In a weak factorization system, the left class is left cancellative iff the right class is what?

Question

Let $\mathcal K$ be a locally presentable category, and let $(\mathcal L, \mathcal R)$ be a cofibrantly-generated weak factorization system thereon. We say that $\mathcal L$ is left cancellative if $gf \in \mathcal L \Rightarrow f \in \mathcal L$.

Question 1: What is a necessary and sufficient condition on $\mathcal R$ for $\mathcal L$ to be left cancellative?

The condition should somehow say that $\mathcal R$ is "determined by its objects", but I'm not sure exactly how. For instance, if $\mathcal R$ is "fibrantly-generated" by the maps in $\mathcal R$ with codomain the terminal object, then $\mathcal L$ is left cancellative. But I'm not sure how to see that this sufficient condition is necessary.

Question 2: If $\mathcal K$ is locally finitely presentable and $\mathcal L$ is cofibrantly-generated by a left-cancellative set of maps $I$ between finitely-presentable objects which is closed under composition and cobase-change, then is $\mathcal L$ also left-cancellative?

I anticipate that a good answer to Question 1 will allow one to deduce a positive answer to Question 2, since if $I$ is as in (2), then it should be easy to see that $I^\square$ satisfies the condition of (1), so that ${}^\square(I^\square)$ should also be left-cancellative by (1).

Answer 1

$\newcommand\Inj{\mathit{Inj}}$The answer to Question 1 is as anticipated: for a wfs $(\mathcal L, \mathcal R)$, we have that $\mathcal L$ is left cancellable iff $\mathcal R = ({}^\square\{X_i \to 1\}_{i \in I})^\square$ is generated as a right class by maps to the terminal object.

The backward implication is well-known and easy. Conversely, suppose that $\mathcal L$ is left-cancellable, and suppose that $A \to B$ lifts against the maps $R \to 1$ which lie in $\mathcal R$ (say that $A \to B$ is "weakly in $\mathcal L$"); we wish to show that $A \to B$ is in $\mathcal L$. Taking an $(\mathcal L, \mathcal R)$ factorization $B \to \bar B \to 1$, we have that $A \to \bar B$ is weakly in $\mathcal L$, so by left cancellation we reduce to the case where $B = \bar B$, i.e. $B \to 1$ is in $\mathcal R$. In this case, take a $(\mathcal L, \mathcal R)$-factorization $A \to X \to B$. We have $X \to 1 \in \mathcal R$, so because $A \to B$ is weakly in $\mathcal L$, we have a lift $B \to X$, so that $A \to X$ factors through $A \to B$. By left cancellation, we have that $A \to B$ is in $\mathcal L$ as desired.

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The answer to a corrected version of Question 2 is affirmative — this is Thm 3.6 in Adamek, Hebert, Souza : A Logic of Injectivity. The correct statement is that if $I \subseteq \mathcal K_\omega^{[1]}$ is a collection of morphisms between finitely-presentable objects in a locally finitely-presentable $\mathcal K$, and if $I$ is closed under composition, cobase-change, and left-cancellation, then $I = {}^\square(I-\Inj) \cap \mathcal K^{[1]}_\omega$.

This implies an affirmative answer to Question 2: we have $I \subseteq {}^\square(I^\square) \cap \mathcal K_\omega^{[1]} \subseteq {}^\square(I-\Inj) \cap \mathcal K_\omega^{[1]} = I$.

The proof of Adamek, Hebert, and Souza can be simplified using the fat small object argument: As above, let $I$ be a collection of morphisms between finitely-presentable objects in a locally finitely-presentable category $\mathcal K$ which is closed under composition, cobase-change, and left cancellation (i.e. closed under deduction in the "finitary injectivity logic"). Let $h : A \to B$ lie in ${}^\square(I-\Inj)$ where $A$, $B$ are finitely presentable. Let $A \to \bar A$ be a map in ${}^\square(I-\Inj)$ where $\bar A \in I-\Inj$; by the fat small object argument, we may assume that $A \to \bar A$ is a well-founded filtered colimit of a diagram with links that are cobase-changes of morphisms in $I$, and hence the links are in fact in $I$. Then we have a factorization $A \to B \to \bar A$ of $A \to \bar A$ because $A \to B \in {}^\square(I-Inj)$ and $\bar A \in I\-Inj$. By compactness it factors through some stage of the colimit. By left cancellation, this implies that $h \in I$ as desired.