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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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2 Answers 2

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.

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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.

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