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I´d like to obtain an argument to prove that the real plane with two holes, for example $\mathbb{R} \setminus \{p,q\}$ is not homotopy equivalent to the circumference $S^1$.

I know they have different number of holes, but I´d like an argument (without using homology, but homotopy or fundamental group is valid!).

Thanks and regards!

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I got it! (I think). This is an argument:

$\mathbb{R} \setminus \{p,q\}$ is homotopy equivalent to the eight space (indeed, it is a deformation retract). So they have same fundamental group, that is $\mathbb{Z} * \mathbb{Z}$, the free group with two generators.

Since the $\pi(S^1) = \mathbb{Z}$, they can´t be homotopy equivalent.

Now, just a question for the comments. You know $\mathbb{Z} = <x : \emptyset >$, but how you write $\mathbb{Z} * \mathbb{Z}$?

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    $\begingroup$ $\mathbb{Z}\ast\mathbb{Z} = \langle x,y: \emptyset\rangle,$ or, just $\mathbb{Z}\ast\mathbb{Z} = \langle x,y\rangle.$ (Also, note the use of \langle and \rangle to make $\langle$ and $\rangle$, which is prettier than using $<$ and $>$.) $\endgroup$ – Jason DeVito Mar 27 at 21:05

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