weak equivalence of simplicial sets Given a morphism f:X --> Y in sSet, and assume that it induces isomorphisms for \pi_0,\pi_1,\pi_2, and all integral homology groups. Does it imply that f is a weak equivalence?
In Hatcher's Algebraic Topology book, he requires both X and Y simply-connected. 
Here is a possible idea of `proof': we may assume f is a fibration between fibrant objects, and let Z be the fiber of f. The long exact sequence for homotopy groups shows that Z is simply-connected. Then need to use Leray-Serre spectral sequence to see all integral homology of Z vanishes. But since Y may not be simply-connected, it is hard to check the condition of the spectral sequence to hold, and I am not good at the twisted coefficients. Just wondering if there is any counterexample for this question, and hopefully some references as well. Thank you.
 A: The answer is no, and there are plenty of counterexamples. Note that simplicial sets are not relevent here; one can cook up examples with spaces and take their total singular complexes.
For example, there are high dimensional knots $K: S^n \to S^{n+2}$ (i.e., smooth embeddings with $n > 1$) such that the complement $X = S^{n+2} - K(S^n)$ has $\pi_1(X) = \Bbb Z$. A generator is represented by a map $X \to S^1$ which is a both a $\pi_1$- and a homology isomorphism.   This will give examples with the exception of your condition on $\pi_2$.
To get the $\pi_2$ condition on the above consider the subclass of those knots such that $n = 2k+1$ is odd and $\pi_j(X) \cong \pi_j(S^1)$ for $j\le k$ and $\pi_{k+1}(X) \ne 0$. These are called "simple knots." There is a complete classification of these in terms of a certain bilinear form (the Blanchfield pairing).  The classification was announced by Kearton in the paper 
Classification of simple knots by Blanchfield duality. Bull. Amer. Math. Soc. 79 (1973), 952–955
