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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMay 2nd 2012
   • (edited May 2nd 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexquick question, am in the middle of some computation: I know that a "cotower diagram" $$ X_0 \to X_1 \to X_2 \to \cdots $$ in some (cofibrantly generated) model category is projectively cofibrant precisely if each morphism is a cofibration and if the first object, hence all objects, are cofibrant. Can we say something along these lines for general filtered diagrams? (I guess it should be an easy generalization of the above statement, but I am throwing out this question anyway now.)

   quick question, am in the middle of some computation:

   I know that a “cotower diagram”

   $$X_0 \to X_1 \to X_2 \to \cdots$$

   in some (cofibrantly generated) model category is projectively cofibrant precisely if each morphism is a cofibration and if the first object, hence all objects, are cofibrant.

   Can we say something along these lines for general filtered diagrams?

   (I guess it should be an easy generalization of the above statement, but I am throwing out this question anyway now.)

2. • CommentRowNumber2.
   • CommentAuthorTim_Porter
   • CommentTimeMay 3rd 2012
   • (edited May 3rd 2012)
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   Author: Tim_Porter
   Format: MarkdownItexIf I remember rightly the answer is YES. It involves the condition that each time you complete a pair of parallel morphisms by a further (co)refinement then the induced map from the (homotopy) pushout of the two maps to the new object is a cofibration.

   If I remember rightly the answer is YES. It involves the condition that each time you complete a pair of parallel morphisms by a further (co)refinement then the induced map from the (homotopy) pushout of the two maps to the new object is a cofibration.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMay 3rd 2012
   • (edited May 3rd 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI guess you mean the map out of the coproduct of the source with itself? To lift in $$ \array{ && A_0 &\stackrel{\to}{\to}& A_1 \\ && \downarrow && \downarrow \\ && B_0 &\stackrel{\to}{\to}& B_1 \\ & \nearrow && \nearrow \\ X_0 &\stackrel{\to}{\to}& X_1 } $$ we can proceed by first finding $\sigma_0$ in $$ \array{ && A_0 \\ & {}^{\mathllap{\sigma_0}}\nearrow & \downarrow \\ X_0 &\to& B_0 } $$ and then finding $\sigma_1 : X_1 \to A_1$ fitting into $$ \array{ && A_0 \coprod A_0 &\to & A_1 \\ && \downarrow && \downarrow \\ {}^{\mathllap{\sigma_0 \coprod \sigma_0}}\nearrow& & B_0 \coprod B_0 &\to & B_1 \\ & \nearrow && \nearrow \\ X_0 \coprod X_0 &\to& X_1 } \,. $$ So we'd need $X_0$ to be cofibrant and $X_0 \coprod X_0 \to X_1$ to be a cofibration.

   I guess you mean the map out of the coproduct of the source with itself?

   To lift in

   $$\array{ && A_0 &\stackrel{\to}{\to}& A_1 \\ && \downarrow && \downarrow \\ && B_0 &\stackrel{\to}{\to}& B_1 \\ & \nearrow && \nearrow \\ X_0 &\stackrel{\to}{\to}& X_1 }$$

   we can proceed by first finding $\sigma_0$ in

   $$\array{ && A_0 \\ & {}^{\mathllap{\sigma_0}}\nearrow & \downarrow \\ X_0 &\to& B_0 }$$

   and then finding $\sigma_1 : X_1 \to A_1$ fitting into

   $$\array{ && A_0 \coprod A_0 &\to & A_1 \\ && \downarrow && \downarrow \\ {}^{\mathllap{\sigma_0 \coprod \sigma_0}}\nearrow& & B_0 \coprod B_0 &\to & B_1 \\ & \nearrow && \nearrow \\ X_0 \coprod X_0 &\to& X_1 } \,.$$

   So we’d need $X_0$ to be cofibrant and $X_0 \coprod X_0 \to X_1$ to be a cofibration.

4. • CommentRowNumber4.
   • CommentAuthorTim_Porter
   • CommentTimeMay 3rd 2012
   • PermaLink
   Author: Tim_Porter
   Format: MarkdownItexI'm not sure that that is the same as what I was meaning, but have to rush at the moment so will just say: if we have two maps from i to j then there is a k and a map from j to k equalising them (i.e. common corefinement) This gives two maps from $X_i$ to $X_j$ and one from $X_j$ to $X_k$. We need all the maps between things to be cofibrations but ALSO that the unique map induced from the pushout of the two maps with source $X_i$ to the third object $X_k$, is also a cofibration. Without spending time on doing the diagrams ..... I hope that is clearer. There may be further 'higher dimensional' constraints but I don't think so.

   I’m not sure that that is the same as what I was meaning, but have to rush at the moment so will just say: if we have two maps from i to j then there is a k and a map from j to k equalising them (i.e. common corefinement)

   This gives two maps from $X_i$ to $X_j$ and one from $X_j$ to $X_k$. We need all the maps between things to be cofibrations but ALSO that the unique map induced from the pushout of the two maps with source $X_i$ to the third object $X_k$, is also a cofibration.

   Without spending time on doing the diagrams ….. I hope that is clearer. There may be further ’higher dimensional’ constraints but I don’t think so.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMay 3rd 2012
   • PermaLink
   Author: Urs
   Format: MarkdownItexI did understand what you are saying about pushouts. I am just wondering if you really mean to say that. Consider the "walking coequalizing" diagram $$ D = \left\{ \array{ X_0 && \stackrel{\to}{\to} & & X_1 \\ & \searrow && \swarrow \\ && X_2 } \right\} \,. $$ Then it seems to me that my previous message shows that in order for $X_\bullet : D \to C$ to be projectively cofibrant we need 1. $X_0$ is cofibrant; 1. $X_0 \coprod X_0 \to X_1$ is a cofibration; 1. $X_1 \to X_2$ is a cofibration.

   I did understand what you are saying about pushouts. I am just wondering if you really mean to say that.

