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I have noticed that some of the most common topological space have fundamental groups related to $\mathbb{Z}$, as we can see below:

table

Why is this the case?

Is it is because they are all realted to the free group on n generators?

I saw in this question Fundamental group of complement of $n$ lines through the origin in $\mathbb{R}^3$ that the free group on n generators corresponds to spaces which are the complement to lines in the origin.

So the circle relates to the open ball with 1 hole drilled though, $\pi(circle)=\mathbb{Z}$, the figure 8 space with two holes drilled through, $\pi(figure 8)=\mathbb{Z} * \mathbb{Z}$, the sphere with zero holes drilled through, $\pi(sphere)=0$.

How could we relate this to the fundamental group of the Klein bottle $\pi(Klein)=\mathbb{Z} \times \mathbb{Z_2}$?

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amiz9
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  • Isn't the fundamental group of the Klein bottle $\Bbb Z\rtimes \Bbb Z_2$ (not $\Bbb Z\times \Bbb Z_2$)? –  May 24 '16 at 23:56
  • I am not sure of the difference in notation to be honest – amiz9 May 24 '16 at 23:57
  • $\Bbb Z\rtimes \Bbb Z_2$ is a semi-direct product. In this case, I mean it with operation $$(a,,b)\cdot (c,,d)=(a+(-1)^bc,,b+d)$$ For instance, $\Bbb Z\rtimes \Bbb Z_2$ is not abelian. –  May 25 '16 at 00:01
  • Also, the table you linked says that $\pi(\text{figure }\infty)=\Bbb Z\Bbb Z$, not $\Bbb Z\times\Bbb Z$. $H_1$ and $\pi_1$ are two different things: $H_1$ is the first homology* group, while $\pi_1$ is the first homotopy group (aka fundamental group). They are closely related, since $H_1$ is the abelianisation of $\pi_1$, but not the same thing. –  May 25 '16 at 00:06
  • OK thanks. So I am only concerned with $\pi$ right now – amiz9 May 25 '16 at 00:09
  • @G.Sassatelli Your group has $(0,1)\cdot(0,1)=(0,0)$, the identity element, but the fundamental group of the Klein bottle has no element of degree $2$ (it has no torsion at all, in fact). ($H_1(\rm Klein)$ does have torsion, though.) – Akiva Weinberger May 25 '16 at 00:40
  • what is a torsion? – amiz9 May 25 '16 at 01:18

1 Answers1

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For $H_1$, there's a theorem that every finitely generated abelian group is a quotient of a finite rank free abelian group $\underbrace{\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}}_{\text{$n$ times}}$ for some $n$ (the theorem is actually more precise than this, but I'm writing it this way to emphasize $\mathbb{Z}$).

For $\pi_1$, there's a theorem that every finitely generated group is a quotient of a finite rank free group $\underbrace{\mathbb{Z} * \cdots * \mathbb{Z}}_{\text{$n$ times}}$ for some $n$. In this case there is no more precise statement in general. But, there are many special ways to study the kernel of the quotient homomorphism; this comes under the banner of "group presentations".

Lee Mosher
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  • Can we say that a sphere with $n$ holes in has fundamental group $\underbrace{\mathbb{Z} * \cdots * \mathbb{Z}}_{\text{$n$ times}}$? This gives us the fundamental group of the sphere and the torus – amiz9 May 25 '16 at 00:11
  • Is there any way to think of the Klein bottle in a similar way? – amiz9 May 25 '16 at 00:13
  • You can learn the general tool for the $\pi_1$ theorem I wrote, and for the special calculations you ask in your comments, by learning about Van Kampen's Theorem, and/or by learning about how to write a presentation of the fundamental group of a CW complex. – Lee Mosher May 25 '16 at 01:49