From 4a of this released practice final it asks:
Let $f : H^3 \to R^3$ be given by $x^2 + y^2 + xz$. Show that $f^{-1}(1)$ is a manifold with boundary, and determine its boundary.
In the proof she states:
[...] If both [$x$ and $y$] are zero, then $f(x, y, z) = 0$, so $f^{-1}(1)$ is empty. Hence, $f$ is transversal to $\{1\}$.
So, I understand why $f(x, y, z) = 0$. What I don't understand is what it means by "so $f^{-1}(1)$ is empty". Does that mean there is no solution and the solution set of $f^{-1}$ is $\varnothing$? If so:
- I don't follow why $f(x, y, z) = 0 \implies Im(f^{-1}(1)) = \varnothing$.
- Why does this imply that it is transversal with $\{1\}$?
- Finally, how does being transversal to $\{1\}$ imply that $f^{-1}(1)$ is a manifold with boundary? The rest of the proof I have no issue with.
For the record I try solving it using the Preimage Theorem by noting that $1$ was indeed a regular value, but got stumped at showing the resulting manifold had a boundary.
Can someone please elaborate on this proof?
I believe I have the answer:
According to Guillimin and Pollack, even "boundaryless" manifolds have a boundary (the empty set). So using the Preimage theorem you get a manifold. Boundaries map to boundaries in the image. So just consider $H^k$'s boundary of $(x, y, 0)$ and determine the image of that which is $x^2 + y^2$. In this case because we want $f^{-1}(1)$ it ends up being the image of $x^2 + y^2 = 1$.
Lastly, we should just double check that $x^2 + y^2 = 1$ is a manifold. $d \partial f$ is surjective as $[2x \ 2y]$ has at least one non-zero term to satisfy $x^2 + y^2 = 1$. And so by the Preimage theorem again $\partial f^{-1}(1)$ must be a manifold.