Proof all anodyne extensions come from saturating $\Lambda_I^0({}, M)$

Question

In Prop 3.1.2 in Cisinski's Higher Categories and Homotopical Algebra, he claims $l(r(\Lambda_I(\{\}, M)) = l(r(\Lambda_I^0(\{\}, M))$, where $M$ is the cellular model and (defined in 2.4.14),

$\Lambda_I^0(\{\}, M)=\{I\times K \cup \{\epsilon\}\times L\to I\times K | K\to L \in M\}$

$\Lambda_I^n(\{\}, M)=\{I\times K \cup \partial I\times L\to I\times K | K\to L \in \Lambda_I^{n-1}(\{\}, M)\}$

$\Lambda_I(\{\},M) = \bigcup \Lambda_I^n(\{\},M)$

So clearly the thing to prove is that $\Lambda_I(\{\},M)\subset l(r(\Lambda_I^0(\{\},M))$. Cisinski writes

The explicit description of $\Lambda_I(\{\}, M)$ shows that it is sufficient to prove that, for any monomorphism of simplicial sets $K \to L$ and for $\epsilon = 0, 1$, the inclusion $I\times K \cup \{\epsilon\}\times L \to I\times L$ belongs to the class $l(r(\Lambda_I^0(\{\}, M)))$.

This doesn't make sense to me? The explicit description of $\Lambda_I(\{\},M)$ doesn't mention these maps at all, but we do know that $l(r(\Lambda_I(\{\},M)))$ is an anodyne class, which means it contains all such morphisms. However, it also contains morphisms of the form $I\times K \cup \partial I\times L \to I\times L$ for $K\to L$ anodyne, so using either the explicit description I copied above or the fact that its an anodyne class, it seems like we're missing the morphisms involving $\partial I$... If I didn't know better, it would seem like he's just proving $l(r(\Lambda_I^0(\{\}, M)) = l(r(\Lambda_I^0(\{\}, M))$... What am I missing?

Answer 1

I figured it out, at least partially. The interesting thing is that, at least for this specific instance of the $\Lambda_I$ construction (specifically with $S=\{\}$), all maps look like $I×K∪{ϵ}×L→I\times L$. To see this, just notice its true for $\Lambda_I^0(\{\},M)$ and in $\Lambda_I^{n+1}(\{\},M)$ you can write

$I\times (I\times K\cup \{\epsilon\}\times L) \cup \partial I \times I \times L \to I\times I \times L$

flipping the first two coordinates and rearranging, you get

$I\times (I\times K\cup \partial I \times L)\cup \{\epsilon\} \times I \times L \to I \times I \times L$

so.... yeah I was just being a nub, but somehow I wasn't expecting the second axiom to be implied by the part of the construction which looked like the first in this case.