# Homotopy equivalence between n times punctured plane and…

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?

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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.
You can put your $n$ points on a (immaginary) circle. So you can find $n$ distinct loops that surround each circle. So you have $n$ distinct generatos of homotopy and if $n-1$ of these runs over the $n$-th (as a nacklace) you can obtain the bouquet on $n$ circles. So the fundamental group of this space is isomorphic to the free group of $n$ generators.