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1. • CommentRowNumber1.
   • CommentAuthorRichard Williamson
   • CommentTimeMay 27th 2017
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   Author: Richard Williamson
   Format: MarkdownItexDoes anyone know of any results or any references pertaining to whether it is possible to show that if a functor between groupoids is an iso-fibration 'up to homotopy' (the left triangle in the lifting square holds up to equality, and the right triangle holds up to natural isomorphism), then it is in fact an iso-fibration on the nose? My gut feeling is that this is true, but I have not so far been able to prove it. Any results/references in other categories are very welcome too.

   Does anyone know of any results or any references pertaining to whether it is possible to show that if a functor between groupoids is an iso-fibration ’up to homotopy’ (the left triangle in the lifting square holds up to equality, and the right triangle holds up to natural isomorphism), then it is in fact an iso-fibration on the nose? My gut feeling is that this is true, but I have not so far been able to prove it. Any results/references in other categories are very welcome too.

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 28th 2017
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   Author: Mike Shulman
   Format: MarkdownItexI must be misunderstanding; doesn't any functor of groupoids satisfy that up-to-homotopy condition by taking the lift to be an identity?

   I must be misunderstanding; doesn’t any functor of groupoids satisfy that up-to-homotopy condition by taking the lift to be an identity?

3. • CommentRowNumber3.
   • CommentAuthorRichard Williamson
   • CommentTimeMay 28th 2017
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   Author: Richard Williamson
   Format: MarkdownItexHi MIke, the definition of an iso-fibration that I was referring to is the following: for any groupoid $X$, it has the right lifting property with respect to the functor $X \times 0 : X \rightarrow X \times I$, where $I$ is the interval groupoid, and $0$ is the inclusion of $0$. For groupoids, it is known that this is actually equivalent to requiring the same only when $X = 1$. In the situation I am looking at, if $f : A \rightarrow B$ is the functor which we are trying to prove is an isofibration, and if we have any commutative square $f \circ a = b \circ (X \times 0)$ for any $X$, then we know that we can construct a functor $l : X \times I \rightarrow A$ such that $l \circ (X \times 0) = a$ (so this triangle holds strictly) and such that there is a homotopy (i.e. natural isomorphism) from $f \circ l$ to $b$. I am asking whether we can in such a case demonstrate that $f$ satisfies the required right lifting property on the nose, by constructing some other lift.

   Hi MIke, the definition of an iso-fibration that I was referring to is the following: for any groupoid $X$, it has the right lifting property with respect to the functor $X \times 0 : X \rightarrow X \times I$, where $I$ is the interval groupoid, and $0$ is the inclusion of $0$.

   For groupoids, it is known that this is actually equivalent to requiring the same only when $X = 1$.

   In the situation I am looking at, if $f : A \rightarrow B$ is the functor which we are trying to prove is an isofibration, and if we have any commutative square $f \circ a = b \circ (X \times 0)$ for any $X$, then we know that we can construct a functor $l : X \times I \rightarrow A$ such that $l \circ (X \times 0) = a$ (so this triangle holds strictly) and such that there is a homotopy (i.e. natural isomorphism) from $f \circ l$ to $b$. I am asking whether we can in such a case demonstrate that $f$ satisfies the required right lifting property on the nose, by constructing some other lift.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeMay 28th 2017
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   Author: Mike Shulman
   Format: MarkdownItexI'm saying it seems to me that any functor $f:A\to B$ at all has that property. Given $f\circ a = b\circ (X\times 0)$, define $l(x,i) = a(x)$. Then the upper triangle holds strictly, while in the lower triangle the functor $b:X\times I \to B$ yields a natural isomorphism from $f\circ l$ to itself. So it certainly can't imply that $f$ is an isofibration.

   I’m saying it seems to me that any functor $f:A\to B$ at all has that property. Given $f\circ a = b\circ (X\times 0)$, define $l(x,i) = a(x)$. Then the upper triangle holds strictly, while in the lower triangle the functor $b:X\times I \to B$ yields a natural isomorphism from $f\circ l$ to itself. So it certainly can’t imply that $f$ is an isofibration.

5. • CommentRowNumber5.
   • CommentAuthorRichard Williamson
   • CommentTimeMay 28th 2017
   • (edited May 28th 2017)
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   Author: Richard Williamson
   Format: MarkdownItexThanks for thinking about it! But I don't follow you, because I cannot make sense of the following. > while in the lower triangle the functor $b:X\times I \to B$ yields a natural isomorphism from $f\circ l$ to itself. By 'itself', did you mean $f \circ l$? Then there is no reason for what you wrote in this quote to be correct: $b$ is a natural transformation from $f \circ a$ to some unknown functor (i.e. $b \circ (X \times 1)$). There is no reason that I see to expect this unknown functor to be $f \circ l$. Moreover, we are looking for a natural transformation to $b$, not to $f \circ l$. If by 'itself' you meant $b$, then this does not parse, because a natural isomorphism from something to $b$ must be of the form $X \times I^{2} \rightarrow B$, not $X \times I \rightarrow B$. There is a further issue. If we define $p : I \rightarrow 1$ to be the canonical functor, then you wish, I believe, to take $l$ to be $a \circ p$. Certainly I agree that the top triangle then holds strictly. But $b$ is a natural transformation from $f \circ a$, not from $f \circ a \circ p$. I see no reason to expect to be able to find _any_ natural isomorphism between the latter and $b$. Am I completely misunderstanding you somehow? PS - I wrote out the definition because I could not parse "taking the lift to be an identity".

