# Real plane with two holes and circumference are not homotopy equivalent

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!

$$\mathbb{R} \setminus \{p,q\}$$ 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 $$\mathbb{Z} = $$, but how you write $$\mathbb{Z} * \mathbb{Z}$$?
• $\mathbb{Z}\ast\mathbb{Z} = \langle x,y: \emptyset\rangle,$ 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 $>$.) – Jason DeVito Mar 27 at 21:05