I'm new to model category theory, and I find myself confused about the different meanings of cofibrant replacement in literature. The usual definition is that we assign to every object $X$ in our model category $\cal M$ a fixed cofibrant object $QX$ with an acyclic fibration $q_X:QX\to X$. This gives rise to a functor $Q:\mathcal M\to h\mathcal M_c$, the category of right homotopy classes between cofibrant objects. Then $R$, the fibrant replacement, induces a functor $R:h\mathcal M_c\to h\mathcal M_{cf}$. We then define $\text{Ho}\cal M$ to have as morphisms $X\to Y$ the set of homotopy classes $\pi(RQX,RQY)$. One can show that this is a localization of $\cal M$ at the weak equivalences $\cal W$. Sometimes, however, it appears that an author does not require the map $QX\to X$ to be a fibration, but only a weak equivalence, especially when $Q$ is already a functor.
So assume we instead have functors $Q$ and $R$ on $\cal M$ with natural transformations $q:Q\Rightarrow\mathbf1_X,\ r:\mathbf1_X\Rightarrow R$ such that $QX$ is cofibrant, $RX$ is fibrant, and $q_X,r_X$ are weak equivalences for any $X$. This gives a functor $RQ:\mathcal M\to\mathcal M_{cf}$. If we define $\gamma$ to be the identity on objects, and to send a map $f$ to the class $[RQf]$, does this produce a localization of $\cal M$ at $\cal W$?
If $w\in\cal W$, then $RQw$ is a weak equivalence between bifibrant objects, thus a homotopy equivalence, so $[RQw]$ is invertible.
Now if $X$ and $Y$ are bifibrant, then $q_X$ and $r_{QX}$ are homotopy equivalences, and we get a surjection
$\mathcal M(X,Y)\to\pi(X,Y)\to\pi(RQX,RQY)$, sending $f$ to $[RQf]=[r_{QY}][q_Y]^{-1}[f][q_X][r_{QX}]^{-1}$.
For general $X,Y$, we have a bijection $\pi(RQX,RQY)\to \pi(RQRQX,RQRQY)$ sending $[f]$ to $\gamma(r_{QY})\gamma(q_Y)^{-1}[f]\gamma(q_X)\gamma(r_{QX})^{-1}$, and this is then the image of some map $f'\in\mathcal M(RQX,RQY)$ by the surjection mentioned above. Hence $[f]$ can be written as
$\gamma(q_Y)\gamma(r_{QY})^{-1}\gamma(f')\gamma(r_{QX})\gamma(q_X)^{-1}$.
This is actually the same proof as the one in literature, where it shows that every map in $\text{Ho}\cal M$ can be written as composite of images of maps in $\cal M$ and inverses of images of weak equivalences in $\cal M$, and it does not seem to require that $q$ and $r$ are (co)fibrations .
I would be pleased if someone could clarify if this works this way. Thanks.
I doubt that claim, given an object $X\in\mathcal{M}$, your functor $Q$ yields a factorization of the initial map $$ \emptyset\stackrel{c}{\to}QX\stackrel{q_X}{\to}X $$ Where $c$ is a cofibration, and $q_x$ a weak equivalence. Applying $R$ to $QX$ yields a factorization of the terminal map $$ \emptyset\stackrel{c}{\to}QX\stackrel{r_{QX}}{\to}RQX\stackrel{f}{\to}* $$ where $f$ is a fibration and $r_{QX}$ is a weak equivalence. But then $RQX$ isn't fibrant-cofibrant, since $r_{QX}\circ c$ isn't necessarily a cofibration.