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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 30th 2010
   • (edited Jun 30th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexLet $X\to \Delta^n$ be a right fibration (has the right lifting property with respect to right horn inclusions), and let $$\Delta^{\{n-i\}}\hookrightarrow \Delta^{\{n-i,\dots, n\}}\hookrightarrow \Delta^n$$ (for a fixed $i: 0\leq i\leq n$) be the obvious inclusion maps. Then why is the induced map: $$X\times_{\Delta^n} \Delta^{\{n-i\}} \hookrightarrow X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}}$$ a deformation retract? (This is from the end of the proof of proposition 2.2.3.1 of [HTT](http://arxiv.org/pdf/math/0608040v4). The statement should be true out of the context in the book with the hypotheses I've given, but if not, there's the source.) And if you'd like to answer it on [Math Overflow](http://mathoverflow.net/questions/29992/why-is-the-induced-map-between-pullbacks-of-inclusions-by-a-right-fibration-a-d), go ahead.

   Let $X\to \Delta^n$ be a right fibration (has the right lifting property with respect to right horn inclusions), and let

   $$\Delta^{\{n-i\}}\hookrightarrow \Delta^{\{n-i,\dots, n\}}\hookrightarrow \Delta^n$$

   (for a fixed $i: 0\leq i\leq n$) be the obvious inclusion maps.

   Then why is the induced map:

   $$X\times_{\Delta^n} \Delta^{\{n-i\}} \hookrightarrow X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}}$$

   a deformation retract?

   (This is from the end of the proof of proposition 2.2.3.1 of HTT. The statement should be true out of the context in the book with the hypotheses I’ve given, but if not, there’s the source.)

   And if you’d like to answer it on Math Overflow, go ahead.

2. • CommentRowNumber2.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 30th 2010
   • (edited Jun 30th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexHere's hoping someone will give it a shot. *Bump* In all seriousness, Tom Goodwillie gave an answer on MO, but I can't see how it works, since it doesn't seem like the inclusion of the point into the n-simplex is actually a deformation retract (although it is in fact anodyne). The trouble is that the n-simplex is not a kan complex.

   Here’s hoping someone will give it a shot.

   Bump

   In all seriousness, Tom Goodwillie gave an answer on MO, but I can’t see how it works, since it doesn’t seem like the inclusion of the point into the n-simplex is actually a deformation retract (although it is in fact anodyne). The trouble is that the n-simplex is not a kan complex.

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 30th 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexIt's fine: it is a deformation retract. Let's take the inclusion of the zeroth vertex (as Tom Goodwillie did), and look at it this way: starting off in $Cat$, we have a natural transformation from the composite $[n] \to [1] \to [n]$ to the identity ($[n]$ is the $n$-element total order), where $[1] \to [n]$ names the initial object. This natural transformation can be re-expressed as a functor $[2] \times [n] \to [n]$. Take the nerve of this functor; it gives $$\hom(-, [2]) \times \hom(-, [n]) \to \hom(-, [n])$$ which is a simplicial homotopy from $i r$ to $id$ ($i$ the inclusion of the point, $r$ the retraction).

   It’s fine: it is a deformation retract. Let’s take the inclusion of the zeroth vertex (as Tom Goodwillie did), and look at it this way: starting off in $Cat$, we have a natural transformation from the composite $[n] \to [1] \to [n]$ to the identity ($[n]$ is the $n$-element total order), where $[1] \to [n]$ names the initial object. This natural transformation can be re-expressed as a functor $[2] \times [n] \to [n]$. Take the nerve of this functor; it gives

   $$\hom(-, [2]) \times \hom(-, [n]) \to \hom(-, [n])$$

   which is a simplicial homotopy from $i r$ to $id$ ($i$ the inclusion of the point, $r$ the retraction).

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 30th 2010
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexThanks!

   Thanks!

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeJun 30th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexAnd it is the fact that we are dealing with right fibrations that allows to lift this to a retract of the objects in question: the composition cells that are guaranteed to exist allow to lift the simplicial homotopy that Todd indicated.

