Which is the fundamental group of $\mathbb{R}^2-\{(0,0),(0,1)\}?$

Question

Making a picture of this space with two closed curves with point basis on $(-1,0)$ and both disjoint each one involving $(0,0)$ and $(0,1)$, I can see that these curves are not homotopic. Even more, they are not contractible to a point. So my group has two classes of equivalence plus the constant curve, that are not homotopic of none of these curves. Then, can I afirme that this fundamental group is $\mathbb{Z}_3?$

It looks more like the free group on two generators... not finite and ceratinly not abelian. — DonAntonio, Jul 09 '16 at 17:32
So, you are saying that if $\alpha$ and $\beta$ are such curves, then $\pi_1(X,(-1,0) = F(\alpha,\beta)$, denoting $F$ de free group generated by such curves? Is this?? @DonAntonio Why? — L.F. Cavenaghi, Jul 09 '16 at 17:35

Mar · Accepted Answer · 2016-07-09T17:41:07.423

1

Your space is homotopically equivalent to the double circle. Just like how $\mathbb{R}^2\setminus\{0\}$ is homotopically equivalent to the circle. As a result, the fundamental group is $\mathbb{Z}\coprod\mathbb{Z} $, the nonabelian coproduct of $\mathbb{Z}$ with itself.

edited Jul 09 '16 at 17:41

answered Jul 09 '16 at 17:37

Mar

1,133

By "coproduct" do you mean the free product? – DonAntonio Jul 09 '16 at 17:38
I've seen free product denoted as $$ so $\Bbb Z\Bbb Z$. – snulty Jul 09 '16 at 17:47
Yes! That is indeed what I mean. But I like saying coproduct as that is the coproduct in the category of groups. – Mar Jul 09 '16 at 17:47
I cannot be sure as I don't know category theory (and you don't want to know what I think, in general, of it...), but take into account that $;\Bbb Z*\Bbb Z=F(a,b);$ , and we get again the free group in two generators. – DonAntonio Jul 09 '16 at 17:58

Which is the fundamental group of $\mathbb{R}^2-\{(0,0),(0,1)\}?$

1 Answers1