TITLE: 4-dimensional cohomology $\mathbb{CP}^2$'s
QUESTION [10 upvotes]: Let $M$ be a closed, smooth $4$-manifold with integral cohomology ring isomorphic to that of $\mathbb{CP}^2$, is it diffeomorphic to it?

REPLY [5 votes]: This was originally going to be a comment to Marco Gallo's answer, but got too long. I figured since this provides an alternative solution to the problem, I'll post it as an answer.
An alternative answer to this question are the fake projective planes, these are algebraic surfaces which have the same cohomology as $\mathbb{CP}^2$, but nontrivial fundamental groups. They have universal cover the complex unit ball which topologically is just $\mathbb{R}^4$.
I offer this additional answer since they provide examples very different to Marco Galla's answer and answer a question of Anubhav Mukherjee's in the comment. Either they are irreducible or have an irreducible summand that is a cohomology $\mathbb{CP}^2$.
Suppose $X$ is one of the fake projective planes. If $X$ is irreducible, we're done. If $X$ is reducible, then it admits a connect sum decomposition $X = Y \# \Sigma$ where $Y$ is a cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ that is not a homotopy $S^4$. Again if $Y$ is irreducible and not homeomorphic to $\mathbb{CP}^2$, we're done. If $Y$ is reducible, repeat this process again until you get either an irreducible cohomology $\mathbb{CP}^2$. This process must terminate by compactness.
This gives a connect sum decomposition which we'll write as $X = Y \# \Sigma$ (slightly abusing notation here) where $Y$ is an irreducible cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ which is the connect sum of all of the homology $S^4$'s we found. If $Y$ is not $\mathbb{CP}^2$, we're done.
Now we see that $Y$ can not be $\mathbb{CP}^2$. Suppose $Y = \mathbb{CP}^2$ and consider the universal cover $f: \tilde{\Sigma} \rightarrow \Sigma$. Take a point $p \in \Sigma$ and it's inverse image $f^{-1}(p)$. We can now construct a cover $f^*: Y \# |f^{-1}(p)|\mathbb{CP}^2 \rightarrow \Sigma \# \mathbb{CP}^2 = X$ by connect summing a $\mathbb{CP}^2$ at each point in $f^{-1}(p)$. This covering space is simply connected since it's summands are simply connected and so it is the universal cover. This universal covering space has 2nd homology coming from the $\mathbb{CP}^2$ summands. This contradicts $X$ having universal cover $\mathbb{R}^4$. Thus $Y$ is not $\mathbb{CP}^2$ and is an irreducible cohomology $\mathbb{CP}^2$.