The relation between $S^3$, $\mathbb{R}P^3$, and solid 2-tori I am reading up on Heegaard splittings and need some conceptual help. $S^3$ may be decomposed into two 3-balls with the boundaries of the 3-balls identified. $S^3$ may be also decomposed into two solid 2-tori with boundary points identified. 
$\mathbb{R}P^3$ has half as many points as $S^3$. This leads me to suspect that $\mathbb{R}P^3$ may be considered as a solid 2-torus with antipodal points on its boundary identified. Is this interpretation correct?
Thanks.
 A: $\newcommand{\Cpx}{\mathbf{C}}$Think of the unit $3$-sphere sitting inside $\Cpx^{2}$ as the set
$$
S^{3} = \{(z_{1}, z_{2}) : |z_{1}|^{2} + |z_{2}|^{2} = 1\}.
$$
The standard decomposition into two tori is via the Clifford torus $C$ defined by $|z_{1}|^{2} = |z_{2}|^{2} = 1/2$. The resulting handlebodies are $T_{1} = \{(z_{1}, z_{2}) : |z_{1}| \leq 1/2\}$ and $T_{2} = \{(z_{1}, z_{2}) : |z_{2}| \leq 1/2\}$.
The antipodal map of $S^{3}$ does send $C$ to itself, but the handlebodies are as far as possible from being fundamental domains of this action: Each handlebody is preserved by the antipodal map, and hence double-covers its own image in the quotient space. Particularly, the quotient space is not obtained from one handlebody by boundary identification.
A: If you think that $RP^3$ is the lie group $SO(3)$, it is easy to describe it as the union of two tori. Note that $SO(3)$ acts on the two sphere, with isotropy group $SO(2)=S^1$, and $RP^3$ is therefor a circle bundle over the two sphere. Writting this 2-sphere as a union of two discs glued along there boundary, we get a decomposition of $RP^3$ as a union of two solid tori (a circle bundle over a disc is trivial). The same argument proves that every circle bundle over the sphere is made by gluing two solid tori along there boundaries.
