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.

  • $\begingroup$ It would be good if you gave a precise reference for the quote. (It's Proposition 3.3 of Chapter II.) Since the context is liftings, I would guess surjectivity refers to componentwise surjectivity. $\endgroup$ – Zhen Lin Jun 14 '12 at 12:10
  • $\begingroup$ Thanks, I just edited it. I didn't mentioned it since I wasn't particularly interested in this precise example, but I am rather interested in a more general statement about the relation cofibration =how= surjectivity. $\endgroup$ – Bogdan Jun 15 '12 at 11:23
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    $\begingroup$ It probably means surjective on all simplices ("each level"). In general, given presheaves on a category, set-theoretic notions are translated to notions on the presheaf category by working levelwise. $\endgroup$ – Dylan Wilson Oct 17 '12 at 23:08
  • $\begingroup$ Maybe. I think I have to try and prove it, then I'll be happy :) $\endgroup$ – Bogdan Oct 17 '12 at 23:27

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!).

  • $\begingroup$ That's funny that the map $\emptyset \to \Delta^1$ is a fibration. Is this the only counter-example ? I should try to work it out myself, and try to prove that if $X \to Y$ is a Kan fibration with $X \neq \emptyset$ then it is level-wise surjective. $\endgroup$ – Bogdan Nov 29 '12 at 0:27
  • $\begingroup$ As hinted by Aaron, you can also get counterexamples if $Y$ is not connected, e.g. the coproduct insertion $\Delta^0 \to \Delta^0 \amalg \Delta^0$. $\endgroup$ – Zhen Lin Nov 29 '12 at 7:45

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.

  • $\begingroup$ Thanks ! I don't know if $\Lambda_0^0 = \emptyset$ is allowed, Hirtschorn doesn't seem to allow it, neither does the ncatlab. It's weird, because maybe $\emptyset \to X$ are then the only Kan fibrations which are not levelwise surjective... $\endgroup$ – Bogdan Nov 29 '12 at 0:31
  • $\begingroup$ $\Lambda^0_0$ is not defined as a simplicial set. $\endgroup$ – Ma Ming Feb 26 '14 at 1:21
  • $\begingroup$ Well, I'd argue that it's certainly defined, exactly as how all the other horns are -- it's just that it's empty! $\endgroup$ – Aaron Mazel-Gee Feb 26 '14 at 17:36
  • $\begingroup$ @AaronMazel-Gee I could argue that $\Lambda^0_0$ should be smaller than empty. See the question I asked mathoverflow.net/questions/156902/… $\endgroup$ – Ma Ming Feb 26 '14 at 19:42
  • $\begingroup$ Perhaps, if you embed ssets into augmented ssets. Anyways, no matter how you slice it I feel like this is mostly a matter of convention -- although I do like your observation in the comments on the question to which you linked. $\endgroup$ – Aaron Mazel-Gee Feb 27 '14 at 2:48

Let me just record here for clarity the actual conclusion from Aaron's answer.

If $f: X \to Y$ is a Kan fibration and $\sigma \in Y_n$ is a simplex in $Y$, then $\sigma$ is in the image of $f$ if and only if the path component of $\sigma$ is in the image of $f$.

Of course, there are plenty of maps satisfying the surjectivity condition which are not Kan fibrations.


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