TITLE: Show the projective space $RP^2$ minus a point is homotopy equivalent to the unit circle $S^1$
QUESTION [10 upvotes]: 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!

REPLY [4 votes]: Note the CW-complex structure of $RP^2$ is $e^0\cup_{c}e^1\cup_{f}e^2$,where $c$is the constant map and $f$is the antipodal map.If you remove a point from $RP^2$ then upto homotopy the cell structure of $RP^2 - \lbrace x\rbrace$  is exactly  $e^0\cup_{c}e^1$.Which is nothing but $S^1.$