   Consider the “walking coequalizing” diagram

   $$D = \left\{ \array{ X_0 && \stackrel{\to}{\to} & & X_1 \\ & \searrow && \swarrow \\ && X_2 } \right\} \,.$$

   Then it seems to me that my previous message shows that in order for $X_\bullet : D \to C$ to be projectively cofibrant we need

   1. $X_0$ is cofibrant;

   2. $X_0 \coprod X_0 \to X_1$ is a cofibration;

   3. $X_1 \to X_2$ is a cofibration.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeMay 3rd 2012
   • (edited May 3rd 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexOf course pushout-morphisms being cofibrations becomes relevant for describing general cofibrations in the diagram category (as spelled out for example at _[[Reedy model structure]]_ [here](http://ncatlab.org/nlab/show/Reedy%20model%20structure#OverTheTowerCategory), maybe that's what you are thinking of?). But not for the characterization of cofibrant objects, I think.

   Of course pushout-morphisms being cofibrations becomes relevant for describing general cofibrations in the diagram category (as spelled out for example at Reedy model structure here, maybe that’s what you are thinking of?). But not for the characterization of cofibrant objects, I think.

7. • CommentRowNumber7.
   • CommentAuthorStephan A Spahn
   • CommentTimeMay 7th 2012
   • (edited May 7th 2012)
   • PermaLink
   Author: Stephan A Spahn
   Format: MarkdownItex> If I remember rightly the answer is YES. So is the statement you indicate is that for all filtered diagrams $X$ we have: $X$ is projectively cofibrant if it is a diagram of cofibrations (=all objects are cofibrant and all arrows are cofibrations)? The argument given in [Example 2.3.15, page 130 here](http://ncatlab.org/schreiber/files/cohesivedocumentv032.pdf#page=130) works for any diagram $X$ where each $X_i$ has an incoming morphism from some other $X_j$ for which a lifting has been shown.

   If I remember rightly the answer is YES.

   So is the statement you indicate is that for all filtered diagrams $X$ we have: $X$ is projectively cofibrant if it is a diagram of cofibrations (=all objects are cofibrant and all arrows are cofibrations)?

   The argument given in Example 2.3.15, page 130 here works for any diagram $X$ where each $X_i$ has an incoming morphism from some other $X_j$ for which a lifting has been shown.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeMay 7th 2012
   • (edited May 7th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexHi Stephan, no, this can't be right. Look at the example in #3, the simple case of a diagram consisting of two parallel morphisms. It is not sufficient that the two morphisms $X_0 \to X_1$ are cofibrations. You need $X_0 \coprod X_0 \to X_1$ to be a cofibration.

   Hi Stephan,

   no, this can’t be right. Look at the example in #3, the simple case of a diagram consisting of two parallel morphisms. It is not sufficient that the two morphisms $X_0 \to X_1$ are cofibrations. You need $X_0 \coprod X_0 \to X_1$ to be a cofibration.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeMay 7th 2012
   • (edited May 7th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexYou can check that this is not only sufficient, which the argument in #3 shows, but also necessary: for let $$ \array{ X_0 \coprod X_0 &\stackrel{(a,b)}{\to}& A_1 \\ {}^{\mathllap{(f,g)}}\downarrow && \downarrow^{\mathrlap{p}} \\ X_1 &\stackrel{q}{\to}& B_1 } $$ be a commuting diagram, where $p : A_1 \to B_1$ is an acyclic fibration. Using this we can build a lifting problem in the category of parallel morphisms, by setting $$ \array{ && X_0 &\stackrel{\overset{a}{\to}}{\underset{b}{\to}}& A_1 \\ && \downarrow && \downarrow^{\mathrlap{p}} \\ && X_0 & \stackrel{\stackrel{p \circ a = q \circ f}{\to}}{\underset{p \circ b = q \circ g}{\to}} & B_1 \\ & \nearrow && \nearrow_{\mathrlap{q}} \\ X_0 &\stackrel{\overset{f}{\to}}{\underset{g}{\to}}& X_1 } \,. $$ Notice that the top square is an acyclic fibration in $[=, \mathcal{C}]_{proj}$. You see that a lift in the diagram category exists precisely if the original square has a lift, hence precisely if $(f,g)$ is a cofibration. So for $X_0 \stackrel{\overset{f}{\to}}{\underset{g}{\to}} X_1 $ to be projectively cofibrant, it is necessary that $(f,g): X_0 \coprod X_0 \to X_1$ is a cofibration.

   You can check that this is not only sufficient, which the argument in #3 shows, but also necessary:

   for let

   $$\array{ X_0 \coprod X_0 &\stackrel{(a,b)}{\to}& A_1 \\ {}^{\mathllap{(f,g)}}\downarrow && \downarrow^{\mathrlap{p}} \\ X_1 &\stackrel{q}{\to}& B_1 }$$

   be a commuting diagram, where $p : A_1 \to B_1$ is an acyclic fibration. Using this we can build a lifting problem in the category of parallel morphisms, by setting

   $$\array{ && X_0 &\stackrel{\overset{a}{\to}}{\underset{b}{\to}}& A_1 \\ && \downarrow && \downarrow^{\mathrlap{p}} \\ && X_0 & \stackrel{\stackrel{p \circ a = q \circ f}{\to}}{\underset{p \circ b = q \circ g}{\to}} & B_1 \\ & \nearrow && \nearrow_{\mathrlap{q}} \\ X_0 &\stackrel{\overset{f}{\to}}{\underset{g}{\to}}& X_1 } \,.$$

   Notice that the top square is an acyclic fibration in $[=, \mathcal{C}]_{proj}$.

   You see that a lift in the diagram category exists precisely if the original square has a lift, hence precisely if $(f,g)$ is a cofibration. So for $X_0 \stackrel{\overset{f}{\to}}{\underset{g}{\to}} X_1$ to be projectively cofibrant, it is necessary that $(f,g): X_0 \coprod X_0 \to X_1$ is a cofibration.

