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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.

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    $\begingroup$ It seems that this question (and questions of this kind) can be answered generally using usual topological spaces. Using words like simplicial sets, morphisms and fibrant objects may be overkill! $\endgroup$
    – Somnath Basu
    Commented Apr 28, 2011 at 4:04
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    $\begingroup$ In my answer to this question: mathoverflow.net/questions/53399/…; I gave an example of a map which is a homology equivalence and an isomorphism on $\pi_i$ for $i < n$ (for a fixed $n$ that can be arbitrary large). Moreover, all homotopy groups of these spaces are abstractly isomorphic. You can see what goes wrong in the Leray-Serre spectral sequence. Talking about fibrant simplicial sets in this situation is a distraction. $\endgroup$
    – Johannes Ebert
    Commented Apr 28, 2011 at 8:21
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    $\begingroup$ A counterexample is given in Example 4.35 in the textbook that's mentioned in the question. It's a pretty simple construction: Start with $S^1\vee S^n$, $n >1$, and attach an $(n+1)$-cell by a map $S^n \to S^1 \vee S^n$ representing the element $2t-1$ in $\pi_n(S^1 \vee S^n) = {\Bbb Z}[t,t^{-1}]$. Then the inclusion of $S^1$ into the resulting space is an isomorphism on all homology groups and on $\pi_i$ for $i < n$ but not on $\pi_n$. $\endgroup$
    – Allen Hatcher
    Commented Apr 29, 2011 at 16:14

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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

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