   And it is the fact that we are dealing with right fibrations that allows to lift this to a retract of the objects in question: the composition cells that are guaranteed to exist allow to lift the simplicial homotopy that Todd indicated.

6. • CommentRowNumber6.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 30th 2010
   • (edited Jun 30th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexHey Todd, is your indexing one different from the usual indexing (where [0] is usually taken to be the terminal category and [1] is taken to be the interval category)? @Urs: I still don't really follow what's lifting what, but I'm going to give it a shot.

   Hey Todd, is your indexing one different from the usual indexing (where [0] is usually taken to be the terminal category and [1] is taken to be the interval category)?

   @Urs: I still don’t really follow what’s lifting what, but I’m going to give it a shot.

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 30th 2010
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexHarry: yes (which is why I bothered to defined what $[n]$ meant (-: ).

   Harry: yes (which is why I bothered to defined what $[n]$ meant (-: ).

8. • CommentRowNumber8.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 30th 2010
   • (edited Jun 30th 2010)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexCould someone help me set up the lifting problem? I don't see what exactly we're trying to lift. We can easily reduce to the case where it's the inclusion of the initial vertex (right fibrations are stable under pullback), so we've got a pullback diagram $X\to \Delta^n \leftarrow \{0\}$, where the first map is a right fibration and the second map is a deformation retract. Call the pullback of this diagram $P$. Then we want to show that the projection $P\to X$ is a deformation retraction. Now, I don't see where we should set up the lifting problem.

   Could someone help me set up the lifting problem? I don’t see what exactly we’re trying to lift.

   We can easily reduce to the case where it’s the inclusion of the initial vertex (right fibrations are stable under pullback), so we’ve got a pullback diagram

   $X\to \Delta^n \leftarrow \{0\}$, where the first map is a right fibration and the second map is a deformation retract.

   Call the pullback of this diagram $P$. Then we want to show that the projection $P\to X$ is a deformation retraction.

   Now, I don’t see where we should set up the lifting problem.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeJun 30th 2010
   • (edited Jun 30th 2010)
   • PermaLink
   Author: Urs
   Format: MarkdownItexWe look at the diagram $$ \array{ X\times_{\Delta^n} \Delta^{\{n-i\}} &\hookrightarrow& X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}} \\ \downarrow && \downarrow \\ \Delta^{\{n-i\}} &\hookrightarrow& \Delta^{\{n-i,\dots, n\}} } \,. $$ We want to show that the top morphism is a deformation retract. We know that the bottom morphism is. We want to see that this lifts from the bottom to the top. What we know is that the two vertical morphisms are right fibrations, because they are both pullbacks of right fibrations and these are stable under pullback. I think one can -- but I haven't taken pen and paper to actually do -- see that using the right horn fillers that the right fibratio property guarantees, we ca build the simplicial map $$ \Delta[2] \times X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}} \to X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}} $$ that covers the one that Todd describes ad exhibits the desired deformation retract. I don't feel I have the leisure to work this out further, I have other exercises that I badly need to be looking into myself, but this is what I am envisioning the strategy here would be.

   We look at the diagram

   $$\array{ X\times_{\Delta^n} \Delta^{\{n-i\}} &\hookrightarrow& X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}} \\ \downarrow && \downarrow \\ \Delta^{\{n-i\}} &\hookrightarrow& \Delta^{\{n-i,\dots, n\}} } \,.$$

   We want to show that the top morphism is a deformation retract. We know that the bottom morphism is. We want to see that this lifts from the bottom to the top.

   What we know is that the two vertical morphisms are right fibrations, because they are both pullbacks of right fibrations and these are stable under pullback.

   I think one can – but I haven’t taken pen and paper to actually do – see that using the right horn fillers that the right fibratio property guarantees, we ca build the simplicial map

   $$\Delta[2] \times X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}} \to X\times_{\Delta^n}\Delta^{\{n-i,\dots, n\}}$$

   that covers the one that Todd describes ad exhibits the desired deformation retract.

