Compact surface homotopy equivalent to $\mathbb{CP}^2 \setminus \{p\}$ Let $p$ be a point $\in \mathbb{CP}^2$. Is there a compact surface which is homotopy equivalent to $\mathbb{CP}^2 \setminus \{p\}$ ?
I know the homology of $\mathbb{CP}^2$, but I'm not sure about the homology of $\mathbb{CP}^2 \setminus \{p\}$. I suppose there is a deformation retraction of $\mathbb{CP}^2 \setminus \{p\}$ to $\mathbb{CP}^1$ and then the homology would be clear.
 A: The space $\mathbb{CP}^2 \setminus \{p\}$ is isomorphic as an algebraic variety, and thus a fortiori as a topological space, to the total space of the complex line bundle $\mathcal O_{\mathbb P^1_\mathbb C}(1)$ defined on $\mathbb P^1_\mathbb C$.
Hence it is, like all vector bundles, homotopically equivalent to its base space $\mathbb P^1_\mathbb C$.
A: There is a CW structure on $\mathbb{CP}^2$ with one cell in each dimension $0,2,4$, the $2$-skeleton being $\mathbb{CP}^1 = S^2$. Thus $\mathbb{CP}^2 = S^2 \cup_f e^4$ where $e^4 \cong D^4$ is a $4$-cell, and $f : \partial e^4 \to S^2$ is some map (the precise expression of $f$ is not relevant right now).
It's possible to arrange things so that $p$ is in the interior of the $4$-cell $e^4$ ($\mathbb{CP}^2$ is a $4$-manifold, we can just move the point a little and get the same thing; more precisely there's an isotopy to a configuration where $p$ is in the interior of the cell). So the space $\mathbb{CP}^2 \setminus p$ becomes homotopy equivalent to the mapping cylinder of $f$, so it's homotopy equivalent to $S^2$, which is of course a compact surface.
