Complex projective manifolds are homeomorphic if homotopy equivalent If two complex projective manifolds are homotopy equivalent are they homeomorphic?
 A: EDIT: Oops, I just remembered that you're asking the manifolds to be projective, which these aren't. Still, it's an example for just complex manifolds.
The Calabi-Eckman manifold (https://en.wikipedia.org/wiki/Calabi%E2%80%93Eckmann_manifold) is the quotient of $\mathbb{C}^m \setminus 0 \times \mathbb{C}^n \setminus 0$ by the holomorphic $\mathbb{C}$-action $t(x,y) = (e^t x, e^{\alpha t}y)$ for some fixed non-real $\alpha$. This quotient is a complex manifold diffeomorphic to $S^{2m-1} \times S^{2n-1}$. It is clear that the usual Lens space action on each of the factors commutes with this $\mathbb{C}$-action, and so we obtain a complex structure on products of Lens spaces. As mentioned in the comments, there are examples of homotopy-equivalent non-diffeomorphic Lens spaces, so this should furnish an example. (I believe Lens spaces are not so pathological that they could be non-diffeomorphic but become diffeomorphic after taking a product with e.g. $S^1$.)
A: Edit: I misread the question. The statement below explains only, that if a homotopy complex projective space other than $\mathbb{CP}^3$ supports a complex projective structure, then the answer would be no. As far as I know, it is not known if such spaces
support even a symplectic structure.
Let us call a manifold which is homotopy equivalent to a complex projective space a homotopy complex projective space (HCP). In dimension 6 there are $\mathbb Z$ many manifolds (up to diffeomorphism) with homotopy type of $\mathbb{CP}^3$. They are distinguished by their first Pontryagin class. In dimension $6$ we have that (under certain conditions, which are fullfilled for HCPs) if a topological manifold admits a smooth structure, then this structure is unique. Hence if two HCPs would be homeomorphic, they would be diffeomorphic, hence they would have the same first Pontryagin class. But as I mentioned above there are $\mathbb Z$ many HCPs with pairwise different first Pontryagin classes.
A: For curves this follows from the classification of (2-dimensional topological) surfaces, and for simply-connected surfaces this follows from Freedman's theorem.
My former colleagues Anatoly Libgober and John Wood found examples of pairs of 3-folds which are complete intersections and are homotopy equivalent but not diffeomorphic, in fact have distinct Pontryagin classes. See Example 9.2. Since in this case $H^4(M;\mathbb{Z})\cong \mathbb{Z}$, this implies that the manifolds are not homeomorphic by the topological invariance of rational Pontryagin classes (see Ben Wieland's comment).
For the higher dimensional case see:
Fang, Fuquan, Topology of complete intersections, Comment. Math. Helv. 72, No. 3, 466-480 (1997). ZBL0896.14028.
