I've learned the theorem when reading a comment by Vidit Nanda to my question see here. Here is the (simplified) version of the theorem for topological spaces:

Vietoris-Begle Theorem

Let $f:X\rightarrow Y$ be a surjective closed continuous map between paracompact Hausdorff spaces. Assume that for any $y\in Y$, $f^{-1}(y)$ is weakly contractible, then the induced map $$H^{\ast}(f): H^{\ast}(Y,G)\rightarrow H^{\ast}(X,G)$$ is an isomorphism (reduced cohomology) for any abelian group $G$.

It seems that the assumption that $f$ is closed is essential.

My question is the following: Do we have a simplicial version of Vietoris-Begle Theorem ? i.e.,

simplicial Vietoris-Begle Theorem ?

Let $f:X\rightarrow Y$ be a surjective morphism of simplicial sets (Kan complexes). Assume that for any $y\in Y_{0}$, $f^{-1}(y)$ is weakly contractible simplicial sets, then the induced map $$H^{\ast}(f): H^{\ast}(Y,G)\rightarrow H^{\ast}(X,G)$$ is an isomorphism (reduced cohomology) for any abelian group $G$.

reference for the topological case: reference

Edit: Please feel free to add more conditions on $f,X ,Y$ but not the trivial ones such as $f$ is a fibration or something similar...
There is for example a simplicial (strong) version here (lemma 26, case (3))

  • $\begingroup$ If you treat your simplicial sets as categories (whose objects are simplices, and where morphisms are generated by face + degeneracy maps modulated by simplicial identities), then your simplicial map becomes a functor and your desired result is Quillen's Theorem A: en.wikipedia.org/wiki/Quillen%27s_theorems_A_and_B Liviu's answer reduces to the case of a simplicial complex while various other counterexamples show why you need homotopy fibers to be contractible (rather than straight-up fibers over vertices). $\endgroup$ – Vidit Nanda Oct 23 '15 at 14:32

(1) As stated, the answer to the question is "no".

Let $A = \Delta[1] \cup_{\partial\Delta[1]} \Delta[1]$ be the union of two copies of $\Delta[1]$ along their common boundary, let $g \colon A \to \Delta[1]$ be the identity on each copy of $\Delta[1]$, and let $f = Ex^\infty g$ map $X = Ex^\infty A$ to $Y = Ex^\infty \Delta[1]$. Here $Ex^\infty = colim_n Ex^n$ is Kan's functor, with $Ex^n$ right adjoint to $Sd^n$. Then $f$ is (split) surjective, $X$ and $Y$ are Kan complexes, and $X_0 = A_0$ maps identically to $Y_0 = \Delta[1]_0$. Hence your hypotheses are satisfied. However, $H^1(Y) = 0$ is not isomorphic to $H^1(X) \cong \mathbb{Z}$.

(2) You might strengthen the hypothesis to assume that the preimage of each simplex of $Y$ is weakly contractible, i.e., that for each (non-degenerate) simplex $y \in Y_n$ with representing map $\bar y \colon \Delta[n] \to Y$ the fibre product $\Delta[n] \times_Y X$ is weakly contractible (or has trivial cohomology). No Kan condition is then needed to inductively prove that $f^{-1}(Z) \to Z$ is a cohomology equivalence for each subcomplex $Z \subseteq Y$. It does not, however, follow that $f \colon X \to Y$ is a homotopy equivalence.

(3) If you want to conclude that $f \colon X \to Y$ is a homotopy equivalence, you can assume that each point inverse $|f|^{-1}(y)$, of the topological realization $|f| \colon |X| \to |Y|$, is contractible. Here $y$ ranges over $|Y|$. For finite simplicial sets $X$ and $Y$ this implies that $f$ is a simple-homotopy equivalence. See e.g. Proposition 2.1.8 in my book with Waldhausen and Jahren.


Here is a theorem due (I think) to E. Dror Farjoun (it is Theorem 9.A.11 in his book on homotopy localization).