10. • CommentRowNumber10.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 7th 2012
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItex> the statement you indicate is that for all filtered diagrams X we have > no, this can't be right Yes, I thought it isn't true that's why I provoked this clear statement. Also the word ''filtered '' from the [other thread](http://nforum.mathforge.org/discussion/3761/compact-objects-in-the-topos-over-all-manifolds/#Item_28) is wrong here; e.g. a span is cofiltered but not filtered.

    the statement you indicate is that for all filtered diagrams X we have

    no, this can’t be right

    Yes, I thought it isn’t true that’s why I provoked this clear statement. Also the word ”filtered ” from the other thread is wrong here; e.g. a span is cofiltered but not filtered.

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItexOkay, so if this is clear, here is an exercise: what's sufficient and necessary for a [[direct category|direct diagram]] to be projectively cofibrant?

    Okay, so if this is clear, here is an exercise:

    what’s sufficient and necessary for a direct diagram to be projectively cofibrant?

12. • CommentRowNumber12.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 7th 2012
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItexFor $D$ a direct category this amounts to determine the cofibrant objects in the [[Reedy model structure]] where $D=D_+$ and $D_-$ are the identities...I will try to dualize the exmple given there.

    For $D$ a direct category this amounts to determine the cofibrant objects in the Reedy model structure where $D=D_+$ and $D_-$ are the identities…I will try to dualize the exmple given there.

13. • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItexRight, for direct categories the projective and the Reedy structure coincide. You might even want to think about cofibrant diagrams in the Reedy model structure over any [[elegant Reedy category]]. (Mike will have some lemmas about that out, soon.) But for this you have to do a little bit of genuine work. My "exercise" was meant as an immediate corollary, solvable essentially without work, just by looking at what we have so far. As a warmup, consider a category with morphisms $$ 0 \stackrel{\to}{\stackrel{\to}{\to}} 1 \stackrel{\to}{\to} 2 $$ and their composites (anything will do). So a category with three objects, with three morphisms from the 0th to the first, with two morphism from the second to the third, and with any choice of composition on these. When is a diagram of this shape projectively cofibrant? Hint: if this isn't clear yet, try induction on $n$ in the followig simple sense: write down a lifting problem. Then think about conditions you need to lift just over 0. After that, think about conditions to lift next just over 1. Use the previous discussion for that. After that, think about conditions to lift next over 2. Again use the previous discussion for that. Once you have done this, you'll see how it works generally for direct categories. If you get stuck, you must let me know. Don't hesitate to throw out questions.

    Right, for direct categories the projective and the Reedy structure coincide. You might even want to think about cofibrant diagrams in the Reedy model structure over any elegant Reedy category. (Mike will have some lemmas about that out, soon.)

    But for this you have to do a little bit of genuine work. My “exercise” was meant as an immediate corollary, solvable essentially without work, just by looking at what we have so far.

    As a warmup, consider a category with morphisms

    $$0 \stackrel{\to}{\stackrel{\to}{\to}} 1 \stackrel{\to}{\to} 2$$

    and their composites (anything will do). So a category with three objects, with three morphisms from the 0th to the first, with two morphism from the second to the third, and with any choice of composition on these.

    When is a diagram of this shape projectively cofibrant?

    Hint: if this isn’t clear yet, try induction on $n$ in the followig simple sense: write down a lifting problem. Then think about conditions you need to lift just over 0. After that, think about conditions to lift next just over 1. Use the previous discussion for that. After that, think about conditions to lift next over 2. Again use the previous discussion for that.

    Once you have done this, you’ll see how it works generally for direct categories.

    If you get stuck, you must let me know. Don’t hesitate to throw out questions.

14. • CommentRowNumber14.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItexThanks, Urs! > As a warmup, consider a category with morphisms I think for this example $$ 0 \stackrel{\to}{\stackrel{\to}{\to}} 1 \stackrel{\to}{\to} 2 $$ the missing commuting diagrams (which look better in instiki than here) are $$ \array{ X_0 \coprod X_0 \coprod X_0&\stackrel{(a,b,c)}{\to}& A_1 \\ {}^{\mathllap{(f,g,h)}}\downarrow && \downarrow^{\mathrlap{p}} \\ X_1 &\stackrel{q}{\to}& B_1 } $$ where $p : A_1 \to B_1$ is an acyclic fibration. And the corresponding lifting problem $$ \array{ && X_0 &\xrightarrow{a,b,c}& A_1 \\ && \downarrow && \downarrow^{\mathrlap{p}} \\ && X_0 &\xrightarrow{p a = q f,p b = qg,pc=qh} & B_1 \\ & \nearrow && \nearrow_{\mathrlap{q}} \\ X_0 &\xrightarrow{f,g,h}& X_1 } \,. $$ and hence the relevant necessary and sufficient conditions would be: $(f_1,g_1): X_1 \coprod X_1 \to X_2$ and $(f,g,h)=(f_0,g_0,h_0): X_0 \coprod X_0\coprod X_0 \to X_1$ ) shall be cofibrations. For a general set of $n$ parallel morphisms we should require $(pr_{i=0,...,n+2}): X_n^{\times_{n+2}}\to X_{n+1}$ to be a cofibration. This means that for any direct diagram to be projectively cofibrant it suffices indeed -as Tim says above- that it is a ''diagram of cofibrations'' and all occurring $(pr_{i=0,...,n+2})$ are cofibrations.

    Thanks, Urs!

