# Is this kind of simplicial complex necessarily homotopy equivalent to a wedge of spheres?

Suppose $d \ge 2$ and $S$ is a finite simplicial complex of dimension $2d$, such that

(1) $S$ is simply connected, i.e. $\pi_1 ( S) = 0$, and

(2) all the homology of $S$ is in middle degree, i.e. $\widetilde{H}_i ( S, \mathbb{Z}) = 0,$ unless $i = d$, and

(3) homology $\widetilde{H}_d(S, \mathbb{Z})$ is torsion-free.

Does it necessarily follow that $S$ is homotopy equivalent to a wedge of spheres of dimension $d$?

If $S$ can be shown to be homotopy equivalent to a $d$-dimensional cell complex, for example, this would follow by standard results.

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Your space is what is called a Moore space (a space with a unique nontrivial homology group), and these are determined up to homotopy by the nontrivial homology group $G$ and the nontrivial dimension (see Hatcher Example 4.34). It follows that your space is homotopy equivalent to a wedge of spheres whenever its nontrivial homology group is a direct sum of $\mathbb Z$'s.
And since $H_d$ is torsion-free and finitely generated, it is indeed a direct sum of $\mathbb{Z}$'s. –  Dan Ramras Jun 30 '12 at 17:57
I will also assume that $S$ is connected, as is usually implied by simply connected, though your definition does not explicitly state this. Since $S$ is a finite simplicial complex, $H_d(S, \mathbb Z) \cong \bigoplus_{i=1}^k \mathbb Z$. By the Hurewicz theorem, the Hurewicz map $\pi_d(S) \to H_d(S)$ is an isomorphism, so we can pick maps $p_i: S^d \to S$ such that $P = \bigvee_i p_i : H_d(\bigvee_i S^d) \to H_d(S)$ is an isomorphism, of course, we also see that $P : H_k(\bigvee_i S^d) \to H_k(S)$ is an isomorphism for $k\neq d$. Since both spaces are simply connected, this implies by the Hurewicz theorem again that $P$ is a weak equivalence, and by the Whitehead Theorem, $P$ is a homotopy equivalence.