I've learned the theorem when reading a comment by Vidit Nanda to my question [see here](https://mathoverflow.net/questions/220930/fiber-homotopy-fiber-of-spaces).
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](http://ac.els-cdn.com/0166864186900180/1-s2.0-0166864186900180-main.pdf?_tid=413cfece-735f-11e5-8bbb-00000aab0f01&acdnat=1444929045_1e6b9674d9340b54c48ceb2b5f051c8b)


**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](http://www.di.ens.fr/~colin/textes/08helly.pdf) (lemma 26, case (3))



Vietoris-Begle theorem for simplicial sets
(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.


