Let $\mathcal{C}$ be an $\infty$-category (meaning quasi-category) and $U\colon N(\Delta)^\text{op}\rightarrow\mathcal{C}$ a simplicial object of $\mathcal{C}$. For a simplicial set $K$, let $U[K]$ denote the composite of $U$ and the projection $N(\Delta_{/K})^\text{op}\rightarrow N(\Delta)^\text{op}$. Furthermore, let $A$ denote the class of maps $f\colon K\rightarrow K^{\prime}$ s.t. the induced map $\mathcal{C}_{/U[K^{\prime}]}\rightarrow\mathcal{C}_{/U[K]}$ is a categorical equivalence. In *Higher Topos Theory*, Proposition 6.1.2.6., Lurie claims (among other things) that TFAE:

- $A$ contains all weak homotopy equivalences that are bijective on vertices,
- $A$ contains all horn inclusions $\Lambda_i^n\hookrightarrow\Delta^n$ for $0\le i\le n$ and $n\ge2$.

It is not hard to see that the class $A^{\prime}$ consisting of the cofibrations in $A$ is saturated. Now, suppose that 2. is satisfied. If $A^{\prime\prime}$ denotes the saturated class generated by horn inclusions in dimension $\ge2$, we have $A^{\prime\prime}\subseteq A^{\prime}\subseteq A$. Now, suppose $f\colon K\rightarrow K^{\prime}$ is a weak homotopy equivalence that is bijective on vertices. Lurie uses the small object argument to construct a commutative square $$\require{AMScd} \begin{CD} K @>{h}>> M\\ @V{f}VV @V{g}VV\\ K^{\prime} @>{h^{\prime}}>> M^{\prime}, \end{CD}$$ where $h,h^{\prime}\in A^{\prime\prime}$, $M^{\prime}$ is a Kan complex and $g$ has the RLP w.r.t. $A^{\prime\prime}$. Since $f$ and maps in $A^{\prime\prime}$ are weak homotopy equivalences that are bijective on vertices, the same holds for $g$ by $2$-out-of-$3$. Thus, the hypotheses of Lemma 6.1.2.4. are satisfied, which implies that $g$ is a trivial fibration. Now, Lurie proceeds

"It follows that $g$ has the right lifting property with respect to the cofibration $g\circ h\colon K\rightarrow M^{\prime}$,…."

This fundamentally does not make sense to me. The cofibrations in this context are simply monomorphisms, but if $g\circ h=h^{\prime}\circ f$ were a monomorphism, then $f$ *had* to be one as well, yet we are expressly trying to generalize to a case where this is not assumed.

*Question 1:* Is this proof erroneous or am I being silly?

I tried for an alternative argument, e.g. $A$ satisfies $2$-out-of-$3$, so it suffices to prove that $g\in A$, but then I had no idea why that should be the case. The map $N(\Delta_{/M^{\prime}})^\text{op}\rightarrow N(\Delta_{/M})^\text{op}$ is coinitial iff it is an isomorphism iff $g$ is an isomorphism, so the obvious route is a dead end. This raises

*Question 2:* Can the proof be salvaged? Or is the statement incorrect?