DStevenson Limits and colimits in the quasicategory of spaces

$\mathcal{S}$ is complete and cocomplete.

Lemma Let $X\to B$ be a left fibration. Then $X\star 1\to B\star 1$ is a left fibration.

Proof: Observe that the natural functor $i_0^*\colon \mathbf{S}/I \to \mathbf{S}$ has a right adjoint $(i_0)_*\colon \mathbf{S}\to \mathbf{S}/I$. This right adjoint is the functor which sends $X\mapsto X\star 1$.

Let us say that a map $X\to Y$ in $\mathbf{S}/I$ is a left fibration if the underlying map in $\mathbf{S}$ is a left fibration. Likewise we will say that a map $A\to B$ in $\mathbf{S}/I$ is left anodyne if the underlying map in $\mathbf{S}$ is left anodyne. Observe that a map in $\mathbf{S}/I$ is a left fibration iff it has the LLP with respect to all left anodyne maps in $\mathbf{S}/I$. Therefore our task is to prove that $(i_0)_*$ sends left fibrations to left fibrations in $\mathbf{S}/I$, or equivalently that $i_0^*$ sends left anodyne maps in $\mathbf{S}/I$ to left anodyne maps. Hence the result follows from the following result of Joyal.

Lemma (Joyal) The functor

preserves left anodyne maps.

Here is another way to think about the first lemma above. We want to prove that the map $X\star 1\to B\star 1$ has the RLP against all horn inclusions $\Lambda^k[n]\subset \Delta[n]$ where $0\leq k\lt n$ and $n\geq 1$. Suppose given a commutative diagram

There are two things that can happen: either $f(n)\in X$, or $f(n)$ is the cone point of $X\star 1$. In the first case, the image of $f$ lies entirely inside $X\subset X\star 1$, and the desired lift can be constructed since $X\to B$ is a left fibration. In the second case, there is a unique way to extend the map $\partial_n \Delta[n]\to X$ to a map $\partial_n \Delta[n]\star 1\to X\star 1$, i.e.\ a map $\Delta[n]\to X\star 1$. This map is a diagonal filler for the diagram above.