TITLE: Homotopy equivalence between n times punctured plane and...
QUESTION [6 upvotes]: how to prove that $\mathbb{R}^2$ without $n$ points is homotopy equivalent to $S^1 \vee ... \vee S^1 $ which means a bouquet of $n$ circles? 
It's easy for $n=1$ and $n=2$ but how to generalize it? By induction? How?

REPLY [4 votes]: Since you asked for homotopy equivalence, not just isomorphism of fundamental groups, let me add a little to ArthurStuart's answer.  Embed the bouquet of circles into the plane so that  each of your $n$ points has exactly one of the circles going around it.  Then do a deformation retraction of the plane to the bouquet of circles by deforming the inside of each circle, minus the one of your $n$ points that lies in that interior, outward to the circle, and deforming the rest of the plane (the part outside all your circles) inward to the circles.