   Thanks for thinking about it! But I don’t follow you, because I cannot make sense of the following.

   while in the lower triangle the functor $b:X\times I \to B$ yields a natural isomorphism from $f\circ l$ to itself.

   By ’itself’, did you mean $f \circ l$? Then there is no reason for what you wrote in this quote to be correct: $b$ is a natural transformation from $f \circ a$ to some unknown functor (i.e. $b \circ (X \times 1)$). There is no reason that I see to expect this unknown functor to be $f \circ l$. Moreover, we are looking for a natural transformation to $b$, not to $f \circ l$.

   If by ’itself’ you meant $b$, then this does not parse, because a natural isomorphism from something to $b$ must be of the form $X \times I^{2} \rightarrow B$, not $X \times I \rightarrow B$.

   There is a further issue. If we define $p : I \rightarrow 1$ to be the canonical functor, then you wish, I believe, to take $l$ to be $a \circ p$. Certainly I agree that the top triangle then holds strictly. But $b$ is a natural transformation from $f \circ a$, not from $f \circ a \circ p$. I see no reason to expect to be able to find any natural isomorphism between the latter and $b$.

   Am I completely misunderstanding you somehow?

   PS - I wrote out the definition because I could not parse “taking the lift to be an identity”.

6. • CommentRowNumber6.
   • CommentAuthorRichard Williamson
   • CommentTimeMay 28th 2017
   • (edited May 28th 2017)
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   Author: Richard Williamson
   Format: MarkdownItexPPS - I do actually have some further structure available in the setting I am looking for, which might be able to be used to construct a strictification. But I'll refrain from mentioning it for now, because I am still somewhat optimistic that my original feeling is correct. This is mostly based on experience: one often encounters this kind of issue when working in 'cylindrical homotopy theory', and usually there is a way to strictify if one's cylinder/co-cylinder is nice enough, perhaps with a clever use of double homotopies/connections/etc,

   PPS - I do actually have some further structure available in the setting I am looking for, which might be able to be used to construct a strictification. But I’ll refrain from mentioning it for now, because I am still somewhat optimistic that my original feeling is correct. This is mostly based on experience: one often encounters this kind of issue when working in ’cylindrical homotopy theory’, and usually there is a way to strictify if one’s cylinder/co-cylinder is nice enough, perhaps with a clever use of double homotopies/connections/etc,

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeMay 29th 2017
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   Author: Mike Shulman
   Format: MarkdownItexWe have $a:X\to A$, $b:X\times I \to B$, $i_0,i_1 : X\to X\times I$, and $p:X\times I\to X$, with $f\circ a = b\circ i_0$. I say take $l :X\times I \to A$ to be $a\circ p$, as you guessed, so that $l \circ i_0 = a\circ p\circ i_0 = a$. Now $f\circ l = f\circ a \circ p = b\circ i_0 \circ p$. But $i_0 \circ p : I \to I$ is isomorphic to the identity functor, so $b\circ i_0\circ p \cong b$.

   We have $a:X\to A$, $b:X\times I \to B$, $i_0,i_1 : X\to X\times I$, and $p:X\times I\to X$, with $f\circ a = b\circ i_0$. I say take $l :X\times I \to A$ to be $a\circ p$, as you guessed, so that $l \circ i_0 = a\circ p\circ i_0 = a$. Now $f\circ l = f\circ a \circ p = b\circ i_0 \circ p$. But $i_0 \circ p : I \to I$ is isomorphic to the identity functor, so $b\circ i_0\circ p \cong b$.

