In the snippet below (taken from Hovey's MC book) why is $$X\mapsto QX$$ but then the direction is reversed:$$QX\to X$$?
Also, I would like to understand in that paragraph how $\alpha$ and $\beta$ are used to get these
replacement functors: are they both used for cofibrant replacement functor, say ?
The functorial factorization axiom asserts that every map factors by a cofibration followed by an acyclic (trivial) fibration (and you can also scoot the "trivial" part over to the cofibration factor). Given the map $\varnothing \to X$, which always exists as $\varnothing$ is initial, we apply this axiom to get a factorization $$ \varnothing \to QX \to X $$ where the first is a cofibration and the second a trivial fibration. This is your map $QX \to X$. The process of doing this, however, is a functor that acts as $X \mapsto QX$. That this is a functor is the "functorial" in functorial factorization. This does not mean there is a map $X \to QX$. In other words, cofibrant replacement gives a functor $Q$ with $Q(X)=QX$. He simply wrote this as $X \mapsto QX$ (and not $X \to QX$, which would mean something different and incorrect).