Kan fibrations and surjectivity

Question

I have a basic question on the usual model structure on simplicial sets.

What is the relation between being a Kan (trivial maybe ?) fibration and surjectivity ?

Surjectivity here means either surjectivity on the components, or surjectivity at each level of the simplicial set, or other interesting notions.

In Simplicial homotopy theory of Goerss and Jardine, they see at a moment, "since trivial fibrations are surjective, the result follows" (Proposition 3.3 of Chapter II). Is this surjectivity on the components ?

Also, if you have a reference to point me too that would be great too, I haven't found much neither in Simplicial homotopy theory nor in others similar books.

Answer 1

First, note that $\emptyset \hookrightarrow \Delta^n$ is a cofibration (because any monomorphism is a cofibration in $\textbf{sSet}$), so the right lifting property of an acyclic fibration $p : X \to Y$ implies $p_n : X_n \to Y_n$ must be a surjection.

In general, if you know a Kan fibration $p : X \to Y$ restricts to a surjection $p_0 : X_0 \to Y_0$, then you can use the fact that the inclusion $\Delta^0 \hookrightarrow \Delta^n$ is an acyclic cofibration to lift any $n$-simplex in $Y$ through $p$ to an $n$-simplex in $X$. However, there are Kan fibrations that are not surjective: if we follow Hirschhorn [Model categories and their localizations, Dfn 7.10.8], the inclusion $\emptyset \hookrightarrow \Delta^1$ is a Kan fibration (for trivial reasons) but not surjective (in any sense!).

Answer 2

The geometric realization of a Kan fibration is a Serre fibration, and these are certainly surjective (unless we're in the trivial case that the total space is empty, or more generally the preimage of some path component of the target is empty).

But it shouldn't be hard to see directly that a Kan fibration $f:X \rightarrow Y$ must be levelwise surjective, either, under the same assumptions. So, assume $Y$ is connected. Then $X$ must be nonempty (say it contains the vertex $x$), using the Kan condition for the horn $\Lambda^0_0=\emptyset$ of $\Delta^0$. Then, you can see that $f:0:X_0\rightarrow Y_0$ must be surjective by starting from $f(x)$ and moving outwards using the horns $\Lambda^1_i=\mbox{pt}$ of $\Delta^1$. From here, you should be able to induct on dimension to show that the Kan condition always holds.