8. • CommentRowNumber8.
   • CommentAuthorRichard Williamson
   • CommentTimeMay 29th 2017
   • (edited May 29th 2017)
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   Author: Richard Williamson
   Format: MarkdownItexThanks! This argument will go through whenever we have a cylinder with a 'lower right connection structure'. I did have some vague premonition of this kind of observation, which was one reason why I asked my question originally. I intend to add your observation to the nLab, perhaps tomorrow. I will have a think about whether the extra structure that I have able in the case I am interested in allows for a strictification. In case anybody's interested, I can describe briefly what I am really interested in. John Baez asked me whether, given an endofunctor $f : S \rightarrow S$ of groupoids, there is an endofunctor $f' : S' \rightarrow S'$ which is equivalent to $f$ in the obvious sense and which is an iso-fibration. He actually had a specific endo-functor in mind, but I have been considering the question in general. What I can show is that if we take $S'$ to be the mapping co-cylinder of $f$, and take $f'$ to be the composition of the iso-fibration $g : S' \rightarrow S$ which occurs in the mapping co-cylinder factorisation of $f$ with the functor $i: S \rightarrow S'$ which is the other part of this factorisation, then we have an 'up to homotopy' iso-fibration which is equivalent to $f$. This is because $i$ is in fact a section of a strong deformation retraction. What I would like to know is whether $i \circ g$ is in fact an iso-fibration, not only an up-to-homotopy one.

   Thanks! This argument will go through whenever we have a cylinder with a ’lower right connection structure’. I did have some vague premonition of this kind of observation, which was one reason why I asked my question originally.

   I intend to add your observation to the nLab, perhaps tomorrow. I will have a think about whether the extra structure that I have able in the case I am interested in allows for a strictification.

   In case anybody’s interested, I can describe briefly what I am really interested in. John Baez asked me whether, given an endofunctor $f : S \rightarrow S$ of groupoids, there is an endofunctor $f' : S' \rightarrow S'$ which is equivalent to $f$ in the obvious sense and which is an iso-fibration. He actually had a specific endo-functor in mind, but I have been considering the question in general. What I can show is that if we take $S'$ to be the mapping co-cylinder of $f$, and take $f'$ to be the composition of the iso-fibration $g : S' \rightarrow S$ which occurs in the mapping co-cylinder factorisation of $f$ with the functor $i: S \rightarrow S'$ which is the other part of this factorisation, then we have an ’up to homotopy’ iso-fibration which is equivalent to $f$. This is because $i$ is in fact a section of a strong deformation retraction. What I would like to know is whether $i \circ g$ is in fact an iso-fibration, not only an up-to-homotopy one.

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeMay 30th 2017
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   Author: Mike Shulman
   Format: MarkdownItexWhat about the following construction? Factor $f$ through its mapping cocyclinder $S \xrightarrow{i} S_1 \xrightarrow{g_1} S$ as usual; then $g$ is an isofibration, and $i$ has a retraction $p_1:S_1 \to S$ that is an acyclic isofibration. Now construct recursively a sequence of pullback squares: $$\array{ \cdots &\to& S_3 &\xrightarrow{p_3}& S_2 &\xrightarrow{p_2}& S_1 \\ && \downarrow^{g_3} && \downarrow^{g_2} && \downarrow^{g_1} \\ \cdots &\to& S_2 &\xrightarrow{p_2}& S_1 &\xrightarrow{p_1}& S }$$ and take the limit. Since the top and bottom sequences are cofinally the same, they have the same limit $S_\infty$, and since each $p_n$ is an acyclic isosfibration, so is the limit projection $p_\infty : S_\infty \to S$. Also, since the commutative squares are pullbacks, each $g_n$ is an isofibration, and moreover the whole transformation $g$ is a Reedy isofibration; thus the limit map $S_{\infty} \xrightarrow{g_\infty} S_{\infty}$ is also an isofibration, and it is equivalent to $g_1$, hence also to $f$.

   What about the following construction? Factor $f$ through its mapping cocyclinder $S \xrightarrow{i} S_1 \xrightarrow{g_1} S$ as usual; then $g$ is an isofibration, and $i$ has a retraction $p_1:S_1 \to S$ that is an acyclic isofibration. Now construct recursively a sequence of pullback squares:

   $$\array{ \cdots &\to& S_3 &\xrightarrow{p_3}& S_2 &\xrightarrow{p_2}& S_1 \\ && \downarrow^{g_3} && \downarrow^{g_2} && \downarrow^{g_1} \\ \cdots &\to& S_2 &\xrightarrow{p_2}& S_1 &\xrightarrow{p_1}& S }$$

   and take the limit. Since the top and bottom sequences are cofinally the same, they have the same limit $S_\infty$, and since each $p_n$ is an acyclic isosfibration, so is the limit projection $p_\infty : S_\infty \to S$. Also, since the commutative squares are pullbacks, each $g_n$ is an isofibration, and moreover the whole transformation $g$ is a Reedy isofibration; thus the limit map $S_{\infty} \xrightarrow{g_\infty} S_{\infty}$ is also an isofibration, and it is equivalent to $g_1$, hence also to $f$.

10. • CommentRowNumber10.
    • CommentAuthorRichard Williamson
    • CommentTimeMay 30th 2017
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    Author: Richard Williamson
    Format: MarkdownItexThat's a fascinating argument, Mike! I agree that it works! I also had the idea to pass to infinity, but was trying it with colimits instead. I'm not quite sure what to make of the argument, yet! I need to reflect upon it some more. It would be very interesting if there were some finite way to do it.