    As a warmup, consider a category with morphisms

    I think for this example

    $$0 \stackrel{\to}{\stackrel{\to}{\to}} 1 \stackrel{\to}{\to} 2$$

    the missing commuting diagrams (which look better in instiki than here) are

    $$\array{ X_0 \coprod X_0 \coprod X_0&\stackrel{(a,b,c)}{\to}& A_1 \\ {}^{\mathllap{(f,g,h)}}\downarrow && \downarrow^{\mathrlap{p}} \\ X_1 &\stackrel{q}{\to}& B_1 }$$

    where $p : A_1 \to B_1$ is an acyclic fibration. And the corresponding lifting problem

    $$\array{ && X_0 &\xrightarrow{a,b,c}& A_1 \\ && \downarrow && \downarrow^{\mathrlap{p}} \\ && X_0 &\xrightarrow{p a = q f,p b = qg,pc=qh} & B_1 \\ & \nearrow && \nearrow_{\mathrlap{q}} \\ X_0 &\xrightarrow{f,g,h}& X_1 } \,.$$

    and hence the relevant necessary and sufficient conditions would be: $(f_1,g_1): X_1 \coprod X_1 \to X_2$ and $(f,g,h)=(f_0,g_0,h_0): X_0 \coprod X_0\coprod X_0 \to X_1$ ) shall be cofibrations. For a general set of $n$ parallel morphisms we should require $(pr_{i=0,...,n+2}): X_n^{\times_{n+2}}\to X_{n+1}$ to be a cofibration. This means that for any direct diagram to be projectively cofibrant it suffices indeed -as Tim says above- that it is a ”diagram of cofibrations” and all occurring $(pr_{i=0,...,n+2})$ are cofibrations.

15. • CommentRowNumber15.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 7th 2012
    • (edited May 7th 2012)
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItex> Right, for direct categories the projective and the Reedy structure coincide. You might even want to think about cofibrant diagrams in the Reedy model structure over > any [[elegant Reedy category]]. (Mike will have some lemmas about that out, soon.) Yes, if what I wrote above is correct so far, I would like to look into that, too.

    Right, for direct categories the projective and the Reedy structure coincide. You might even want to think about cofibrant diagrams in the Reedy model structure over > any elegant Reedy category. (Mike will have some lemmas about that out, soon.)

    Yes, if what I wrote above is correct so far, I would like to look into that, too.

16. • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeMay 7th 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Stephan, that looks right, though I thought you might want to give the argument for _why_ this is the answer. If you want to proceed to the general Reedy case, a next good exercise would be to look at the previous example again, by now formally computing the latching objects and the morphisms out of them and checking that this reproduces what you have above.

    Hi Stephan,

    that looks right, though I thought you might want to give the argument for why this is the answer.

    If you want to proceed to the general Reedy case, a next good exercise would be to look at the previous example again, by now formally computing the latching objects and the morphisms out of them and checking that this reproduces what you have above.

17. • CommentRowNumber17.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 8th 2012
    • (edited May 8th 2012)
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItex> I thought you might want to give the argument for _why_ this is the answer. Before doing this I will write down what (I think) comes directly out of the definition of the latching objects: As said above a direct category $R$ becomes a Reedy category by choosing $R=R_+$ and $R_-=\{r,id_r|r\in R\}$ (the discrete subcategory on all objects of $R$). For $X\in Fun(R,C)$ and $r\in R$ we have by definition the latching object $$L_r X:=colim_{s\xrightarrow{+}r}X_s:=colim_{s\in R_+/r -\{id_r\}}X(s)$$ (the matching object $M_r X:=colim_{s\in r/R_- -\{id_r\}}X(s)$ we do not need this for the cofibrancy condition) In the example $$X_0\stackrel{f_0,g_0,h_0}{\to}X_1\stackrel{f_1,g_1}{\to}X_2\stackrel{f_2}{\to}X_3$$ we have $L_0 X=colim_{s\in R_+/0 -\{id_0\}}X(s)=0$ $L_1 X=colim_{s\in R_+/1 -\{id_1\}}X(s)=coeq(f_0,g_0,h_0)$ $L_2 X=colim_{s\in R_+/2 -\{id_1\}}X(s)=coeq(f_0,g_0,h_0)\coprod coeq(f_1,g_1)$ $L_3 X=colim_{s\in R_+/3 -\{id_3\}}X(s)=coeq(f_0,g_0,h_0)\coprod coeq(f_1,g_1)\coprod (X_3,f_2,id)$ where the last cofactor is the couniversal cocone of the rightest subdiagram. For $r\in R$ we have $X$ is cofibrant iff $L_r X\to X_r$ is a cofibration. Now the statement from above giving necessary- and sufficient conditions for a diagram of two (respectively more) parallel morphisms to be Reedy-cofibrant should enter. For this we could write the colimit in question instead as the colimit of $$\array{X_0 &\stackrel{i_0}{\to}& X_0\coprod X_0 \\ \downarrow^{i_1}&&\downarrow{(f,g)} \\ X_0\coprod X_0 &\stackrel{(f,g)}{\to}& X_1 }$$ (I have to interrupt now to catch a train)

    I thought you might want to give the argument for why this is the answer.

    Before doing this I will write down what (I think) comes directly out of the definition of the latching objects:

    As said above a direct category $R$ becomes a Reedy category by choosing $R=R_+$ and $R_-=\{r,id_r|r\in R\}$ (the discrete subcategory on all objects of $R$).

    For $X\in Fun(R,C)$ and $r\in R$ we have by definition the latching object

    $$L_r X:=colim_{s\xrightarrow{+}r}X_s:=colim_{s\in R_+/r -\{id_r\}}X(s)$$

    (the matching object $M_r X:=colim_{s\in r/R_- -\{id_r\}}X(s)$ we do not need this for the cofibrancy condition)

    In the example

    $$X_0\stackrel{f_0,g_0,h_0}{\to}X_1\stackrel{f_1,g_1}{\to}X_2\stackrel{f_2}{\to}X_3$$

    we have

    $L_0 X=colim_{s\in R_+/0 -\{id_0\}}X(s)=0$

    $L_1 X=colim_{s\in R_+/1 -\{id_1\}}X(s)=coeq(f_0,g_0,h_0)$

    $L_2 X=colim_{s\in R_+/2 -\{id_1\}}X(s)=coeq(f_0,g_0,h_0)\coprod coeq(f_1,g_1)$

    $L_3 X=colim_{s\in R_+/3 -\{id_3\}}X(s)=coeq(f_0,g_0,h_0)\coprod coeq(f_1,g_1)\coprod (X_3,f_2,id)$

    where the last cofactor is the couniversal cocone of the rightest subdiagram. For $r\in R$ we have $X$ is cofibrant iff $L_r X\to X_r$ is a cofibration.