   I don’t feel I have the leisure to work this out further, I have other exercises that I badly need to be looking into myself, but this is what I am envisioning the strategy here would be.

10. • CommentRowNumber10.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 30th 2010
    • (edited Jun 30th 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexI think I've got it, but it's pretty complicated. @Urs: I don't think that will work.

    I think I’ve got it, but it’s pretty complicated.

    @Urs: I don’t think that will work.

11. • CommentRowNumber11.
    • CommentAuthorHarry Gindi
    • CommentTimeJul 1st 2010
    • (edited Jul 1st 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexAlright, the trouble is, I can't show that my new homotopy sends the 0 projection to the idempotent composite. What I did was notice that in the diagram where we're trying to show a lifting ($X\times \Delta^1\leftarrow X\times \partial \Delta^1 \stackrel{(fg,id_x)}{\to} X$), we can factor the (implied) terminal maps as $X\to \Delta^n\to *$ and $X\times \Delta^1\to\Delta^n\times\Delta^1\to *$ because $X$ lives over $\Delta^n$. It's easy to check that the factorization is compatible with the the map (fg,id_X), which gives us a lift $X\times \Delta^1 \to \Delta^n$, making a commutative square. Now it looks like we're in business. Using the thing Tom Goodwillie mentioned, we can look at the union and induce maps on the righthand inclusion, which we choose to be the identity and the projection composed with the map f, which makes the diagram commute. We now have by the lifting property a homotopy H such that $H|X\times \{1\}=id_X$ and also that H behaves as a projection composed with the inclusion $f: P\to X$ on $P\times I$ where $P=X\times_{\Delta^n} \{0\}$ from the beginning. Well it looks like we're finally on to something, but now we must verify that $H|X\times\{0\}$ is the composite $fg$, where $g$ is the map splitting $f$. It's not clear to me how we can show this. The composite of the restriction with the map $X\to \Delta^n$ is equal to the composite of $fg$ with the map $X\to \Delta^n$, but I can't seem to show that the restriction itself is actually equal to $fg$. Do we need to do some kind of double homotopy trick? Maybe it has to do with the uniqueness of the maps f and g, since they are induced by the pullback?

    Alright, the trouble is, I can’t show that my new homotopy sends the 0 projection to the idempotent composite.

    What I did was notice that in the diagram where we’re trying to show a lifting ($X\times \Delta^1\leftarrow X\times \partial \Delta^1 \stackrel{(fg,id_x)}{\to} X$), we can factor the (implied) terminal maps as $X\to \Delta^n\to *$ and $X\times \Delta^1\to\Delta^n\times\Delta^1\to *$ because $X$ lives over $\Delta^n$. It’s easy to check that the factorization is compatible with the the map (fg,id_X), which gives us a lift $X\times \Delta^1 \to \Delta^n$, making a commutative square.

    Now it looks like we’re in business. Using the thing Tom Goodwillie mentioned, we can look at the union and induce maps on the righthand inclusion, which we choose to be the identity and the projection composed with the map f, which makes the diagram commute. We now have by the lifting property a homotopy H such that $H|X\times \{1\}=id_X$ and also that H behaves as a projection composed with the inclusion $f: P\to X$ on $P\times I$ where $P=X\times_{\Delta^n} \{0\}$ from the beginning.

    Well it looks like we’re finally on to something, but now we must verify that $H|X\times\{0\}$ is the composite $fg$, where $g$ is the map splitting $f$. It’s not clear to me how we can show this. The composite of the restriction with the map $X\to \Delta^n$ is equal to the composite of $fg$ with the map $X\to \Delta^n$, but I can’t seem to show that the restriction itself is actually equal to $fg$.

    Do we need to do some kind of double homotopy trick?

    Maybe it has to do with the uniqueness of the maps f and g, since they are induced by the pullback?

