Understanding the proof of Proposition 1.2.13.8 in Lurie's Higher Topos Theory 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?
 A: 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).
