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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:

  1. $A$ contains all weak homotopy equivalences that are bijective on vertices,
  2. $A$ 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?

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    $\begingroup$ Reading the relevant part of HTT, this seems more like an argument that 3 implies 2 than 3 implies 1. And it seems one can prove that 2 implies 1 by reducing to 1 to the case of a trivial fibration and noting that any such map has a section to which 2 then applies and so 3 for 2 finishes things off $\endgroup$ Commented Sep 4 at 23:36
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    $\begingroup$ @danielgratzer Thanks, I somehow missed that applying (2) to the section of a trivial fibration that is bijective on vertices finishes the argument. If you want to post this as an answer, I'll gladly accept. $\endgroup$
    – Thorgott
    Commented Sep 5 at 11:26

1 Answer 1

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I agree that this proof is erroneous; it looks like this is a bit of an editing error rather than just a mathematical one. The theorem statement proves the equivalence of 8 separate conditions (labeled 1, 2, 2', 3, 3', 4, 4', 4'') and the implication given above is that condition (3) implies condition (1). Inspecting the proof, however, it appears to prove that condition (3) implies condition (2) and there is an entirely separate (though fortunately simple) argument that condition 2 implies 1.

For concreteness, these are conditions 1, 2, and 3

  1. For every weak homotopy equivalence $f : K \to K'$ such that $K_0 \to K_0'$ is a bijection, $\mathcal{C}_{/U[K']} \to \mathcal{C}_{/U[K]}$ is a categorical equivalence.

  2. For every weak homotopy equivalence $f : K \to K'$ such that $K_0 \to K_0'$ is a bijection and $f$ is a monomorphism, $\mathcal{C}_{/U[K']} \to \mathcal{C}_{/U[K]}$ is a categorical equivalence.

  3. For ever $n \ge 2$ and $0 \le i \le n$, $\mathcal{C}_{/U[\Delta^n]} \to \mathcal{C}_{/U[\Lambda_i^n]}$ is a categorical equivalence.

Conditions 2' and 3' claim that the induced map is a trivial fibration, not just a weak homotopy equivalence. Proving that (2) = (2') and (3) = (3') is straightforward. Lurie's argument makes sense for proving that (3) implies (2) because this is exactly the situation where $f$ can be assumed to be a monomorphism (something that does not make sense if we are instead proving (3) implies (1) as noted in the question).

It remains then to show that (2) implies (1). Let us therefore suppose we have $f$ a weak homotopy equivalence which is bijective on objects. We can factor $f$ into $p \circ i$ where $i : K \to K_0$ is a (Kan-Quillen) trivial cofibration and $p : K_0 \to K'$ is a trivial fibration, both bijective on objects. We next note that the canonical map $\mathcal{C}_{/U[K']} \to \mathcal{C}_{/U[K]}$ is then the composite of $\mathcal{C}_{/U[K']} \to \mathcal{C}_{/U[K_0]} \to \mathcal{C}_{/U[K]}$. Condition (2) tells us that the second portion of this composite is a weak categorical equivalence, so it suffices to prove $\mathcal{C}_{/U[K']} \to \mathcal{C}_{/U[K_0]}$ is one as well.

For this, we note that $p$ has a section $s$ as a trivial fibration between cofibrant objects and that (2) applies to $s$ to show that $\mathcal{C}_{/U[K_0]} \to \mathcal{C}_{/U[K']}$ is a weak equivalence, but this map is a retract of the morphism $\mathcal{C}_{/U[K']} \to \mathcal{C}_{/U[K_0]}$ induced by $p$ so 2-of-3 yields the conclusion.

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