12. • CommentRowNumber12.
    • CommentAuthorHarry Gindi
    • CommentTimeJul 1st 2010
    • (edited Jul 1st 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexLet $X$ be a simplicial set, and let $\ell:X\to \Delta^n$ be a right fibration. Consider the inclusion of the $0^{th}$ vertex $\{0\}\hookrightarrow \Delta^n$. This is a deformation retract, since we have a homotopy $h:\Delta^n\times \Delta^1 \to \Delta^n$, where $h|_{\Delta^n\times \{0\}}=\Delta^n\to \{0\} \hookrightarrow \Delta^n$ is the composite of the terminal map and the inclusion. and $h|_{\Delta^n\times \{1\}}=id_{\Delta^n}$. Taking the pullback $X\times_{\Delta^n}(-)$, we would like to show that the induced map $f:X\times_{\Delta^n}\{0\}\hookrightarrow X\times_{\Delta^n}\Delta^n=X$ is also a deformation retract. It's clear that $f$ admits a retraction, $g$, so we'd like to give a homotopy $fg\to id_X$. We have the obvious setup to construct the homotopy as the span $X\times \Delta^1 \hookleftarrow X\times \partial\Delta^1\stackrel{(fg,id_X)}{\to} X$. We notice that the implied terminal maps of the diagram can be factored into $X\to \Delta^n\to *$ and $X\times \Delta^1 \to \Delta^n\times \Delta^1 \to *$. We notice further that we can factor the the composite $X\times \partial\Delta^1\hookrightarrow X\times \Delta^1\to \Delta^n \times \Delta^1$ instead as $$X\times \partial\Delta^1\to \Delta^n\times \partial\Delta^1 \to \Delta^n\times \Delta^1,$$ and the composite $X\times \partial\Delta^1 \to X\to \Delta^n$ as $$X\times \partial\Delta^1\to \Delta^n\times \partial\Delta^1 \to \Delta^n.$$ These factorizations let us fit the homotopy $h$ into the diagram, since it is easy to see that these maps are exactly the ones that $h$ lifts. This gives a commutative square for our span with $\Delta^n$ in the bottom right corner, where the maps into it are the obvious ones induced by the homotopy. We'd now like to give the homotopy we've been looking for. Since $\ell$ is a right fibration, we can lift maps of the form $K\times \Delta^1 \cup L\times \{1\}\to L\times \Delta^1$ for any inclusion $K\hookrightarrow L$. Let $A:=X\times_{\Delta^n}\{0\}$. We have the inclusion $f:A\hookrightarrow X$, so applying the pushout above, we have the map $A\times \Delta^1 \cup X\times \{1\}\to X\times \Delta^1$. We give a map from the pushout to $X$ by specifying it on the components. The map $A\times \Delta^1 \to X$ is given by the composite of the projection onto $A$ with the map $f$. The map $X\times \{1\}\to X$ is just the identity on $X$. Then we have a commutative square with the pushout at the top left, $X\times \Delta^1$ at the bottom left, $X$ at the top right, and $\Delta^n$ in the bottom right. The by the lifting property, we have a lift $H:X\times\Delta^1\to X$ making the whole diagram commute. We'd like to show that $H$ is the required homotopy. By construction, $H|_{X\times \{1\}}$ gives the identity and $H\circ (f\times \Delta^1)$ is equal to the first projection composed with $f$. Then we need only show that $H|_{X\times \{0\}}=fg$. It's easy to show that the composite $\ell H|_{X\times \{0\}}=\ell fg$ by the commutativity of the diagrams, and also that $H|_{X\times \{0\}} f=f$ by the last sentence of the previous paragraph. Can we prove that $H$ is the required homotopy, or do we need to do another construction and another homotopy? If this is the wrong way to prove this, let me know. Also, I'm really stuck, so if you have time, I entreat you, please help. I've been stuck for 28 hours, and I'm now obsessing over it. =(.

    Let $X$ be a simplicial set, and let $\ell:X\to \Delta^n$ be a right fibration.

