# Equivalent definition of a Kan fibration

It is known (follows for example from Proposition 4.2 of Simplicial Homotopy Theory by Goerss and Jardine) that a Kan-fibration can be defined as a map having the right lifting property with respect to all maps of the form $$(\Delta^1\times\partial\Delta^n)\coprod_{\Delta^{\{i\}}\times\partial\Delta^n}(\Delta^{\{i\}}\times\Delta^n)\hookrightarrow \Delta^1\times\Delta^n$$ for $n\geq 0,\:i=0,1$.

Moving to topological spaces, a Serre-fibration is defined to be a map having the right lifting property with respect to all maps of the form $$([0,1]\times\partial D^n)\coprod_{\{0\}\times\partial D^n}(\{0\}\times D^n)\hookrightarrow [0,1]\times D^n$$ for $n\geq 0$. This resembles the definition of Kan-fibration given above, but we need not take $i=0,1$ as in the simplicial case, because of the inherent symmetry of topological spaces. Now in topological spaces, the above maps are isomorphic (homeomorphic) to the maps
$$\{0\}\times D^n\hookrightarrow [0,1]\times D^n,$$ so a Serre-fibration can be defined as a map having the right lifting property with respect to this last class of maps. Taking into account the nonsymmetry of simplicial sets, my question is

Can we define a Kan fibration to be a map having the right lifting property with respect to all maps of the form $$\Delta^{\{i\}}\times\Delta^n\hookrightarrow \Delta^1\times\Delta^n$$ for $n\geq 0,\:i=0,1$?

• Am I right that when $i=1$, your lifting condition is trivial? So then you are asking whether a Kan fibration is characterized by having the right lifting property with respect to the maps $\Delta^n \hookrightarrow \Delta^1 \times \Delta^n$. – Theo Johnson-Freyd Jul 13 '15 at 0:34
• The question is what is the map $\Delta^n\hookrightarrow \Delta^1\times\Delta^n$? If it is the embedding "on the floor" I denote it by $\Delta^{\{0\}}\times\Delta^n\hookrightarrow \Delta^1\times\Delta^n$ and if it is the embedding "on the roof" I denote it by $\Delta^{\{1\}}\times\Delta^n\hookrightarrow \Delta^1\times\Delta^n$. – Ilan Barnea Jul 13 '15 at 0:45
• Oh, I misunderstood. My apologies. – Theo Johnson-Freyd Jul 13 '15 at 20:16

The class of morphisms having the right lifting property with respect to $\Lambda^1_k \times \Delta^n \hookrightarrow \Delta^1 \times \Delta^n$ (for all $k \in \{ 0, 1 \}$ and all $n \ge 0$) is strictly larger than the class of Kan fibrations.

Indeed, $\Lambda^1_k \times \Delta^n \hookrightarrow \Delta^1 \times \Delta^n$ is a split monomorphism, so for any $X$, the unique morphism $X \to \Delta^0$ has the right lifting property with respect to $\Lambda^1_k \times \Delta^n \hookrightarrow \Delta^1 \times \Delta^n$.