This is a question regarding Differential Geometry.
[TL;DR] What are the different implications of the differential of a map being (i) 1-1, versus the differential of a map being (ii) onto?
Background: In determining whether a subset $S \subset R^3$ is a regular surface or not, we must check that for each point $p \in S$ there exists a neighbourhood $V$ in $R^3$ and a map $x:U \rightarrow V \cap S$ of an open set $U \subset R^3$ onto $V \cap S \subset R^3$ that satisfy three conditions, including (1) $\boldsymbol x$ is differentiable, (2) $\boldsymbol x$ is a homeomorphism, and (3) For each $q \in U$ the differential $d \boldsymbol x_q : R^2 \rightarrow R^3$ is 1-1.
When I encountered the question: Show that the two-sheeted cone with vertex at the origin ${(x,y,z) \in R^3; x^2+y^2+z^2 =0}$ is not a regular surface, my first thought process was to let $f(x,y,z) = x^2+y^2-z^2=0$ where $f^{-1}(0)$ is our surface. Then the critical point is where $f_x$, $f_y$, $f_z$ are simultaneously zero, hence $2x = 2y = 2z = 0$ which happens at the point $(x,y,z) = (0,0,0)$. So our critical value would therefore be $f(0,0,0) =0$. Now since 0 is a critical value of $f$, its preimage $f^{-1}(0)$ is a critical point.
At this point, I found that in the book* that I'm studying, a critical point of a map $f$ is defined to be where the differential $df_p$ is not onto. However, arriving at this conclusion would not be enough to conclude that our subset above, $f^{-1}(0)$, is not a regular surface, because this is different from saying that our differential is not 1-1 (condition 3 as above).
How are the two conditions different, and in what way are each important in looking at the regularity of parametrisations of surfaces? Is there a relationship between the differential of a map being onto and 1-1?
Thanks!
* Manfredo Do Carmo, Differential Geometry of Curves and Surfaces 2nd edition.