If we have a fibration $f:X\rightarrow Y$ in a model category $C$, where $Y$ is cofibrant and both $X, Y$ are fibrant.

Does f admit a section (right inverse)?. If it does not work in general, actually I saw this claimed when $C$ is the category of $\mathbb{k}$-dg algebras, where fibrations are morphisms such that $f^n$ are epimorphisms for all $n\in\mathbb{Z}$ (so, every object is fibrant) and $Y$ is (forgetting differentials) a quasi-free graded algebra, these restrictions may be helpful but I cannot conclude yet.

Thank you.

Edit: I realized this not work in a general model category, please focus only on the particular case of dg algebras. In this case, the morphism takes the form $f:A\rightarrow\Omega D$, I obtained a section (only as a morphism of graded algebras) $s:\Omega D\rightarrow A$, but I do not know why it should respect the differentials.

  • $\begingroup$ This shall not work in every model category, for example, Set has several (two) model structures where every map is a fibration. In your case I don't know how it could be solved. $\endgroup$ – elidiot Apr 23 at 16:02
  • $\begingroup$ Indeed, a counterexample in Top is a non trivial fiber bundle with CW spaces as base and total space. I just realised it. Thanks. $\endgroup$ – Victor TC Apr 23 at 17:03

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