4
$\begingroup$

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.)

$\endgroup$

1 Answer 1

7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.