Proposition A.3.3.9. in Higher Topos Theory is as follows:

Let $S$ be an excellent model category and let $f:C\rightarrow C'$ be a cofibration of small $S$-enriched categories. Then (1) for every combinatorial $S$-enriched model category $A$, the pullback $f^*:A^{C'}\rightarrow A^C$ preserves projective cofibrations and (2) for every projectively cofibrant object $F\in S^C,$ the unit map $F\rightarrow f^*f_!F$ is a projective cofibration.

I am OK with all of the proof after the first sentence, which claims that these two properties are clearly invariant under retracts of $f$. After that, the argument proceeds as usual, by proving the claim for transfinite compositions of pushouts of generating cofibrations.

But it is not at all clear to me why these two properties persist after a retract. For instance, take a retract $D$ of $C'$, given by functors $p:C'\rightarrow D$ and $q:D\rightarrow C'$ such that $pq$ is the identity functor. Say that $f$ factors through a map $g:C\rightarrow D$, so $g$ is a retract of $f$.

Then to show $g^*\cong f^*\circ p^*$ preserves projective cofibrations, it would suffice to show that $p^*$ has the same property. Unfortunately this is usually false. There are a few other ideas that don't seem to work either - is there some nice argument that I'm missing?

  • $\begingroup$ Maybe it's obvious, but which to version of HTT are you referring? $\endgroup$ – David Roberts Aug 19 '18 at 11:50
  • $\begingroup$ Could it be that everything to be preserved by $g^*$ has form $q^*$ of something that is preserved by $f^*$? $\endgroup$ – მამუკა ჯიბლაძე Aug 19 '18 at 12:01
  • $\begingroup$ E. g. if, say, $p^*q^*$ preserves projective cofibrations even though $p^*$ might not $\endgroup$ – მამუკა ჯიბლაძე Aug 19 '18 at 12:06
  • 2
    $\begingroup$ @DavidRoberts I'm using what I believe is the most recent version (April 2017) at math.harvard.edu/~lurie/papers/HTT.pdf $\endgroup$ – dhy Aug 19 '18 at 12:34
  • $\begingroup$ @dhy ok, good, so not the arXiv or the paper versions ;-) $\endgroup$ – David Roberts Aug 19 '18 at 21:00

You can argue as follows. Suppose that $g: D \to D'$ is a retract of $f: C \to C'$ (in the category of $S$-enriched categories) via maps $D \stackrel{i}{\to} C \stackrel{r}{\to} D$ and $D' \stackrel{i'}{\to} C' \stackrel{r'}{\to} D'$. Assume that $g^*: A^{C'} \to A^C$ preserves projective cofibrations and that the unit map $F \to f^*f_!F$ is a projective cofibration for every projectively cofibrant $F \in A^C$. We wish to show that $g^*:A^{D'} \to A^D$ preserves cofibrations and that $G \to g^*g_!G$ is a projective cofibration for every projectively cofibrant $G \in A^D$.

1) Consider the Beck-Chevalley transformations $\tau: i_!g^* \Rightarrow f^*i'_!$ and $\sigma: r_!f^* \Rightarrow g^*r'_!$. Then the composed transformation $$ g^* \cong r_!i_!g^* \stackrel{r_!\tau}{\Rightarrow} r_!f^*i'_! \stackrel{\sigma}{\Rightarrow} g^*r'_!i'_! \cong g^* $$ is the identity (indeed, it is the Beck-Chevalley transformation of a square in which two parallel legs are isomorphisms), and hence $g^*$ is a retract of $r_!f^*i'_!$ (in the category of functors $A^{D'} \to A^{D}$). Since $r_!f^*i'_!$ preserves projective cofibrations, so does $g^*$.

2) Consider again the retract diagram $g^* \Rightarrow r_!f^*i'_! \Rightarrow g^*$ above. Pre-composing with $g_!$ we obtain a retract diagram $$g^*g_! \Rightarrow r_!f^*i'_!g_! \cong r_!f^*f_!i_! \Rightarrow g^*g_! ,$$ and one can check that it is compatible with the unit maps $u_g: Id \Rightarrow g^*g_!$ and $u_f: Id \Rightarrow f^*f_!$, i.e., that it renders $u_g:Id \Rightarrow g^*g_!$ a retract of $r_!u_fi_!:Id \cong r_!i_! \Rightarrow r_!f^*f_!i_!$ (in the category of functors $A^{D} \to A^{D}$). It then follow that if $u_f$ is a projective cofibration on every projectively cofibrant functor then $u_g$ a projective cofibration on every projectively cofibrant functor, as desired.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.