    That’s a fascinating argument, Mike! I agree that it works! I also had the idea to pass to infinity, but was trying it with colimits instead.

    I’m not quite sure what to make of the argument, yet! I need to reflect upon it some more. It would be very interesting if there were some finite way to do it.

11. • CommentRowNumber11.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • (edited Jun 8th 2017)
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    Author: John Baez
    Format: TextI should add that I asked Dan Christensen about the same puzzle, and he seems to have solved it. I'll take the liberty of posting some of our correspondence here. My student Kenny Courser and Daniel Cicala are currently filling in the details (with Dan's permission) as part of a paper on Morton and Vicary's approach to Khovanov's categorified Heisenberg algebra. Here's my initial question: ------------ I hope you're doing fine! I bumped into a question that's in your bailiwick. Suppose I have an infinite sequence of topological spaces and maps: X_0 -> X_1 -> X_2 -> .... Is there a way to do a "fibrant replacement" to get a weakly equivalent sequence Y_0 -> Y_1 -> Y_2 -> .... where now each map is a fibration? I'm hoping you can guess what I mean by weakly equivalent sequence: I'm hoping for weak equivalences X_n -> Y_n such that all the resulting squares commute - perhaps if necessary up to homotopy. This might follow from the existence of a suitably well-behaved model structure on these sequences, but I'm hoping there might be a fairly low-brow way to construct Y_0 -> Y_1 -> Y_2 -> .... by first replacing X_0 -> X_1 with a fibration X'_0 -> X_1, then replacing X_0 -> X_1 -> X_2 with fibrations X''_0 -> X''_1 -> X''_2, etc, and then taking some sort of (co)limit of X_i, X'_i, X''_i, etc. But this is just a rough sketch of an idea at this point!

    I should add that I asked Dan Christensen about the same puzzle, and he seems to have solved it. I'll take the liberty of posting some of our correspondence here. My student Kenny Courser and Daniel Cicala are currently filling in the details (with Dan's permission) as part of a paper on Morton and Vicary's approach to Khovanov's categorified Heisenberg algebra.
    
    Here's my initial question:
    
    ------------
    
    I hope you're doing fine! I bumped into a question that's in your bailiwick. Suppose I have an infinite sequence of topological spaces and maps:
    
    X_0 -> X_1 -> X_2 -> ....
    
    Is there a way to do a "fibrant replacement" to get a weakly equivalent sequence
    
    Y_0 -> Y_1 -> Y_2 -> ....
    
    where now each map is a fibration? I'm hoping you can guess what I mean by weakly equivalent sequence: I'm hoping for weak equivalences X_n -> Y_n such that all the resulting squares commute - perhaps if necessary up to homotopy.
    
    This might follow from the existence of a suitably well-behaved model structure on these sequences, but I'm hoping there might be a fairly low-brow way to construct
    
    Y_0 -> Y_1 -> Y_2 -> ....
    
    by first replacing X_0 -> X_1 with a fibration X'_0 -> X_1, then replacing X_0 -> X_1 -> X_2 with fibrations X''_0 -> X''_1 -> X''_2, etc, and then taking some sort of (co)limit of X_i, X'_i, X''_i, etc. But this is just a rough sketch of an idea at this point!
12. • CommentRowNumber12.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
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    Author: John Baez
    Format: TextAfter some discussion Dan Christensen wrote: ------------ On further thought, I think your argument just works. You need the fact that a sequential colimit of fibrations is a fibrations, which follows from the fact that the generating acyclic cofibrations have omega-small domains and codomains. (A map f is a fibration iff it has the right lifting property with respect to these maps.) It's a fairly simple diagram chase, but I'm on my way out right now so I can't write it down. Let me know if you'd like me to. You also need that the transfinite composite of the maps X_0 --> X_0' --> is a weak equivalence, but this follows from the fact that each of those maps can be chosen to be a trivial cofibration. To check that I'm not confused about the first part, I tried searching for "colimit of fibrations is a fibration", but I didn't find a match. I hope I'm not making a mistake!

    After some discussion Dan Christensen wrote:
    
    ------------
    
    On further thought, I think your argument just works. You need the fact
    that a sequential colimit of fibrations is a fibrations, which follows
    from the fact that the generating acyclic cofibrations have omega-small
    domains and codomains. (A map f is a fibration iff it has the right
    lifting property with respect to these maps.) It's a fairly simple
    diagram chase, but I'm on my way out right now so I can't write it down.
    Let me know if you'd like me to.
    
    You also need that the transfinite composite of the maps X_0 --> X_0'
    --> is a weak equivalence, but this follows from the fact that each
    of those maps can be chosen to be a trivial cofibration.
    
