# Projective Geometry: Prove that the mapping in P2R is not well-defined.

Prove that the mapping F: P2(R) to P2(R) given by F(x1,x2,x3) = (x1x2, x2, x3) is not well-defined.

I know that to determine whether a mapping is well-defined, you should pick two points that are the same in P2(R) and show that the mapping transforms them the same way. So, would the points (1,2,3) and (2,4,6), which are scalar multiples of each other so are the same point in P2(R), an example showing why it is not well-defined? Because (1,2,3) is mapped to (2,2,3) and (2,4,6) is mapped to (8,4,6) and (2,2,3) and (8,4,6) are not scalar multiples of each other. Any help is appreciated.

• This looks like a fine example to me. (Somewhat more formally, elements of $P_2(\mathbb{R})$ are (generally defined as) equivalence classes of points in $\mathbb{R}^3$, so by showing that two elements of $\mathbb{R}^3$ that were in the same equivalence class do not map to the same equivalence class under the mapping $F$, you've shown what you need to.) – Steven Stadnicki Feb 27 at 23:26

To put your example in context, you can define a map $$F:\Bbb P^2\longrightarrow \Bbb P^2$$ setting $$F([x_1\colon x_2\colon x_3])=[y_1\colon y_2\colon y_3]$$ where each $$y_i$$ is a homogeneous polynomial in $$x_1, x_2, x_3$$ of the same degree $$d$$.
Indeed, if so, multiplying each of the $$x$$'s by $$\lambda\neq0$$ the $$y$$'s get modified by the same factor $$\lambda^d$$ leaving the image point unchanged.
But if the degrees of the $$y$$'s are not equal, as in your example, the map $$F$$ is not well-defined.