    Now the statement from above giving necessary- and sufficient conditions for a diagram of two (respectively more) parallel morphisms to be Reedy-cofibrant should enter. For this we could write the colimit in question instead as the colimit of

    $$\array{X_0 &\stackrel{i_0}{\to}& X_0\coprod X_0 \\ \downarrow^{i_1}&&\downarrow{(f,g)} \\ X_0\coprod X_0 &\stackrel{(f,g)}{\to}& X_1 }$$

    (I have to interrupt now to catch a train)

18. • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeMay 8th 2012
    • (edited May 8th 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Stephan, this is roughly going in the right direction, but needs some corrections: there are too many _coeqs_. Let's break this down in substeps and walk through it in small steps. Consider just the category $$ R_+ := \left\{ 0 \stackrel{\overset{f}{\to}}{\underset{g}{\to}} 1 \right\} $$ Then what is the category that we take the colimit over to produce the latching object at $1$? It is defined to be the full subcategory of the [[overcategory]] $(R_+)_{/1}$ on the non-identity elements. So first of all, what is that overcategory $(R_+)_{/1}$? Its objects are the morphisms into "1", so $$ Obj((R_+)_{/1}) = \{f,g, id_1\} \,. $$ Its morphisms are the commuting triangles over "1". There are precisely two non-identity triangles namely $$ \array{ 0 &&\stackrel{f}{\to}&& 1 \\ & {}_{\mathllap{f}}\searrow && \swarrow_{\mathrlap{id_1}} \\ && 1 } \;\;\;\;and \;\;\; \array{ 0 &&\stackrel{g}{\to}&& 1 \\ & {}_{\mathllap{g}}\searrow && \swarrow_{\mathrlap{id_1}} \\ && 1 } $$ Agreed? So this means there are precisely two non-identity morphisms in $(R_+)_{/1}$, and so this category looks as follows: $$ (R_+)_{/1} = \left\{ \array{ f \to id_1 \leftarrow g } \right\} \,. $$ Okay? But now the category we are after is supposed to be the [[full subcategory]] of this on all objects except $id_1$. So we are to throw away that object together with all morphisms to and from it, but keep everything else. The result of that is $$ \left\{ \array{ f , g } \right\} \,, $$ which is the category with just two objects and _no_ nontrivial morphisms. Now for $$ X_0 \stackrel{\to}{\to} X_1 $$ a diagram in some model category $\mathcal{C}$, the latching object at "1" is defined to be the colimit over the functor $$ \left\{ \array{ f,g } \right\} \to \mathcal{C} $$ which takes a morphism in $R_+$ to the value $X_{dom}$ of the above diagram at the domain of this morphism. So this functor is the functor on the category with just two objects called $f$ and $g$ and no non-identity morphisms which assigns $$ \array{ f & \mapsto X_{dom(f)} = X_0 \\ g & \mapsto X_{dom(g)} = X_0 } $$ Okay? What is the colimit over such a diagram, which just consists of two copies of the object $X_0$? It is the [[coproduct]] of these objects, hence $$ \underset{\longrightarrow}{\lim} \left( \left\{ \array{ f,g } \right\} \to \mathcal{C} \right) = X_0 \coprod X_0 . $$ Hence this is the latching object $L_1 X$ of our diagram. Convince yourself that the canonical morphism $L_1 X \to X_1$ out of the latching object into $X_1$ is $$ (f,g ) : X_0 \coprod X_0 \to X_1 \,. $$ So a necessary condition for Reedy cofibrancy = projective cofibrancy of the diagram $X_\bullet$ is that _this_ is a cofibration. (No coequalizers anywhere.) Notice that this reproduces what we found before in #3! :-) Let me know if this is clear so far. If not, let me know which step is unclear. If it is clear, redo the previous exercise! :-)

    Hi Stephan,

    this is roughly going in the right direction, but needs some corrections: there are too many coeqs.

    Let’s break this down in substeps and walk through it in small steps.

    Consider just the category

    $$R_+ := \left\{ 0 \stackrel{\overset{f}{\to}}{\underset{g}{\to}} 1 \right\}$$

    Then what is the category that we take the colimit over to produce the latching object at $1$?

    It is defined to be the full subcategory of the overcategory $(R_+)_{/1}$ on the non-identity elements.

    So first of all, what is that overcategory $(R_+)_{/1}$? Its objects are the morphisms into “1”, so

    $$Obj((R_+)_{/1}) = \{f,g, id_1\} \,.$$

    Its morphisms are the commuting triangles over “1”. There are precisely two non-identity triangles namely

    $$\array{ 0 &&\stackrel{f}{\to}&& 1 \\ & {}_{\mathllap{f}}\searrow && \swarrow_{\mathrlap{id_1}} \\ && 1 } \;\;\;\;and \;\;\; \array{ 0 &&\stackrel{g}{\to}&& 1 \\ & {}_{\mathllap{g}}\searrow && \swarrow_{\mathrlap{id_1}} \\ && 1 }$$

    Agreed? So this means there are precisely two non-identity morphisms in $(R_+)_{/1}$, and so this category looks as follows:

    $$(R_+)_{/1} = \left\{ \array{ f \to id_1 \leftarrow g } \right\} \,.$$

    Okay?

    But now the category we are after is supposed to be the full subcategory of this on all objects except $id_1$. So we are to throw away that object together with all morphisms to and from it, but keep everything else. The result of that is

    $$\left\{ \array{ f , g } \right\} \,,$$

    which is the category with just two objects and no nontrivial morphisms.