    To check that I'm not confused about the first part, I tried searching
    for "colimit of fibrations is a fibration", but I didn't find a match.
    I hope I'm not making a mistake!
13. • CommentRowNumber13.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • (edited Jun 8th 2017)
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    Author: John Baez
    Format: TextI replied: ------------ > On further thought, I think your argument just works. Wow! > You need the fact that a sequential colimit of fibrations is a fibrations, which follows > from the fact that the generating acyclic cofibrations have omega-small > domains and codomains. (A map f is a fibration iff it has the right > lifting property with respect to these maps.) It's a fairly simple > diagram chase, but I'm on my way out right now so I can't write it down. > Let me know if you'd like me to. I might at some point request that. Let me give away what I was secretly thinking about. Let S be the groupoid of finite sets and bijections. There's a functor +1 : S -> S sending each set to its disjoint union with a one-element set. I would like to find a groupoid equivalent to S, say S', and a functor equivalent to +1, say F, such that F: S' -> S' is a fibration of groupoids. (I.e., given any morphism f: x -> x' in S' and any object y with F(y) = x, there's a morphism g: y -> y' with F(g) = f.) Of course it's easy to get a functor G: S' -> S equivalent to F that's a fibration of groupoids, but for some reason I want to replace both the domain and codomain with S'. One approach is to note that S is the coproduct of groupoids S_n, where S_n is the groupoid of n-element sets, and note that we have functors S_0 -> S_1 -> S_2 -> ... where each arrow is the restriction of +1 to S_n. If we could find an equivalent sequence of fibrations S'_0 --F_0--> S'_1 --F_1--> S'_2 --F_2--> ... then we could let S' be the coproduct of the groupoids S'_n and let F: S' -> S' be the coproduct of the F_n. I translated this question into topology so you might like it more. It's possible that there's a much simpler way to solve my original puzzle. Trying to think about it in simpler ways seemed to push me in this direction. But maybe now that I've revealed my secret motivation you'll see a better solution.

    I replied:
    
    ------------
    
    > On further thought, I think your argument just works.
    
    Wow!
    
    > You need the fact that a sequential colimit of fibrations is a fibrations, which follows
    > from the fact that the generating acyclic cofibrations have omega-small
    > domains and codomains. (A map f is a fibration iff it has the right
    > lifting property with respect to these maps.) It's a fairly simple
    > diagram chase, but I'm on my way out right now so I can't write it down.
    > Let me know if you'd like me to.
    
    I might at some point request that. Let me give away what I was secretly thinking
    about.
    
    Let S be the groupoid of finite sets and bijections. There's a functor
    
    +1 : S -> S
    
    sending each set to its disjoint union with a one-element set. I would like
    to find a groupoid equivalent to S, say S', and a functor equivalent to +1,
    say F, such that
    
    F: S' -> S'
    
    is a fibration of groupoids. (I.e., given any morphism f: x -> x' in S' and any
    object y with F(y) = x, there's a morphism g: y -> y' with F(g) = f.)
    
    Of course it's easy to get a functor G: S' -> S equivalent to F that's a fibration
    of groupoids, but for some reason I want to replace both the domain and
    codomain with S'.
    
    One approach is to note that S is the coproduct of groupoids S_n, where S_n is
    the groupoid of n-element sets, and note that we have functors
    
    S_0 -> S_1 -> S_2 -> ...
    
    where each arrow is the restriction of +1 to S_n. If we could find an equivalent
    sequence of fibrations
    
    S'_0 --F_0--> S'_1 --F_1--> S'_2 --F_2--> ...
    
    then we could let S' be the coproduct of the groupoids S'_n and let F: S' -> S'
    be the coproduct of the F_n.
    
    I translated this question into topology so you might like it more.
    
    It's possible that there's a much simpler way to solve my original puzzle.
    Trying to think about it in simpler ways seemed to push me in this direction.
    But maybe now that I've revealed my secret motivation you'll see a better
    solution.
14. • CommentRowNumber14.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • (edited Jun 8th 2017)
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    Author: John Baez
    Format: TextDan replied: ------------ Hi John, > I might at some point request that. Let me give away what I was > secretly thinking about. Ah, groupoids will be a different (and probably easier) story. For some strange reason, questions like this tend to depend delicately on non-homotopical properties like being locally presentable, etc. And groupoids are closer to simplicial sets, which are better behaved than spaces in this respect. The idea you sketched sounds plausible. I suspect there's a concrete way to make this work, and I'll think about it.

    Dan replied:
    
    ------------
    
    Hi John,
    
    > I might at some point request that. Let me give away what I was
    > secretly thinking about.
    
    Ah, groupoids will be a different (and probably easier) story. For
    some strange reason, questions like this tend to depend delicately
    on non-homotopical properties like being locally presentable, etc.
    And groupoids are closer to simplicial sets, which are better behaved
    than spaces in this respect.
    
