punctured real projective space

Question

Let $\mathbb{R}P^m$ be the real projective space and $X=\mathbb{R}P^m\setminus \{*\}$ be the punctured space by removing one point. How to get the cohomology ring of $X$ with integer coefficient? Is $X$ always homotopic to $\mathbb{R}P^{m-1}$?

Answer 1

You can describe a deformation retraction from $X$ to $\mathbb{R}P^{m-1}$ very simply. Think of $\mathbb{R}P^{m}$ as a quotient of $Y=\mathbb{R}^{m+1}\setminus\{0_{m+1}\}$, consider $\mathbb{R}P^{m-1}$ as the quotient of $Z=(\mathbb{R}^m\setminus\{0_m\})\times\{0\}\subset Y$, and let the removed point $*$ be the quotient of the vertical line $L=\{0_m\}\times\mathbb{R}\setminus\{0\}$ (here $0_m$ denotes the origin in $\mathbb{R}^m$). Then you can deformation-retract $Y\setminus L$ down to $Z$ by just a vertical linear homotopy (i.e., $((x_1,\dots,x_m,x_{m+1}),t)\mapsto (x_1,\dots,x_m,tx_{m+1})$). It is easy to see that this deformation retraction is compatible with the quotient maps, so it descends to a deformation retraction from $X$ to $\mathbb{R}P^{m-1}$.

Or, more briefly: $\mathbb{R}P^m$ can be constructed as $\mathbb{R}P^{m-1}$ with an $m$-cell attached. If you remove a point in the interior of the $m$-cell, you can deform the rest of the $m$-cell to its boundary, leaving you with just $\mathbb{R}P^{m-1}$.