# 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.

• 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! Apr 28 '11 at 4:04
• 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. Apr 28 '11 at 8:21
• 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 $\pi_n$. Apr 29 '11 at 16:14

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