$S=\{(x,y,z)\in\Bbb{R}^3: xyz=1\}$ has $4$ connected components isomorphic to $\Bbb{R}^2$ I have the following set $S=\{(x,y,z)\in\Bbb{R}^3: xyz=1\}$, I already proved that the set is a two dimensional submanifold, now it's asking to find the connected components and to prove they're isomorphic to $\Bbb{R}^2$. 
I am not sure how to do it, I can "easily" imagine that the set has $4$ connected components which correspond to a "division" of the space.
Question: How can I prove that there is only $4$ connected components ?
 A: I do not really understand your comments. You have successfully proven that your space has at least four connected components: if the various open orthants of $\Bbb R^3$ are labeled $O_i$, then $O_i \cap S$ is open in $S$, and because clearly the complement of $O_i$ (the coordinate planes) do not intersect $S$, this gives a decomposition of $S$ into the open sets $O_i \cap S$. Precisely four of these intersections are nonempty.
Now you just need to verify what the components of these four sets are. Let's work with the positive orthant $x,y,z>0$ for convenience. I claim $O_1 \cap S \cong (\Bbb R^+)^2$ (which is then homeomorphic to $\Bbb R^2$). This is because we can just write down the homeomorphism $(x,y) \mapsto (x,y,1/xy)$. This is clearly a continuous bijection with continuous inverse (the inverse is just projection), as desired.
A: @Mike Miller This is not a new solution, but it takes a volume that is not compatible with a simple comment.
I just added a modest figure (see below) with level lines materializing horizontal cuts on the surface with equation $xyz=1$, materializing the four "inhabited" octants.
Too few geometrical issues are associated with a graphical representation...
In particular, this issue is "visibly" an immediate 3D extension of the 2D case where (equilateral) hyperbolas with equation $xy=k$ (in fact, here, $k=1/z$) have two connected components (their two branches) occupying the 1st and 3rd quadrant. 

