Show the projective space $RP^2$ minus a point is homotopy equivalent to the unit circle $S^1$.
I can image how to do by the graph,as think of $RP^2$ as the unit disk with opposite boundary points identified, and remove one point $x$ from this disk, then remaining points can be gradually pushed out to the boundary, and the boundary is a strong deformation retract of $RP^2-\{x\}$, also the the boundary is homeomorphic to $S^1$.
However, I would like to do this by formally defining a homotopy that show $RP^2-\{x\}$ are homotopy equivalent to $S^1$. How would the function looks like?
Any help is appreciated, thanks a lot!
Best Answer
The easiest way to define a function precisely is via coordinates, and $RP^2$ has a very convenient system of coordinates, called homogeneous. Identifying a point $\ell$ in $RP^2$ with a line $\{v:v\in \ell\}$ in $\mathbb R^3$, take some $v=(v_1,v_2,v_3)\in \ell$. Then $[v_1:v_2:v_3]$ are coordinates of $\ell$ as a point in the projective plane, unique up to multiplication by a nonzero constant.
The unit circle in $RP^2$ is naturally identified with the lines in $\mathbb{R}^3$ lying in the $xy$-plane, i.e. those $[v_1:v_2:v_3]$ with $v_3=0$. So we want a continuous map from $RP^2\setminus \{*\}$ to points with zero third coordinate. If we send $[v_1:v_2:v_3]$ to $[v_1:v_2:0]$, we certainly have a continuous map, with one problem: if $v_1=v_2=0,$ then we've sent our point to $[0:0:0]$, which are not the coordinates of any point of $RP^2$. That's why we have to rule out a single point, namely $[0:0:1]$.
Elsewhere our map is well defined (as you should check) and it's not hard to see it's homotopic to the identity: the homotopy is $H_t([v_1:v_2:v_3])=[v_1:v_2:tv_3]$, which steadily tilts a line down into the $xy$-plane. On the other hand our map is the identity when restricted to the circle in $RP^2$, so we have a deformation retract and in particular a homotopy equivalence as desired.