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I am working through Hatcher's Algebraic Topology and don't get what he means by "standard decomposition of $S^3$ into two solid tori",

$$S^3= S^1 \times D^2 \cup D^2 \times S^1$$

Can someone explain (or visualize) what the union looks like? First, I thought it looked like a genus-$2$-torus:

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but I guess that's not what he means, actually... I have basic knowledge in algebraic topology and I am not familiar with Heegaard Splittings.

To obtain an embedding of $X$ in $S^3-K$ as a deformation retract we will use the standard decomposition of $S^3$ into two solid tori $S^1\times D^2$ and $D^2\times S^1$, the results of regarding $S^2$ as $\delta D^4=\delta(D^2\times D^2)=\delta D^2\times D^2\cup D^2\times\delta D^2$. Geometrically, the first solid torus $S^1\times D^2$ can be identified with the compact region in $\Bbb R^3$ bounded by the standard torus $S^1\times S^1$ containing $K$, and the second solid torus $D^2\times S^1$ is then the closure of the complement of the first solid torus, together with the compactification point at infinity. Notice that meridional circles in $S^1\times S^1$ bound disks in the first solid torus, while it is longitudinal circles that bound disks in the second solid torus.

I have read many similar questions and answers, but none I found could help me out, so I started a new one.

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    $\begingroup$ While the answers here are quite nice, the one line version is perhaps useful to hear: "if $X$ is the (closure of the) complement of a solid torus sitting in $\mathbb R^3$ as thickened unknot, then the one-point compactification of $X$ is another solid torus." We are, of course, using here that $S^3$ is the one-point compactification of $\mathbb R^3$. $\endgroup$ Commented Feb 6, 2021 at 7:52

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Here's one way of visualizing it.

In the same way that $S^2$ can be mapped to the extended plane $\mathbb R^2+\infty$ via the stereoscopic projection, use stereoscopic projection to map $S^3$ to $\mathbb R^3+\infty$.

Now imagine a solid torus $T$ and its complement $T'$ in $\mathbb R^3$. $T$ is evidently $S^1\times D^2$. But so is $T'$, because it can be decomposed into circles that go through the disk spanned by the smallest circle in $T$ (i.e. the circle that travels around the "hole" in the torus) and travel around $T$. This decomposition will be preserved topologically when the inverse stereographic projection takes it back to $S^3$.

(Lest the verbal description above doesn't make sense, a cross section would look like the following picture from wikipedia)

enter image description here

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  • $\begingroup$ Very helpful picture! $\endgroup$
    – melmo99
    Commented Feb 6, 2021 at 12:09
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if we write the 3 sphere as $w^2 +x^2 +y^2 +z^2 =1 $ the torus of intersection is $$ w^2 +x^2 = \frac{1}{2} \; , \; \; \; y^2+z^2 = \frac{1}{2} $$

One solid torus has $w^2 + x^2$ from $0$ to $1/2$ but $y^2 + z^2$ from $1/2$ to $1,$ so that each fixed value of $y^2 + z^2$ gives a disc

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You can view the 3-sphere are the points $(x,y)$ in $\mathbb C^2$ such that $|x|^2+|y|^2=1$. Can you see what the set of points in the sphere with $|x|^2\leq 1/2$ is? Reversing the inequality gives you the other one.

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