Is the complex projective plane a compact manifold with or without boundary (closed manifold)? my question is the one in the title. (My motivation is to understand in which way Freedman's classification of compact simply-connected 4-manifolds implies the Poincare conjecture for 4-manifolds, as the complex projective plane should be a counter example, i.e. compact, simply-connected ,(closed or not? that is the question), non-homeomorphic to the 4-sphere.)
Thank you for help
I wish a nice day
 A: $\mathbb{CP}^2$ is a compact manifold without boundary (equivalently, a closed manifold). But the (topological) $4$-dimensional Poincare conjecture does not state that every closed simply connected $4$-manifold without boundary is homeomorphic to the $4$-sphere (because, as you observe, this statement is clearly false). It states that every closed $4$-manifold which is homotopy equivalent to the $4$-sphere is homeomorphic to the $4$-sphere. 
Freedman's classification proves this because a closed $4$-manifold which is homotopy equivalent to the $4$-sphere must both be simply connected and have trivial intersection form (this is necessary and sufficient), and as the trivial intersection form is even there is a unique corresponding homeomorphism class of $4$-manifolds which must be the $4$-sphere. (Freedman's classification also implies that the $4$-sphere has vanishing Kirby-Siebenmann invariant, but of course we already knew that because we already knew that the $4$-sphere admits a smooth structure.) 
It's a nice exercise to show that a closed $n$-manifold, $n \ge 2$, is homotopy equivalent to the $n$-sphere iff it is simply connected and has the same integral homology as the $n$-sphere. When $n = 3$ the second condition is redundant, but in general it's not. 
