus -$ and everything else about them follows from that. (but I do think that it is worth exploring them as the left and right cones on the augmentation. (There is a lovely paper by Eric Goubault and his brother which uses dec and its geometric meaning in the context of intuitionistic modal logic. I keep on meaning to write up an account for the lab but there are so many things I should do!!!)