    Consider the inclusion of the $0^{th}$ vertex $\{0\}\hookrightarrow \Delta^n$. This is a deformation retract, since we have a homotopy $h:\Delta^n\times \Delta^1 \to \Delta^n$, where $h|_{\Delta^n\times \{0\}}=\Delta^n\to \{0\} \hookrightarrow \Delta^n$ is the composite of the terminal map and the inclusion. and $h|_{\Delta^n\times \{1\}}=id_{\Delta^n}$.

    Taking the pullback $X\times_{\Delta^n}(-)$, we would like to show that the induced map $f:X\times_{\Delta^n}\{0\}\hookrightarrow X\times_{\Delta^n}\Delta^n=X$ is also a deformation retract. It’s clear that $f$ admits a retraction, $g$, so we’d like to give a homotopy $fg\to id_X$.

    We have the obvious setup to construct the homotopy as the span $X\times \Delta^1 \hookleftarrow X\times \partial\Delta^1\stackrel{(fg,id_X)}{\to} X$. We notice that the implied terminal maps of the diagram can be factored into $X\to \Delta^n\to *$ and $X\times \Delta^1 \to \Delta^n\times \Delta^1 \to *$. We notice further that we can factor the the composite $X\times \partial\Delta^1\hookrightarrow X\times \Delta^1\to \Delta^n \times \Delta^1$ instead as

    $$X\times \partial\Delta^1\to \Delta^n\times \partial\Delta^1 \to \Delta^n\times \Delta^1,$$

    and the composite $X\times \partial\Delta^1 \to X\to \Delta^n$ as

    $$X\times \partial\Delta^1\to \Delta^n\times \partial\Delta^1 \to \Delta^n.$$

    These factorizations let us fit the homotopy $h$ into the diagram, since it is easy to see that these maps are exactly the ones that $h$ lifts. This gives a commutative square for our span with $\Delta^n$ in the bottom right corner, where the maps into it are the obvious ones induced by the homotopy.

    We’d now like to give the homotopy we’ve been looking for. Since $\ell$ is a right fibration, we can lift maps of the form $K\times \Delta^1 \cup L\times \{1\}\to L\times \Delta^1$ for any inclusion $K\hookrightarrow L$.

    Let $A:=X\times_{\Delta^n}\{0\}$. We have the inclusion $f:A\hookrightarrow X$, so applying the pushout above, we have the map $A\times \Delta^1 \cup X\times \{1\}\to X\times \Delta^1$. We give a map from the pushout to $X$ by specifying it on the components. The map $A\times \Delta^1 \to X$ is given by the composite of the projection onto $A$ with the map $f$. The map $X\times \{1\}\to X$ is just the identity on $X$. Then we have a commutative square with the pushout at the top left, $X\times \Delta^1$ at the bottom left, $X$ at the top right, and $\Delta^n$ in the bottom right. The by the lifting property, we have a lift $H:X\times\Delta^1\to X$ making the whole diagram commute.

    We’d like to show that $H$ is the required homotopy. By construction, $H|_{X\times \{1\}}$ gives the identity and $H\circ (f\times \Delta^1)$ is equal to the first projection composed with $f$.

    Then we need only show that $H|_{X\times \{0\}}=fg$. It’s easy to show that the composite $\ell H|_{X\times \{0\}}=\ell fg$ by the commutativity of the diagrams, and also that $H|_{X\times \{0\}} f=f$ by the last sentence of the previous paragraph. Can we prove that $H$ is the required homotopy, or do we need to do another construction and another homotopy?

    If this is the wrong way to prove this, let me know.

    Also, I’m really stuck, so if you have time, I entreat you, please help. I’ve been stuck for 28 hours, and I’m now obsessing over it. =(.

