Real plane with two holes and circumference are not homotopy equivalent

Question

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!

Answer 1

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 $\mathbb{Z} * \mathbb{Z}$?