    The idea you sketched sounds plausible. I suspect there's a concrete
    way to make this work, and I'll think about it.
15. • CommentRowNumber15.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • (edited Jun 8th 2017)
    • PermaLink
    Author: John Baez
    Format: TextDan wrote: ------------ > Let S be the groupoid of finite sets and bijections. There's a functor > > +1 : S -> S > > sending each set to its disjoint union with a one-element set. I > would like to find a groupoid equivalent to S, say S', and a functor > equivalent to +1, say F, such that > > F: S' -> S' > > is a fibration of groupoids. (I.e., given any morphism f: x -> x' in > S' and any object y with F(y) = x, there's a morphism g: y -> y' with > F(g) = f.) Does this work? Define a category S' whose objects are tuples (X, m, f), where X is a set, m is a natural number, and f is an injection from {i : i >= m} to X such that the complement of the image has cardinality m. Hom((X, f, m), (Y, g, n)) is empty if m != n and otherwise consists of bijections h : X --> Y such that h(f(i)) = g(i) for each i >= m. Intuitively, (X, f, m) should be thought of as a set of size m (the complement of the image) plus a countable sequence of elements that are "on deck" to be added in. The morphisms are determined on the on deck elements, so this category should be equivalent to the category of finite sets. The +1 functor should send (X, f, m) to (X, f, m+1), where the second f is really the restriction of the first f. I think this is an isofibration. Given (X, f, m) and a morphism h : (X, f, m+1) --> (Y, g, m+1), define g(m) = h(f(m)), so we have an object (Y, g, m) whose image is (Y, g, m+1). The same h defines a map (X, f, m) --> (Y, g, m) whose image is h : (X, f, m+1) --> (Y, g, m+1). I came up with this by thinking about how to find a fibration equivalent to S_3 --> S_4. Let me know if you think it works! I'm off to bed, and haven't checked the details carefully, but it feels like something like this should work.

    Dan wrote:
    
    ------------
    
    > Let S be the groupoid of finite sets and bijections. There's a functor
    >
    > +1 : S -> S
    >
    > sending each set to its disjoint union with a one-element set. I
    > would like to find a groupoid equivalent to S, say S', and a functor
    > equivalent to +1, say F, such that
    >
    > F: S' -> S'
    >
    > is a fibration of groupoids. (I.e., given any morphism f: x -> x' in
    > S' and any object y with F(y) = x, there's a morphism g: y -> y' with
    > F(g) = f.)
    
    Does this work? Define a category S' whose objects are tuples (X, m, f),
    where X is a set, m is a natural number, and f is an injection from
    {i : i >= m} to X such that the complement of the image has cardinality m.
    
    Hom((X, f, m), (Y, g, n)) is empty if m != n and otherwise consists of
    bijections h : X --> Y such that h(f(i)) = g(i) for each i >= m.
    
    Intuitively, (X, f, m) should be thought of as a set of size m (the
    complement of the image) plus a countable sequence of elements that
    are "on deck" to be added in. The morphisms are determined on the on
    deck elements, so this category should be equivalent to the category
    of finite sets.
    
    The +1 functor should send (X, f, m) to (X, f, m+1), where the second
    f is really the restriction of the first f. I think this is an
    isofibration. Given (X, f, m) and a morphism
    
    h : (X, f, m+1) --> (Y, g, m+1),
    
    define g(m) = h(f(m)), so we have an object (Y, g, m) whose image is
    (Y, g, m+1). The same h defines a map (X, f, m) --> (Y, g, m) whose
    image is h : (X, f, m+1) --> (Y, g, m+1).
    
    I came up with this by thinking about how to find a fibration equivalent
    to S_3 --> S_4. Let me know if you think it works! I'm off to bed, and
    haven't checked the details carefully, but it feels like something like
    this should work.
16. • CommentRowNumber16.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • (edited Jun 8th 2017)
    • PermaLink
    Author: John Baez
    Format: TextI wrote: ------------ I roughly get the drift of your concrete solution to my puzzle, but I'll think it about more when I'm feeling a bit more peppy. Thanks! I'm curious about the phrase "on deck". I'm not familiar with it. It sounds nautical... do people keep stuff "on deck" and bring it down as needed? While I'm at it, I should mention some more general questions. Suppose you have any endomorphism of groupoids A: X -> X. Can we always find an equivalent A': X' -> X' that's a fibration? I also have the same question for spaces... or for that matter, objects in any model category! In that most general case, I imagine the answer must be "no" or someone would have told me. But I have no idea for how to find counterexamples. It just feels sort of lucky that in my example X is a coproduct and A maps the nth piece to the (n+1)st piece.

    I wrote:
    
    ------------
    
    I roughly get the drift of your concrete solution to my puzzle, but I'll think
    it about more when I'm feeling a bit more peppy. Thanks! I'm curious
    about the phrase "on deck". I'm not familiar with it. It sounds nautical...
    do people keep stuff "on deck" and bring it down as needed?
    