    Now for

    $$X_0 \stackrel{\to}{\to} X_1$$

    a diagram in some model category $\mathcal{C}$, the latching object at “1” is defined to be the colimit over the functor

    $$\left\{ \array{ f,g } \right\} \to \mathcal{C}$$

    which takes a morphism in $R_+$ to the value $X_{dom}$ of the above diagram at the domain of this morphism.

    So this functor is the functor on the category with just two objects called $f$ and $g$ and no non-identity morphisms which assigns

    $$\array{ f & \mapsto X_{dom(f)} = X_0 \\ g & \mapsto X_{dom(g)} = X_0 }$$

    Okay?

    What is the colimit over such a diagram, which just consists of two copies of the object $X_0$? It is the coproduct of these objects, hence

    $$\underset{\longrightarrow}{\lim} \left( \left\{ \array{ f,g } \right\} \to \mathcal{C} \right) = X_0 \coprod X_0 .$$

    Hence this is the latching object $L_1 X$ of our diagram.

    Convince yourself that the canonical morphism $L_1 X \to X_1$ out of the latching object into $X_1$ is

    $$(f,g ) : X_0 \coprod X_0 \to X_1 \,.$$

    So a necessary condition for Reedy cofibrancy = projective cofibrancy of the diagram $X_\bullet$ is that this is a cofibration. (No coequalizers anywhere.)

    Notice that this reproduces what we found before in #3! :-)

    Let me know if this is clear so far. If not, let me know which step is unclear. If it is clear, redo the previous exercise! :-)

19. • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeMay 8th 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItexAnother good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is: 1. write down $\Delta_+/[2]$. 1. Write down the full subcategory of it that is the shape for the colimit that gives the latching object at $[2]$. (Hint: it has six objects. A cartoon of it is depicted at _[[Kan fibrant replacement]]_.)

    Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is:

    1. write down $\Delta_+/[2]$.

    2. Write down the full subcategory of it that is the shape for the colimit that gives the latching object at $[2]$.

       (Hint: it has six objects. A cartoon of it is depicted at Kan fibrant replacement.)

20. • CommentRowNumber20.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItexHi Urs, > Let me know if this is clear so far Yes it is. The error in my above writing was that I didn't really exclude the identities. > Convince yourself that the canonical morphism $L_1 X \to X_1$ out of the latching object into $X_1$ is $ (f,g ) : X_0 \coprod X_0 \to X_1$. This is clear by the universal property of the colimiting cocone since the canonical morphism $can:X_0 \coprod X_0 \to X_1$ satisfies $can\circ i_1 =f $ and $can\circ i_2=g$ for the canonical inclusions $i_1$,$i_2$ and by the unicity of $can$ and the fact that $(f,g)$ solves this equalities. > redo the previous exercise! :-) For more than two parallel morphisms your explanation with $\{f,g,h\}$ instead of $\{f,g\}$ applies mutatis mutandis and I guess in my above ''solution'' the part where I combine the diagrams ''with the coproduct between the colimits'' was right (by the universal properties of the colimits of the sub diagrams). > Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is: Yes, I will do this. > You might even want to think about cofibrant diagrams in the Reedy model structure over any [[elegant Reedy category]]. (Mike will have some lemmas about that out, soon.) I suppose you think of a sequel of material in [The univalence axiom for inverse diagrams](http://arxiv.org/abs/1203.3253)?

    Hi Urs,

    Let me know if this is clear so far

    Yes it is. The error in my above writing was that I didn’t really exclude the identities.

    Convince yourself that the canonical morphism $L_1 X \to X_1$ out of the latching object into $X_1$ is $(f,g ) : X_0 \coprod X_0 \to X_1$.

    This is clear by the universal property of the colimiting cocone since the canonical morphism $can:X_0 \coprod X_0 \to X_1$ satisfies $can\circ i_1 =f$ and $can\circ i_2=g$ for the canonical inclusions $i_1$,$i_2$ and by the unicity of $can$ and the fact that $(f,g)$ solves this equalities.

    redo the previous exercise! :-)

    For more than two parallel morphisms your explanation with $\{f,g,h\}$ instead of $\{f,g\}$ applies mutatis mutandis and I guess in my above ”solution” the part where I combine the diagrams ”with the coproduct between the colimits” was right (by the universal properties of the colimits of the sub diagrams).

    Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is:

    Yes, I will do this.

    You might even want to think about cofibrant diagrams in the Reedy model structure over any elegant Reedy category. (Mike will have some lemmas about that out, soon.)

    I suppose you think of a sequel of material in The univalence axiom for inverse diagrams?

21. • CommentRowNumber21.
    • CommentAuthorUrs
    • CommentTimeMay 9th 2012
    • (edited May 9th 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Stephan, yes, okay. You might want to look at the diagram category $$ \array{ 0 &&\stackrel{\to}{\to}&& 1 \\ & \searrow && \swarrow \\ && 2 } $$ just once more, to cross check your understanding (where the diagram is shorthand for saying that the composite of either of the two horizontal morphisms with the right one is equal to the left morphism). Here, the latching object at 2 does involve a coequalizer. > I suppose you think of a sequel of material in _The univalence axiom for inverse diagrams_? Yes, there is a sequel in the making.

    Hi Stephan,

    yes, okay. You might want to look at the diagram category

    $$\array{ 0 &&\stackrel{\to}{\to}&& 1 \\ & \searrow && \swarrow \\ && 2 }$$

    just once more, to cross check your understanding (where the diagram is shorthand for saying that the composite of either of the two horizontal morphisms with the right one is equal to the left morphism). Here, the latching object at 2 does involve a coequalizer.

    I suppose you think of a sequel of material in The univalence axiom for inverse diagrams?

    Yes, there is a sequel in the making.

22. • CommentRowNumber22.
    • CommentAuthorMike Shulman
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> Yes, there is a sequel in the making. But I'm not sure the relevance that it has here. Every presheaf on an elegant Reedy category (in a model category whose cofibrations are the monos) is Reedy cofibrant, since the Reedy cofibrations are then also the (levelwise) monos, but that's already in Julie and Charles' paper. What else are you thinking of?