13. • CommentRowNumber13.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 1st 2010
    • (edited Jul 1st 2010)
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItex> It's clear that $f$ admits a retraction, $g$ You're much deeper into this problem than I am, but even this much isn't obvious to me. In general, it will not be true that given a diagram $$\array{ & & X \\ & & \downarrow h \\ Y & \underset{f}{\to} & Z }$$ the pullback $h^* Y \to X$ of $f$ along $h$ will admit a retraction if $f$ does. For an example in $Top$, take $f = incl: \{0\} \to [0, 1]$ and $h$ to be the projection to the coordinate $z$ of the northern hemisphere $X = \{(x, y, z): x^2 + y^2 + z^2 = 1, z \geq 0\}$. Then the fiber $h^{-1}(0) \hookrightarrow X$ is not a retract of $X$, since $X$ is contractible but the fiber is not. (It is easy to mimic this example in simplicial sets.) Is there some extra hypothesis you're using to get the retraction $g$ of $f$?

    It’s clear that $f$ admits a retraction, $g$

    You’re much deeper into this problem than I am, but even this much isn’t obvious to me. In general, it will not be true that given a diagram

    $$\array{ & & X \\ & & \downarrow h \\ Y & \underset{f}{\to} & Z }$$

    the pullback $h^* Y \to X$ of $f$ along $h$ will admit a retraction if $f$ does. For an example in $Top$, take $f = incl: \{0\} \to [0, 1]$ and $h$ to be the projection to the coordinate $z$ of the northern hemisphere $X = \{(x, y, z): x^2 + y^2 + z^2 = 1, z \geq 0\}$. Then the fiber $h^{-1}(0) \hookrightarrow X$ is not a retract of $X$, since $X$ is contractible but the fiber is not. (It is easy to mimic this example in simplicial sets.)

    Is there some extra hypothesis you’re using to get the retraction $g$ of $f$?

14. • CommentRowNumber14.
    • CommentAuthorHarry Gindi
    • CommentTimeJul 1st 2010
    • (edited Jul 1st 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexNo, I thought that was clear. Hmm... (Although it was bothering me. For some reason, I thought we got this for free but couldn't seem to verify it.) So does being incorrect count as an extra hypothesis? =D! I'm sure Mike (because he's an algebraic topologist) knows the actual answer/could sketch out a proof (for the problem of pulling back the inclusion of the 0-vertex into an n-simplex by a right fibration and showing it's a deformation retract), so Mike, I beseech you, please help.

    No, I thought that was clear. Hmm…

    (Although it was bothering me. For some reason, I thought we got this for free but couldn’t seem to verify it.)

    So does being incorrect count as an extra hypothesis? =D!

    I’m sure Mike (because he’s an algebraic topologist) knows the actual answer/could sketch out a proof (for the problem of pulling back the inclusion of the 0-vertex into an n-simplex by a right fibration and showing it’s a deformation retract), so Mike, I beseech you, please help.

15. • CommentRowNumber15.
    • CommentAuthorMike Shulman
    • CommentTimeJul 1st 2010
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexWhat gave you the idea I was an algebraic topologist? (-: I'm a category theorist who knows (and occasionally does) some algebraic topology. My advisor was an algebraic topologist, but a topological one, not a simplicial one, so my simplicial-fu is not that great.

    What gave you the idea I was an algebraic topologist? (-: I’m a category theorist who knows (and occasionally does) some algebraic topology. My advisor was an algebraic topologist, but a topological one, not a simplicial one, so my simplicial-fu is not that great.

16. • CommentRowNumber16.
    • CommentAuthorHarry Gindi
    • CommentTimeJul 1st 2010
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexDidn't May write a book on simplicial stuff? Also, isn't he famous for inventing some simplicial stuff as well?

    Didn’t May write a book on simplicial stuff? Also, isn’t he famous for inventing some simplicial stuff as well?

17. • CommentRowNumber17.
    • CommentAuthorMike Shulman
    • CommentTimeJul 1st 2010
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    Author: Mike Shulman
    Format: MarkdownItexThe book was a long time ago, methods are a lot different now, and he's working on different things.