    While I'm at it, I should mention some more general questions. Suppose you
    have any endomorphism of groupoids A: X -> X. Can we always find an equivalent
    A': X' -> X' that's a fibration? I also have the same question for spaces... or
    for that matter, objects in any model category!
    
    In that most general case, I imagine the answer must be "no" or someone would
    have told me. But I have no idea for how to find counterexamples. It just feels sort
    of lucky that in my example X is a coproduct and A maps the nth piece to the
    (n+1)st piece.
17. • CommentRowNumber17.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • (edited Jun 10th 2017)
    • PermaLink
    Author: John Baez
    Format: MarkdownItexDan wrote: ------------ > I roughly get the drift of your concrete solution to my puzzle, but > I'll think it about more when I'm feeling a bit more peppy. I didn't have time to motivate it properly last night, but here's a simpler warm-up. Suppose we want to replace +1 : S_3 --> S_4 by an isofibration. Let S'_3 be the groupoid of sets of size 4 with one marked point, with morphisms the bijections preserving the marked point. This is equivalent to S_3, via an equivalence S_3 --> S'_3 that adds a disjoint marked point. The functor S'_3 --> S_4 that forgets the marked point is an isofibration such that the composite S_3 --> S'_3 --> S_4 is the +1 functor. If I also want to replace S_4 --> S_5 with an isofibration, then I'll use S'_4 analogous to S'_3. And I'll need S''_3 with an ordered pair of marked points, so that after forgetting the first one, I land in S'_4. Continuing, you see that what you want for the final S^\infty_n is a countable set, with n unmarked points, and a countable sequence of marked points. Then you let n vary (the coproduct you mentioned) to get the final answer. > I'm curious about the phrase "on deck". I'm not familiar with it. It > sounds nautical... do people keep stuff "on deck" and bring it down > as needed? I'm not a baseball fan, but that's where the phrase is usually used. The batter who is next to play is "on deck" and waits in a certain spot, usually swinging the bat around to warm up a bit. But I looked it up, and it turns out that it originates from aircraft carriers: https://en.wikipedia.org/wiki/On-deck > While I'm at it, I should mention some more general questions. > Suppose you have any endomorphism of groupoids A: X -> X. Can we > always find an equivalent A': X' -> X' that's a fibration? I also > have the same question for spaces... or for that matter, objects in > any model category! > > In that most general case, I imagine the answer must be "no" or > someone would have told me. That's how I feel too. I'm surprised that it works in this case, but as you said, maybe it's just because of how simple the functor is. But thinking about it some more, I think your original trick works. First, factor A as a trivial cofibration followed by a fibration: X >-~-> X_1 | | A| v v v X ===== X Then do the same for the composite X_1 --> X --> X_1: X >-~-> X_1 >-~-> X_2 | | | A| v v v v v X ===== X >-~-> X_1 Continue in this way and take the colimit X' = colim X_n. You end up with a square X >-~-> X' | | A| | v v X >-~-> X' The horizontal maps are still trivial cofibrations, since those are closed under sequential composition. And in many model categories, fibrations are closed under sequential colimits. This happens, for example, in a cofibrantly generated model category where the domains and codomains of the generating trivial cofibrations are omega-small: given a generating trivial cofibration B --> C, consider a square B --> X' | | v v C --> X' By smallness, it factors (as a square) through some finite stage: B --> X_{n+1} --> X' | | | v v v C --> X_n --> X' (there's a little diagram chase hidden in that claim). And the first square has a lift, since X_{n+1} --> X_n is a fibration. So the outer square has a lift, which means X' --> X' is a fibration. If the domains and codomains are only small with respect to a larger ordinal, one can just make the diagram bigger. Here's a more high-powered proof that works if M is a combinatorial model category. Let BN be the free category on an endomorphism, so that a functor BN --> M is an object X of M with a self-map. Then the category M^BN of diagrams has an injective model structure, and I'm pretty sure that the fibrant objects there in particular have the self-map being a fibration. This doesn't include Top, so I think this second argument is strictly less general.

    Dan wrote:

    ---

    I roughly get the drift of your concrete solution to my puzzle, but I’ll think it about more when I’m feeling a bit more peppy.

    I didn’t have time to motivate it properly last night, but here’s a simpler warm-up. Suppose we want to replace +1 : S_3 –> S_4 by an isofibration. Let S’_3 be the groupoid of sets of size 4 with one marked point, with morphisms the bijections preserving the marked point. This is equivalent to S_3, via an equivalence S_3 –> S’_3 that adds a disjoint marked point. The functor S’_3 –> S_4 that forgets the marked point is an isofibration such that the composite S_3 –> S’_3 –> S_4 is the +1 functor.

