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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))

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))

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 version here (lemma 26, case (3))

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))

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...

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 version here (lemma 26, case (3))