Theorem: Let $f:X\to Y$ be a surjective map of connected pointed simplicial sets. Then the homotopy fiber $F_f$ of $f$ is in the smallest closed class containing $f^{-1}(\sigma)$ for every simplex $\sigma$ in $Y$.

A closed class is a class $\mathcal{C}$ of spaces which is closed under weak equivalence and (pointed) homotopy colimit. For pointed spaces use pointed homotopy colimit; for unpointed ones, restrict to homotopy colimits of diagrams whose nerve is contractible.

It follows, in particular, if every $f^{-1}(\sigma)$ is weakly contractible, then $f$ will be a weak homotopy equivalence (Corollary 9.B.3.1).


In the paper Homotopy properties of the poset of nontrivial $p$-subgroups of a group, Adv. in Math., 28(1978), D.Quillen proves a nontrivial version of Vietoris-Begle theorem.

Recall that to a poset $P$ we can associate a simplicial complex, its nerve $|P|$. Any simplicial complex is, canonically, the nerve of a poset, the poset of faces.

If $P, Q$ are posets, and $f: P\to Q$ is a monotone nondecreasing map, then its fibers are defined to be the subsets

$$ f^{-1}(\leq y):=\bigl\{p \in P;\;\;f(p)\leq y\;\bigr\},\;\;y\in Q. $$

In the paper mentioned above Quillen proves that if $f: P\to Q$ is monotone nondecreasing and the fibers are homotopically trivial (resp. acyclic) then $f$ induces a homotopy equivalence between the nerves (resp. an isomorphism between the homology of the nerves).

If $P, Q$ are the posets of faces of simplicial sets, then their nerves coincide with the simplicial sets they were obtained from. A simplicial map $f$ between the simplicial sets induces a monotone map between the corresponding posets of faces, and Quillen's theorem can be rephrased as saying that if the preimage of any face is contractible, then $f$ is a homotopy equivalence.

For more details see this nice survey of Anders Bjorner.

  • $\begingroup$ Thank you, this looks like a strong version with very specific assumptions. I'am looking for a simplicial version which is the analogue of the topological version, the one I have written in the question. I'm juste little bit surprised that there is not such analogue in the literature, as far as I know. But may be I'm wrong. $\endgroup$ – Ilias A. Oct 15 '15 at 19:49

No. Let $Y$ be $\Delta^1$ and let $X$ be the boundary of $\Delta^2$. Map $X$ to $Y$ by the simplicial map taking vertices $0,1,2$ to $0,1,1$.

In effect, by just looking at fibers over vertices you are not getting a grip on fibers over interior points of simplices.

(Edited later: This is wrong: I overlooked the requirement that $X$ and $Y$ should be Kan complexes. But the examples given by others in the comments make the same point without being wrong.)

  • $\begingroup$ in my version, X and Y are Kan complexes. Can you provide a counterexample? $\endgroup$ – Ilias A. Oct 15 '15 at 18:32
  • 2
    $\begingroup$ Let $X$ be the nerve of the groupoid with one object and two automorphisms (also known as the classifying space of $\mathbb{Z}/2$). Then $X$ is a Kan complex with just one vertex but the corresponding map $\Delta^0 \to X$ does not induce an isomorphism on $H^1$ with coefficients in $\mathbb{Z}/2$ $\endgroup$ – Yonatan Harpaz Oct 15 '15 at 20:13
  • 2
    $\begingroup$ The map $\Delta^0 \to B(\mathbb{Z}/2)$ is not surjective, which is one of the requirements in the question, @YonatanHarpaz. $\endgroup$ – Omar Antolín-Camarena Oct 15 '15 at 21:16
  • $\begingroup$ Right. So take instead the map $B(\mathbb{Z}/2) \to \Delta^0$. $\endgroup$ – Yonatan Harpaz Oct 16 '15 at 15:24
  • $\begingroup$ Sorry, then the fibers are not contractible. $\endgroup$ – Yonatan Harpaz Oct 16 '15 at 15:38

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