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Suppose that $G:M\leftrightarrow N: U$ is a Quillen equivalence between two model categories. Suppose that $a\in M$ is a fibrant object, is it true that there exists always a fibrant object $b\in N$ and weak equivalence $a\rightarrow U(b)$ in $M$ ?

We are not assuming cofibrancy property on $a$ and $b$.

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2 Answers 2

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Here is a counter-example to the dual assertion (so that you can get a counter-example to your original question by taking the opposite model categories). Consider the category ${\rm Set_\Delta}$ of simplicial sets with the Kan-Quillen model structure. Since ${\rm Set}_{\Delta}$ is right proper and the inclusion $i_0:\Delta^{\{0\}} \subseteq \Delta^1$ of the $0$-vertex is a weak equivalence the adjunction $$ (i_0)_!:{\rm Set}_{\Delta} = ({\rm Set}_{\Delta})_{/\Delta^{\{0\}}} \rightleftarrows ({\rm Set}_{\Delta})_{/\Delta^1}: i^*_0 $$ is a Quillen equivalence, where the right hand side is endowed with the model structure induced on the slice category by the model structure of ${\rm Set}_{\Delta}$. Here the left adjoint $(i_0)_!$ sends a simplicial set $X \in {\rm Set}_{\Delta}$ to $X \to \Delta^{\{0\}} \subseteq \Delta^1$ and the right adjoint $i_0^*$ takes a simplicial set $Y \to \Delta^1$ over $\Delta^1$ to $Y \times_{\Delta^1} \Delta^{\{0\}}$. Now let $Y$ be any non-empty simplicial set and consider the object of $({\rm Set}_{\Delta})_{/\Delta^1}$ given by $Y \to \Delta^{\{1\}} \subseteq \Delta^1$. Then $Y \to \Delta^1$ admits no maps from objects of the form $(i_0)_!(X)$ unless $X=\emptyset$, and hence no weak equivalences from objects of the form $(i_0)_!(X)$.

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No, the most we can say is that there exists a zig-zag.

$a\leftarrow Qa\rightarrow U(b)$

where the first arrow is the the component of the natural weak equivalence $Q\to Id$ with $Q$ the cofibrant replacement functor.

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  • $\begingroup$ I do not understand your answer! if the answer is no, you need to prove the following: for a fixed fibrant object $a\in M$ there is no fibrant object $b\in N$ such that $a\rightarrow U(b)$ is an equivalence. $\endgroup$
    – tictac
    Jul 25, 2018 at 18:20
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    $\begingroup$ You asked if it is true in general. It is not. I don't have a specific counterexample at the ready. $\endgroup$ Jul 25, 2018 at 18:30

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