How can I calculate the fundamental group of $S^3-S^1$? I believe it might be similar to the fundamental group of $S^2-S^0$ which is $ \Bbb Z$ but I'm having trouble showing it. Is this the right direction (showing it's $\Bbb Z$) or am I way off?
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What direction? – Mariano Suárez-Álvarez Jan 12 '18 at 20:58
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I'm trying to show it's $\Bbb Z$ – Amontillado Jan 12 '18 at 20:59
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It is $\mathbb Z$, but you did not give any hint as to how you think youy are going to prove it. – Mariano Suárez-Álvarez Jan 12 '18 at 21:01
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I wasn't sure it was. I was trying to show it deformation retracts to ${\Bbb R}^2-{x}$ – Amontillado Jan 12 '18 at 21:06
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Since it does not contain $\mathbb R^2-\text{pt}$, it can't! – Mariano Suárez-Álvarez Jan 12 '18 at 21:07
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Poor choice of words on my part. I'm sorry. I wast trying to show it's homeomorphic to the plane minus point. And then deformation retract to the circle – Amontillado Jan 12 '18 at 21:10
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It is not homeomorphic to the plane minus a point: it is a $3$-dimensional manifold. – Mariano Suárez-Álvarez Jan 12 '18 at 21:11
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One can write $S^3$ as the union of two solid tori. Using this, it is easy to see that the complement of a standard $S^1$ in $S^3$ deformation retracts onto a solid torus.
Since a solid torus deformation retracts onto its central $S^1$, the fundamental group of your space is $\mathbb Z$.
Alternatively, if you consider the stereographic projection $S^3-\{P\}\to\mathbb R^3$ from the sphere from a point $P$ in your $S^1$, then the image of the $S^1$ is a line in $\mathbb R^3$. You this get a homeo of $S^3-S^1$ to $\mathbb R^3-\text{line}$. (For this to work, the $S^1$ has to be a maximal circle of $S^3$)
Mariano Suárez-Álvarez
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Here is a cute visualization. You can think of $S^3$ as the one-point compactification of $\mathbb{R}^3$, so as $\mathbb{R}^3$ together with a "point at infinity." So for starters you can visualize a circle in $\mathbb{R}^3$, keeping in mind that there is a point at infinity, and wanting to figure out the fundamental group of its complement.
But now you can send a point on the circle to infinity, since you were going to remove the circle anyway! This straightens out the circle to a line (if it helps, visualize the analogous construction for $S^2$), so your space is now the complement of a line $\mathbb{R}$ in $\mathbb{R}^3$. This is clearly $\mathbb{R}$ times $\mathbb{R}^2 \setminus \{ 0 \}$, so is homotopy equivalent to a circle $S^1$.
Qiaochu Yuan
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