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In the first chapter of Lurie's HTT, Proposition 1.2.13.8 shows that the functors out of over (infinity)-categories reflect infinity-colimits.

For the life of me I cannot follow the proof.

Can anyone help?

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    $\begingroup$ What specific parts are you having trouble with? $\endgroup$ – Théo de Oliveira Santos Mar 8 at 2:32
  • $\begingroup$ First, I am confused about where the first lifting diagram comes into the picture, and I don't follow how the following sentence solves the lifting problem. $\endgroup$ – Patrick Elliott Mar 8 at 2:54
  • $\begingroup$ Next, I am not sure why, in the proof of part (1), having the lifting property "with respect to T" implies that the colimit extends to the over-category. $\endgroup$ – Patrick Elliott Mar 8 at 2:56
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    $\begingroup$ Let $\mathcal{D}=\mathcal{C}_{/q}$. You want to decide whether $\mathcal{D}_{\tilde{p}/} \to \mathcal{D}_{p/}$ is a trivial fibration. So for every inclusion $A \subseteq B$ and square diagram, you need to build a lift. The adjunction defining over and undercategories gets you to the depicted diagram. Now adjoint back over the $K$ and $K\star \Delta^0$ from the left to the right, and you get a lifting diagram involving $A\star T \subseteq B\star T$ and $\mathcal{C}_{\tilde{p}_0/} \to \mathcal{C}_{p_0/}$ as indicated. It all makes more sense if you draw pictures- even of the 1-category case. $\endgroup$ – Dylan Wilson Mar 8 at 5:36
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As a learning exercise, I will try to expand Dylan's comment into an answer. If things are still unclear, please ask in the comments. First, the $\star$ operator is defined in 1.2.8.1. The notation $K^{\triangleright}$ is defined to mean $K \star \Delta^0$ (see 1.2.8.4). The proposition you ask about features $q: T\to \mathcal{C}$. Prop 1.2.9.2 says $\tilde{p}:K^\triangleright \to \mathcal{C}_{/q}$ is adjoint to a map $K^\triangleright \star T \to \mathcal{C}$. The goal is to show that $\tilde{p}$ is a colimit of $p: K\to \mathcal{C}_{/q}$. Restating the universal property of a colimit, we must show that, for any inclusion $A\subset B$ of simplicial sets, the diagonal lift displayed in Lurie's proof (an up and right arrow below) exists:

$\begin{array}{ccc} K \star B \star T \coprod_{K \star A \star T} K\star \Delta^0 \star A\star T & \to & C \\ \downarrow & & \\ K \star \Delta^0 \star B \star T \end{array}$

Let $\pi: \mathcal{C} \to \mathcal{C}_{/q}$. The hypothesis is that the composite $\tilde{p}_0 = \pi \circ p$ is a colimit of $\pi \circ p_0$. We will use this $p_0$.

Following Dylan's hint, adjunction says the lift above exists if and only if the following diagram has a lift

$\begin{array}{ccc} A \star T & \to & C_{\tilde{p_0}/} \\ \downarrow & & \downarrow \\ B \star T & \to & C_{p_0/} \end{array}$

In this square, the map on the left is a cofibration, because $A\subset B$ and the cofibrations are the monomorphisms. The map on the right is a trivial fibration, by the assumption that $\tilde{p}_0$ is a colimit of $p_0$ (trivial fibrations are by definition maps with the right lifting property with respect to monomorphisms, and in this case, that lifting property is an exact restatement of the universal property of the colimit).

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    $\begingroup$ The penultimate sentence should maybe say “The map on the right is a trivial fibration by our assumption that $\tilde{p}_0$ is a colimit diagram in C” instead of what’s currently written, and I think you can omit the comment about categorical equivalences as far as I can tell... $\endgroup$ – Dylan Wilson Mar 9 at 1:38
  • $\begingroup$ @DylanWilson thanks, that's a good point. I was thinking about trivial fibrations in the Joyal model structure, but of course they are the same in as in the Quillen model structure, where this lifting property is even easier to see. $\endgroup$ – David White Mar 9 at 17:42

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