    Yes, there is a sequel in the making.

    But I’m not sure the relevance that it has here. Every presheaf on an elegant Reedy category (in a model category whose cofibrations are the monos) is Reedy cofibrant, since the Reedy cofibrations are then also the (levelwise) monos, but that’s already in Julie and Charles’ paper. What else are you thinking of?

23. • CommentRowNumber23.
    • CommentAuthorUrs
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItexWell, there is an op-ing here changing the variance. We have been discussing co-presheaves here.

    Well, there is an op-ing here changing the variance. We have been discussing co-presheaves here.

24. • CommentRowNumber24.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItex> yes, okay. You might want to look at the diagram category ... If we consider $$ \array{ 0 &&\stackrel{f,g}{\to}&& 1 \\ & \searrow^h && \swarrow^k \\ && 2 } $$ we have $R_+/2=\{ \array{h &\stackrel{f,g}{\to}& k \\ &\searrow^h \swarrow^k \\ & id_2} \}$ and a for a functor $X$ of this shape the latching object over $2$ gives me the opportunity to get the coequalizer off $$L_2 X= colim (R_+/2-\{id_2\}\stackrel{X)}{\to} C)=colim(\{h\stackrel{f,g}{\to}k\}\stackrel{X}{\to} C)=coeq(X(f),X(g))$$ > Then it seems to me that my previous message shows that in order for $X_\bullet : D \to C$ to be projectively cofibrant we need > 1. $X_0$ is cofibrant; > 2. $X_0 \coprod X_0 \to X_1$ is a cofibration; > 3. $X_1 \to X_2$ is a cofibration. The latching object in $0$ is the colimit over the empty diagram aka the initial object. This explains why $X_0$ needs to be cofibrant. Condition 2 has already been clarified. It remains to see that $X_1 \to X_2$ being a cofibration implies that $coeq(X(f),X(g))\to X_2$ is, too (I do not see this instantly and will return to it later). > Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is: Let $\Delta_+$ denote the wide subcategory of $\Delta$ with only surjections as morphisms. These surjections are generated by cofaces only. $$ \array{ [0] &&\stackrel{d^0,d^1}{\to}&& [1] \\ & \searrow&& \swarrow^{d^0,d^1,d^2} \\ && [2] } $$ $$\Delta_+/[2]-\{id_{[2]}\}=\{ \array{ d^0d^0\stackrel{d^0}{\to}d^0 \\ d^0d^1\stackrel{d^1}{\to}d^0 \\ d^1d^0\stackrel{d^0}{\to}d^1 \\ d^1d^1\stackrel{d^1}{\to}d^1 \\ d^2d^0\stackrel{d^2}{\to}d^0 \\ d^2d^1\stackrel{d^1}{\to}d^2 }\}$$ The colimit over this diagram is the coproduct over the colimits of the sub diagrams consisting of just one morphism e.g. $d^0d^0\stackrel{d^0}{\to}d^0$ which are the cocones on the codomain e.g. $(d_0,id_{d_0}, d^0d^0\stackrel{d^0}{\to}d^0)$.

    yes, okay. You might want to look at the diagram category …

    If we consider

    $$\array{ 0 &&\stackrel{f,g}{\to}&& 1 \\ & \searrow^h && \swarrow^k \\ && 2 }$$

    we have

    $R_+/2=\{ \array{h &\stackrel{f,g}{\to}& k \\ &\searrow^h \swarrow^k \\ & id_2} \}$

    and a for a functor $X$ of this shape the latching object over $2$ gives me the opportunity to get the coequalizer off

    $$L_2 X= colim (R_+/2-\{id_2\}\stackrel{X)}{\to} C)=colim(\{h\stackrel{f,g}{\to}k\}\stackrel{X}{\to} C)=coeq(X(f),X(g))$$

    Then it seems to me that my previous message shows that in order for $X_\bullet : D \to C$ to be projectively cofibrant we need

    1. $X_0$ is cofibrant;

    2. $X_0 \coprod X_0 \to X_1$ is a cofibration;

    3. $X_1 \to X_2$ is a cofibration.

    The latching object in $0$ is the colimit over the empty diagram aka the initial object. This explains why $X_0$ needs to be cofibrant. Condition 2 has already been clarified. It remains to see that $X_1 \to X_2$ being a cofibration implies that $coeq(X(f),X(g))\to X_2$ is, too (I do not see this instantly and will return to it later).

    Another good exercise to do to help with the above, which you will need for lots of other things in homotopy theory anyway, is:

    Let $\Delta_+$ denote the wide subcategory of $\Delta$ with only surjections as morphisms. These surjections are generated by cofaces only.

    $$\array{ [0] &&\stackrel{d^0,d^1}{\to}&& [1] \\ & \searrow&& \swarrow^{d^0,d^1,d^2} \\ && [2] }$$ $$\Delta_+/[2]-\{id_{[2]}\}=\{ \array{ d^0d^0\stackrel{d^0}{\to}d^0 \\ d^0d^1\stackrel{d^1}{\to}d^0 \\ d^1d^0\stackrel{d^0}{\to}d^1 \\ d^1d^1\stackrel{d^1}{\to}d^1 \\ d^2d^0\stackrel{d^2}{\to}d^0 \\ d^2d^1\stackrel{d^1}{\to}d^2 }\}$$

    The colimit over this diagram is the coproduct over the colimits of the sub diagrams consisting of just one morphism e.g. $d^0d^0\stackrel{d^0}{\to}d^0$ which are the cocones on the codomain e.g. $(d_0,id_{d_0}, d^0d^0\stackrel{d^0}{\to}d^0)$.

25. • CommentRowNumber25.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItex> Well, there is an op-ing here changing the variance. We have been discussing co-presheaves here. I thought the sequel will be (partly) about this op-ing?. Also in this thread we didn't yet talk about an explicit choice of the model structure on the target category of the functor.

