Suppose that I have a map of simplicial spaces,

$ f: X_* \to Y_*$,

and that I know that the map on zero spaces $f_0: X_0 \to Y_0$ is n-connected. Can I conclude anything about the connectivity of the map of geometric realizations?

$ |f|: |X| \to |Y|$

Are there any reasonable conditions I can place on the simplicial spaces X and Y that would allow me to conclude something along these lines? I'm especially interested in knowing when the map is 0-connected (i.e. a surjection on $\pi_0$).


2 Answers 2


Yes it is a surjection on $\pi_0$, because each component of $|Y|$ has at least one component of $Y_0$.

Beyond that there are no restrictions. For instance, you can get any homotopy type for $|X|$ and $|Y|$ and any homotopy type for the map between them with $X_0$ and $Y_0$ just one point, as long as you ask that $\pi_0(|X|)$ and $\pi_0(|Y|)$ are trivial.

(I'm taking "simplicial space" to mean a simplicial object in the category of topological spaces, say the compact Hausdorff ones.)

  • $\begingroup$ Okay, that's nice. What if you know some additional connectivity for the maps $f_i: X_i \to Y_i$? Would that allow you to conclude more connectivity for the map of geometric realizations? (also, yes, by "simplicial space" I meant simplical object in (nice) topological spaces). $\endgroup$ Dec 4, 2009 at 20:15
  • $\begingroup$ Perhaps a nicer way to look at these is to look at simplicial objects in the category Kan complexes, which is equivalent to looking at nice spaces for all intents and purposes. There is a connection here to quasicategories, if you're familiar with them. $\endgroup$ Dec 4, 2009 at 22:01
  • $\begingroup$ My thoughts on this second question are like Oscar's answer, except that he speaks with more authority. As I was considering this, I found it helpful to think about the simplicial space as a simplicial homotopy space. Then you could imagine a spectral sequence or something to get a clearer picture of computing its homotopy groups from the homotopy groups of its simplicial blocks. $\endgroup$ Dec 4, 2009 at 22:01
  • $\begingroup$ (fpqc's comment, written in parallel, is similar of course.) $\endgroup$ Dec 4, 2009 at 22:03
  • 2
    $\begingroup$ The easiest way to see the connectivity is on homology via the spectral sequence for a simplicial space: here the i-th homology of the space of k-simplices contributes to the (i+k)-th homology of the realisation. You then just have to check the fundamental group is as expected. $\endgroup$ Dec 4, 2009 at 22:14

If you know the map on k-simplices is (n-k)-connected, you can deduce the map on realisations is n-connected. I don't think you can do better in any sort of generality.


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