Unit of a Quillen equivalence and fibration Let $L:\mathcal{M}\leftrightarrows\mathcal{N}:R$ be a Quillen equivalence between combinatorial model categories such that all objects are fibrant. Let $X$ be a cofibrant object of $\mathcal{M}$. Then the unit of the adjunction gives rise to a weak equivalence $X\to RL(X)$.

Is there a known sufficient condition for $X\to RL(X)$ to be a trivial
fibration ?

 A: If we write down the lifting square for an arbitrary cofibration $f\colon A→B$ and the unit map $η\colon X→RLX$ (with the bottom map being $b\colon B→RLX$),
and then use the adjunction to pass to the adjoint square with maps $Lf$ and $\def\id{{\rm id}} \id_{LX}$, the resulting lifting problem has a solution if and only if the new bottom map $β\colon LB→LX$ is in the image of $L$.
Thus, we immediately obtain a necessary and sufficient condition:
for any map $b\colon B→RLX$ that fits into a commutative square with some cofibration $f\colon A→B$ and the unit map $X→RLX$,
the adjoint map $β\colon LB→LX$ must satisfy $β=L(g)$ for some morphism $g$ such that $gf=a$, where $a\colon A→X$ is the top map in the original square.
In particular, this condition must hold for any map $b\colon B→RLX$
whose source $B$ is cofibrant (since in this case we can take $A$ to be the initial object).
The requirement for the adjoint map of $b$ to be in the image of $L$
already seems like an extremely strong condition on a model category, which fails for many (if not most) known examples of model categories, the only example I can think of is when $L$ is a reflection onto a full subcategory.
However, I do not know what kind of model categories are used in the potential application, some additional details would be helpful here.
