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I know that the fundamental group of $X = \mathbb R^2 \setminus \{(0,0)\}$ is the same as the fundamental group of the circle $Y = S^1$, namely $\mathbb Z$.

However, $X$ and $Y$ are not homotopic, i.e. we can't find continuous maps $f:X\to Y, g : Y \to X$ such that $f \circ g$ is homotopic to $id_Y$ and $g \circ f$ is homotopic to $id_X$.

I would like to prove it, but I don't really know how to do it. If such $f$ and $g$ existed, then it would be something like $f : x \mapsto x/\|x\|$, and $g(Y)$ has to be compact... I don't know how to continue.

Any hint would be appreciated. I apologize if this has already been asked.

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    $\begingroup$ Why do you think they aren't homotopy equivalent? $\endgroup$ – Eric Wofsey Jan 30 '16 at 11:40
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    $\begingroup$ Those spaces are homotopy equivalent. $\endgroup$ – Brian M. Scott Jan 30 '16 at 11:40
  • $\begingroup$ The trivial embedding and your $f$ do the trick, I would say. $\endgroup$ – Henno Brandsma Jan 30 '16 at 11:55
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    $\begingroup$ Thank you ! My intuition wasn't correct, then. I will try to finish the proof. $\endgroup$ – Alphonse Jan 30 '16 at 13:04
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Just to answer this question: they are homotopy equivalent. If $g : Y \to X$ is the natural embedding, and $f$ is as above, then

$g \circ f$ is homotopy equivalent to $id_X$ : Let $H_1 : X \times [0,1] \to X$ be defined as $$(x, t) \mapsto tx/\|x\| + (1-t)x.$$ (Even if $X$ is not convex, $H_1(x,t) \in X$ for any values $x,t$ — this is clear geometrically). Then $H_1(x,0)=x,H_1(x,1) = g(f(x))$ for any $x \in X$.

Moreover $f \circ g$ is homotopy equivalent to $id_Y$, because it is equal to $id_Y$.

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