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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 $(E_{01}\wr L, M_{01}\wr R)$?

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    $\begingroup$ What is the meaning of the tilte symbol. Is it the set of compositions ? Is $E_{01}$ the intersection of $E_0$ and $E_1$ ? Actually it is equal to $E_1$. $\endgroup$
    – Philippe Gaucher
    Commented Apr 21, 2016 at 15:27
  • $\begingroup$ In what application does this arise? $\endgroup$
    – Paul Taylor
    Commented Apr 21, 2016 at 16:12
  • $\begingroup$ @PhilippeGaucher $E_{01}$ is not the intersection $E_0\cap E_1$; it is only a shorthand to denote the ternary factorization $\Delta[1]\to FS({\cal C})$ with a single symbol. $\endgroup$
    – fosco
    Commented Apr 21, 2016 at 21:33

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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.

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  • $\begingroup$ I'm not able to see why you define $L', R'$. I also actually realized that the generality I need is $E_1\subset L\subset E_0$, I think that in this case there are simpler proofs. Anyways, your general argument is neat and elegant. Thanks! If you spend two more words on the definition of $L', R'$ I'll be happy to accept the answer. $\endgroup$
    – fosco
    Commented Apr 21, 2016 at 21:37
  • $\begingroup$ (a rapid check after: I'm convinced by your argument; the same proof of yours seems valid in case $R\subset M_1$ ,using $(L,R)$ instead of $(L', R')$, simply because in that case $r,l\in E_0\cap M_1$) $\endgroup$
    – fosco
    Commented Apr 21, 2016 at 21:53
  • $\begingroup$ Ah, now I see the edit; nice. I think I'll try to do the exercise, do you have an answer for it? $\endgroup$
    – fosco
    Commented Apr 22, 2016 at 12:11
  • $\begingroup$ I am a child indeed (-: $\endgroup$
    – fosco
    Commented Apr 22, 2016 at 12:28

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