    The book was a long time ago, methods are a lot different now, and he’s working on different things.

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeJul 1st 2010
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    Author: Urs
    Format: MarkdownItexI still think the idea that I sketched works.

    I still think the idea that I sketched works.

19. • CommentRowNumber19.
    • CommentAuthorHarry Gindi
    • CommentTimeJul 1st 2010
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexI'm not sure how you could build up the map, could you elaborate?

    I’m not sure how you could build up the map, could you elaborate?

20. • CommentRowNumber20.
    • CommentAuthorTodd_Trimble
    • CommentTimeJul 2nd 2010
    • PermaLink
    Author: Todd_Trimble
    Format: MarkdownItexI can't guarantee that I'll find time to think about this, but for my education, could someone explain the notation $\Delta^{\{n-i\}}$, $\Delta^{\{n-i, \ldots, n\}}$, and also explain what a right horn is?

    I can’t guarantee that I’ll find time to think about this, but for my education, could someone explain the notation $\Delta^{\{n-i\}}$, $\Delta^{\{n-i, \ldots, n\}}$, and also explain what a right horn is?

21. • CommentRowNumber21.
    • CommentAuthorHarry Gindi
    • CommentTimeJul 2nd 2010
    • (edited Jul 2nd 2010)
    • PermaLink
    Author: Harry Gindi
    Format: MarkdownItexAlright, first off, you should submit your answer (if you come up with one) on [MO](http://mathoverflow.net/questions/30224/proof-sketch-the-pullback-of-the-inclusion-of-the-0th-vertex-into-the-standard-n), since I'm offering a bounty like last time (by 2AM EST if you'd like to claim it)). The notation $\Delta^{\{n-i,...,n\}}$ is used only in the context of morphisms. That is, the map $\Delta^{\{n-i,...,n\}}\hookrightarrow \Delta^n$ is the inclusion of the n-i simplex spanned by the vertices $\{n-i,...,n\}$ (note that the order here does matter). The $i^{th}$ horn for $0\leq i \leq n$ of an n-simplex is given by the subfunctor of $\Lambda^n_i\subset \Delta^n\coloneqq Hom_\Delta(-,[n])$ where $(\Lambda^n_i)([m])\subseteq Hom([m],[n])$ is subset of all maps $p:[m]\to [n]$ such that $p[m]\cup \{i\}\neq [n]$ (viewed as sets). Geometrically this corresponds to the boundary of the n-simplex without the $i^{th}$ face. A right horn inclusion is just the inclusion $\Lambda^n_i\to \Delta^n$ for a horn $0 < i \leq n$. Edit: My original proof was almost right. Tom Goodwillie has answered it on MO. In particular, we don't need to assume that the new retraction exists. The result follows immediately.

    Alright, first off, you should submit your answer (if you come up with one) on MO, since I’m offering a bounty like last time (by 2AM EST if you’d like to claim it)).

    The notation $\Delta^{\{n-i,...,n\}}$ is used only in the context of morphisms. That is, the map $\Delta^{\{n-i,...,n\}}\hookrightarrow \Delta^n$ is the inclusion of the n-i simplex spanned by the vertices $\{n-i,...,n\}$ (note that the order here does matter).

    The $i^{th}$ horn for $0\leq i \leq n$ of an n-simplex is given by the subfunctor of $\Lambda^n_i\subset \Delta^n\coloneqq Hom_\Delta(-,[n])$ where $(\Lambda^n_i)([m])\subseteq Hom([m],[n])$ is subset of all maps $p:[m]\to [n]$ such that $p[m]\cup \{i\}\neq [n]$ (viewed as sets). Geometrically this corresponds to the boundary of the n-simplex without the $i^{th}$ face.

    A right horn inclusion is just the inclusion $\Lambda^n_i\to \Delta^n$ for a horn $0 < i \leq n$.

    Edit: My original proof was almost right. Tom Goodwillie has answered it on MO. In particular, we don’t need to assume that the new retraction exists. The result follows immediately.