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:

![1](https://i.stack.imgur.com/HT3pDm.png)

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.