# Injective differential in $\mathbb{R}P^{2}$

Consider the map $f: \mathbb{R}P^{2} \rightarrow \mathbb{R}^{3}$ given by $f([a,b,c])=(bc,ac,ab)$ where $\mathbb{R}P^{2}$ denotes the projective plane.

In order to show that $f$ is not an immersion in general, why it suffices to show that the map:

$g: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ given by $g(a,b,c)=(bc,ac,ab)$ is not an immersion in all $\mathbb{S}^{2}$?

I can see that if $g$ is not an immersion at a point then $f$ is not an immersion at the image of that point.

But why is the converse true? how do we know that if $f$ is not an immersion then $g$ is not an immersion? I mean how do we know that by checking those points where $g$ is not an immersion then we know there cannot be other points in $\mathbb{R}P^{2}$ for which $f$ is an immersion?

-
Your map $f$ isn't well-defined. If $a$, $b$, $c$ are multiplied by $u$ then the right side gets multiplied by $u^2$, which isn't the same point as the original (since the right side is not in a projective space). – Ted Oct 6 '11 at 6:35
@Ted: See here: math.niu.edu/~rusin/known-math/98/immersion – user17182 Oct 6 '11 at 6:42
Okay, in that link, we are considering $\mathbb{R}P^2$ as a quotient of $S^2 = \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1\}$ (identifying antipodal points), and $[a,b,c]$ is meant to be the equivalence class of a point on $S^2$. In that case, only an overall sign change is possible within the same equivalence class, so the right side is well-defined. I would mention this in the post though, since normally in the context of projective coordinates, the notation $[a,b,c]$ means the coordinates can be any values that are not all zero. – Ted Oct 6 '11 at 7:12

## 1 Answer

Being an immersion at a point is a local property (depends only on the behavior of the function in a small open neighborhood of the point). If $p \in S^2$ and $U$ is a neighborhood of $p$ small enough that $U$ does not contain any 2 antipodal points of $S^2$, then $g$ restricted to $U$ is the same as $f$ restricted to $[U]$. So $g$ is an immersion at $p$ iff $f$ is an immersion at $[p]$.

(Brackets denote the quotient map from $S^2$ to $P^2$.)

-