Is $\mathbb{CP}^4 \vee S^5$ homotopy equivalent to any compact manifold without boundary?

Question

I don't really know the answer to this question, any help is greatly appreciated:

Is $\mathbb{CP}^4 \vee S^5$ homotopically equivalent to any oriented compact smooth manifold without boundary?

Answer 1

Let call $X = \mathbb{CP}^{4} \vee \mathbb{S}^{5}$. Suppose on a contrary that $X \simeq M$, where $M$ is smooth compact orientable manifold without boundary. Then $dim(M)$ must be $8$. Otherwise if $dim(M) \leq 7$, then $H_{8}(X;\mathbb{Z})$ = $\mathbb{Z}$ but $H_{8}(M;\mathbb{Z})$ = $0$ by Poincare duality. On the other hand if $dim(M) = m$ where $m \geq 9$, then $H_{m}(M;\mathbb{Z})$ = $\mathbb{Z}$ by Poincare duality but $H_{m}(X;\mathbb{Z})$ = $0$. So $dim(M)$ must be $8$. Then again by Poincare duality $H_{3}(X;\mathbb{Z})$ = $H_{3}(M;\mathbb{Z})$ = $H^{5}(M;\mathbb{Z})$ = $H^{5}(X;\mathbb{Z})$, but $H_{3}(X;\mathbb{Z})$ = $0$ but $H^{5}(X;\mathbb{Z}) \neq 0$, which is a contradiction.