LemmaLet $X\to B$ 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

$i_0^*\colon \mathbf{S}/I \to \mathbf{S}$

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.

Revised on October 9, 2013 at 02:15:43
by
Danny Stevenson