TITLE: example of "really" non-existent transferred model structure
QUESTION [8 upvotes]: I am looking for an example where a transferred model structure fails to exist, even if one is willing to work with semi-model category. But let me be more precise:
Let's say I have a combinatorial model category $C$, a locally presentable category $D$ and an adjunction :
$$ L: C \rightleftarrows D : U$$
A classical (at least - mentioned on the nLab) necessary and sufficient condition (in this case) for the existence of a transferred model structure on $D$ is that one has the following two:
(A) For every object $X \in D$ such that $U(X)$ is fibrant, there exists a "path object"
$X \overset{a}{\to} P \overset{p}{\to} X \times X$ such that $U(a)$ is a weak equivalence and $U(p)$ is a fibration.
(B) There exists a "fibrant replacement" functor and natural transformation $X \overset{a_x}{\to} FX$ on $D$, such that $U(FX)$ is fibrant and $U(a_x)$ is a weak equivalence.
I know examples where condition (B) fails, but I can't find an example where (A) fails. Do you know one ?
Some details and motivations:
In practice, it appears that condition (A) is often almost free and condition (B) is the hard one. For example, if $C$ is a simplicial model category, $D$ is simplicially enriched (with cotensor) and the adjunction is a simplicial adjunction, you can take $P$ to be the cotensor $P = X^{\Delta[1]}$. The same applies with other enrichement.
Now, it also appears that condition (A) is sufficient to build a "transferred model structure" on D, at least if one is willing to work with right semi-model category and slightly generalizing what one means by transferred model structure. So failure of condition (B) isn't really a deal breaker, but just an additional hassle.
This being said, I can't find a single example where condition (A) fails.

REPLY [4 votes]: The usual example in operad theory is when $C$ is a combinatorial, monoidal model category and $D$ is the category of commutative monoids in $C$. Unless $C$ satisfies a strong condition (that in my thesis, I called the commutative monoid axiom) guaranteeing symmetric powers are homotopically well-behaved, $D$ won't even have a semi-model structure.
For example, if $C = Ch(\mathbb{F}_p)$ is chain complexes over a field $k$ of characteristic p, then it is easy to show that $D$ can't have a transferred semi-model structure. You know that, if it did, then the generating trivial cofibrations would be of the form $Sym(J)$ where $J$ is the set of generating trivial cofibrations in $C$, and $Sym$ is the free commutative monoid function (L in your notation). Recall that maps in $J$ look like $0\to D(n)$ where $D(n)$ is the chain complex with one copy of $k$ in degrees $n$ and $n-1$, and identity boundary map.
Let's be quite explicit. Take $p=2$. Then $Sym(0)=k \to Sym(D(n))$ is not a weak equivalence, because if $y\in D(n)$ is non-zero then $y^2 \in Sym(D(n))$ is a cycle of degree $2n$ which is not a boundary. This is Example 3.7 in Model Categories and Simplicial Methods.