    If I also want to replace S_4 –> S_5 with an isofibration, then I’ll use S’_4 analogous to S’_3. And I’ll need S”_3 with an ordered pair of marked points, so that after forgetting the first one, I land in S’_4.

    Continuing, you see that what you want for the final S^\infty_n is a countable set, with n unmarked points, and a countable sequence of marked points. Then you let n vary (the coproduct you mentioned) to get the final answer.

    I’m curious about the phrase “on deck”. I’m not familiar with it. It sounds nautical… do people keep stuff “on deck” and bring it down as needed?

    I’m not a baseball fan, but that’s where the phrase is usually used. The batter who is next to play is “on deck” and waits in a certain spot, usually swinging the bat around to warm up a bit. But I looked it up, and it turns out that it originates from aircraft carriers:

    https://en.wikipedia.org/wiki/On-deck

    While I’m at it, I should mention some more general questions. Suppose you have any endomorphism of groupoids A: X -> X. Can we always find an equivalent A’: X’ -> X’ that’s a fibration? I also have the same question for spaces… or for that matter, objects in any model category!

    In that most general case, I imagine the answer must be “no” or someone would have told me.

    That’s how I feel too. I’m surprised that it works in this case, but as you said, maybe it’s just because of how simple the functor is.

    But thinking about it some more, I think your original trick works. First, factor A as a trivial cofibration followed by a fibration:

      X >-~-> X_1
      |       |
     A|       v
      v       v
      X ===== X
    

    Then do the same for the composite X_1 –> X –> X_1:

      X >-~-> X_1 >-~-> X_2
      |           |             |
    A|           v             v
      v           v             v
      X ===== X   >-~-> X_1
    

    Continue in this way and take the colimit X’ = colim X_n. You end up with a square

      X >-~-> X'
      |          |
    A|          |
      v          v
      X >-~-> X'
    

    The horizontal maps are still trivial cofibrations, since those are closed under sequential composition. And in many model categories, fibrations are closed under sequential colimits. This happens, for example, in a cofibrantly generated model category where the domains and codomains of the generating trivial cofibrations are omega-small: given a generating trivial cofibration B –> C, consider a square

    B --> X'
    |      |
    v      v
    C --> X'
    

    By smallness, it factors (as a square) through some finite stage:

    B --> X_{n+1} --> X'
    |          |            |
    v          v            v
    C --> X_n     -->  X'
    

    (there’s a little diagram chase hidden in that claim). And the first square has a lift, since X_{n+1} –> X_n is a fibration. So the outer square has a lift, which means X’ –> X’ is a fibration.

    If the domains and codomains are only small with respect to a larger ordinal, one can just make the diagram bigger.

    Here’s a more high-powered proof that works if M is a combinatorial model category. Let BN be the free category on an endomorphism, so that a functor BN –> M is an object X of M with a self-map. Then the category M^BN of diagrams has an injective model structure, and I’m pretty sure that the fibrant objects there in particular have the self-map being a fibration.

    This doesn’t include Top, so I think this second argument is strictly less general.

18. • CommentRowNumber18.
    • CommentAuthorJohn Baez
    • CommentTimeJun 8th 2017
    • PermaLink
    Author: John Baez
    Format: TextI'm sorry for the munged diagrams in the final email: if anyone knew how to get fixed width space spacing here (without a lot of work), I would use that. But I think the intent is clear.

    I'm sorry for the munged diagrams in the final email: if anyone knew how to get fixed width space spacing here (without a lot of work), I would use that. But I think the intent is clear.
19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeJun 8th 2017
    • (edited Jun 8th 2017)
    • PermaLink
    Author: Urs
    Format: MarkdownItexTo make the software here display ASCII art verbatim you need to indent the relevant lines by (at least) four whitespaces X >-~-> X_1 | | A| v v v X ===== X

    To make the software here display ASCII art verbatim you need to indent the relevant lines by (at least) four whitespaces

      X >-~-> X_1
      |       |
     A|       v
      v       v
      X ===== X
    

20. • CommentRowNumber20.
    • CommentAuthorMike Shulman
    • CommentTimeJun 8th 2017
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexThanks for sharing all that, John! I'm unsure whether any of these sequential colimit arguments work in $Top$, though, because my memory is that compact spaces are not $\omega$-small in $Top$ with respect to *all* sequential colimits, only sequential colimits of something like "closed $T_1$ inclusions". I would also like to see the argument that the injective fibrant objects in $M^{B N}$ are fibrations.

    Thanks for sharing all that, John!

    I’m unsure whether any of these sequential colimit arguments work in $Top$, though, because my memory is that compact spaces are not $\omega$-small in $Top$ with respect to all sequential colimits, only sequential colimits of something like “closed $T_1$ inclusions”.

    I would also like to see the argument that the injective fibrant objects in $M^{B N}$ are fibrations.