Is it true that if a finite CW complex $X$ is simply connected, and $\tilde{H}\_i(X, \mathbb{Q}) =0$ for $i \neq D$, then $X$ is rationally homotopy equivalent to a bouquet of $D$-dimensional spheres?


(In my setting $D \ge 3$, in case that makes any difference.)



sufficient conditions for rational homotopy equivalence
Yes. By a generalization of the Hurewicz theorem (which can be formulated more generally in terms of Serre classes), if $X$ is simply connected and has trivial rational homology below dimension $D$, then $\pi\_i(X)\otimes\mathbb{Q}=\tilde{H}\_i(X,\mathbb{Q})$ via the Hurewicz map for all $i\leq D$. In particular, for $i=D$ this implies we can find elements of $\pi\_D(X)$ whose Hurewicz images form a basis for $\tilde{H}\_D(X,\mathbb{Q})$. These elements together give a map from a wedge of $D$-spheres to $X$ which induces an isomorphism on rational homology and is hence a rational equivalence.


