7
$\begingroup$

In the paper http://arxiv.org/pdf/math/0101162.pdf, the authors claim during the proof of Prop. 4.2 that a functor $F:A \to B$ which preserves fibrations and weak equivalences preserves homotopy cartesian squares. It seems to me that without asking it to commute with ordinary pullbacks the claim is false.

Moreover, it is also stated that the induced functor $A^C \to B^C$, where $C$ is a Reedy category, preserves Reedy fibration. Again, I think it has to commute with pullbacks (e.g. a right Quillen would satisfy everything of the above for fibrant objects).

Am I right or am I missing something? Thanks in advance for any help.

$\endgroup$
  • 1
    $\begingroup$ Concerning the first claim, I don't see any problem, a square is a homotopy pull-back iff the induced map between homotopy fibers of parallel arrows is a weak equivalence. $\endgroup$ – Fernando Muro May 24 '15 at 21:44
  • 2
    $\begingroup$ @FernandoMuro, the proposition in the paper says that the functor preserves fibrations in the model structure, not fibre sequences. $\endgroup$ – AAK May 24 '15 at 21:49
  • 1
    $\begingroup$ @AdeelKhan if it preserves fibrations and weak equivalences, then it preserves homotopy fiber sequences. $\endgroup$ – Fernando Muro May 24 '15 at 21:53
  • 2
    $\begingroup$ @FernandoMuro The last thing you said is false: take a constant function to a non terminal object $\endgroup$ – Denis Nardin May 25 '15 at 4:38
  • 2
    $\begingroup$ Denis thanks, now I understand Edoardo's and Adeel's concern. I agree with Edoardo that I wouldn't be able to prove the claim without asking that the functor preserves pull backs. $\endgroup$ – Fernando Muro May 25 '15 at 6:35

Your Answer

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

Browse other questions tagged or ask your own question.