    Well, there is an op-ing here changing the variance. We have been discussing co-presheaves here.

    I thought the sequel will be (partly) about this op-ing?. Also in this thread we didn’t yet talk about an explicit choice of the model structure on the target category of the functor.

26. • CommentRowNumber26.
    • CommentAuthorUrs
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Urs
    Format: MarkdownItex> the sequel will be (partly) about this op-ing? Yes. But, okay, I guess I misspoke somewhere above. The thing is: we are currently looking at co-presheaf diagram on direct categories. This are presheaf diagrams in inverse categories. So that's what Mike discussed in the earlier article. My fault.

    the sequel will be (partly) about this op-ing?

    Yes. But, okay, I guess I misspoke somewhere above. The thing is:

    we are currently looking at co-presheaf diagram on direct categories. This are presheaf diagrams in inverse categories. So that’s what Mike discussed in the earlier article.

    My fault.

27. • CommentRowNumber27.
    • CommentAuthorUrs
    • CommentTimeMay 9th 2012
    • (edited May 9th 2012)
    • PermaLink
    Author: Urs
    Format: MarkdownItexHi Stephan, re #24: Yes, you've got it! Good! Indeed, the point you say you need to return to I also came across. I found it easiest to see by just writing down a lifting problem and checking that you can immediately find the required lifts that show this.

    Hi Stephan,

    re #24: Yes, you’ve got it! Good!

    Indeed, the point you say you need to return to I also came across. I found it easiest to see by just writing down a lifting problem and checking that you can immediately find the required lifts that show this.

28. • CommentRowNumber28.
    • CommentAuthorMike Shulman
    • CommentTimeMay 9th 2012
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> I thought the sequel will be (partly) about this op-ing? No. The existing paper is about diagrams (or, if you must, "co-presheaves") on inverse categories, which are presheaves on direct ones. The sequel is about presheaves on elegant Reedy categories, which generalize presheaves on direct ones.

    I thought the sequel will be (partly) about this op-ing?

    No. The existing paper is about diagrams (or, if you must, “co-presheaves”) on inverse categories, which are presheaves on direct ones. The sequel is about presheaves on elegant Reedy categories, which generalize presheaves on direct ones.

29. • CommentRowNumber29.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 10th 2012
    • (edited May 10th 2012)
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItexHi Urs, > Indeed, the point you say you need to return to I also came across. I found it easiest to see by just writing down a lifting problem and checking that you can immediately > find the required lifts that show this. Maybe the two lifting problems are linked by the diagram with the sub diagrams bottom: $$\array{ X_1 &\stackrel{}{\to}& X_2 \\ &\searrow^h\nearrow \\&coeq& }$$ top: $$\array{ A &\stackrel{}{\to}& B \\ &\searrow^{id}\nearrow \\&A& }$$ sides:(edited) $$\array{ A &\stackrel{id}{\to}& A &\stackrel{p}{\to}& B \\ \uparrow^{\phi h}&&\uparrow{\phi}&&\uparrow \\ X_1 &\stackrel{h}{\to}& coeq &\stackrel{p}{\to}& X_2 }$$ and $$\array{ A &\stackrel{p}{\to}& B \\ \uparrow&&\uparrow \\ X_1&\stackrel{}{\to}&X_2 }$$ where the letters with a hat are those induced from the ones without hat by the couniversal property of $coeq$. I will check tomorrow if this is correct.

    Hi Urs,

    Indeed, the point you say you need to return to I also came across. I found it easiest to see by just writing down a lifting problem and checking that you can immediately > find the required lifts that show this.

    Maybe the two lifting problems are linked by the diagram with the sub diagrams

    bottom:

    $$\array{ X_1 &\stackrel{}{\to}& X_2 \\ &\searrow^h\nearrow \\&coeq& }$$

    top:

    $$\array{ A &\stackrel{}{\to}& B \\ &\searrow^{id}\nearrow \\&A& }$$

    sides:(edited)

    $$\array{ A &\stackrel{id}{\to}& A &\stackrel{p}{\to}& B \\ \uparrow^{\phi h}&&\uparrow{\phi}&&\uparrow \\ X_1 &\stackrel{h}{\to}& coeq &\stackrel{p}{\to}& X_2 }$$

    and

    $$\array{ A &\stackrel{p}{\to}& B \\ \uparrow&&\uparrow \\ X_1&\stackrel{}{\to}&X_2 }$$

    where the letters with a hat are those induced from the ones without hat by the couniversal property of $coeq$. I will check tomorrow if this is correct.

30. • CommentRowNumber30.
    • CommentAuthorStephan A Spahn
    • CommentTimeMay 10th 2012
    • (edited May 10th 2012)
    • PermaLink
    Author: Stephan A Spahn
    Format: MarkdownItexLet $p$ be an acyclic fibration giving a lifting problem of the putative cofibration $coeq \to X_2$ against an acyclic fibration $p:A\to B$. this gives a lifting problem $$\array{ X_1 &\stackrel{\phi h}{\to}& A \\ \downarrow&&\downarrow \\ X_2&\stackrel{x}{\to}&B }$$ which possesses a solution $l:X_2\to A$ by assumption and by commutativity of the 3d diagram this is also a solution of the original problem. (Note that it is easier to see this if we insert a identity cell $X_2\to X_2$ instead of $A\to A$ in the 3d diagram - which I have edited since my post.)

    Let $p$ be an acyclic fibration giving a lifting problem of the putative cofibration $coeq \to X_2$ against an acyclic fibration $p:A\to B$.

    this gives a lifting problem

    $$\array{ X_1 &\stackrel{\phi h}{\to}& A \\ \downarrow&&\downarrow \\ X_2&\stackrel{x}{\to}&B }$$

    which possesses a solution $l:X_2\to A$ by assumption and by commutativity of the 3d diagram this is also a solution of the original problem. (Note that it is easier to see this if we insert a identity cell $X_2\to X_2$ instead of $A\to A$ in the 3d diagram - which I have edited since my post.)