Factorization system "tilted" by $(L,R)$ Suppose you have a pair of orthogonal factorization systems, $(E_0, M_0), (E_1, M_1)$ in a category $\cal C$ such that $M_0\subseteq M_1$; this entails that there is a ternary factorization
$$
X\xrightarrow{e_1} A\xrightarrow{e_0 m_1} B\xrightarrow{m_0} Y
$$
where each arrow is labeled according to the class it belongs to.
Suppose now to have another OFS $(L,R)$ on $\cal C$, and to factor the middle arrow $A\to B$ into $A\xrightarrow{l} S\xrightarrow{r} B$; this gives a factorization $X\to S\to B$.

Does this define a third factorization system $(E_{01}\wr L, M_{01}\wr R)$?

 A: Yes.
First note that Galois connections are far more common than field theorists would have you believe: any binary relation gives rise to one and the two fixed sets a closed under any appropriate algebraic structure.
Write


*

*$P\perp Q$ for "every member of $P$ is orthogonal (in the sense of factorisation systems) to every member of $Q$";

*$P;Q$ for the set of all composites of members of $P$ followed by members of $Q$;

*$L'=L\cap E_0\cap M_1$ and $R'=R\cap E_0\cap M_1$ (to avoid any ambiguity in this argument); and

*$E=E_1;L'$ and $M=R';M_0$.
Then


*

*$E_1\perp M_1\supset R'$,

*$E_1\perp M_1\supset M_0$,

*$L'\subset L\perp R\supset R'$ and

*$L'\subset E_0\perp M_0$.
Since orthogonality respects composition, it follows that $E\perp M$, whilst by construction $E;M$ is the entire hom-class of the category.
Therefore $(E,M)$ is a factorisation system.
The things that I have assumed are all proved in my book but I don't have a copy to hand to look up the references.
In the question as stated, it is given that $M_0\subset M_1$, but there is no order-relationship between these and $R$. This is why it was necessary to introduce $R'$.  The construction gives a way of handling expressions in the lattice of factorisation systems on a category, so it would be an interesting exercise in lattice theory to find out whether this is modular or even distributive.
