# A question about morphism of projective spaces

Consider the morphism: $$f: (\mathbb{P}^2 -\{(0:0:1),(0:1:0) \} )\to \mathbb{P}^3$$ Given by $$f((x:y:z))=(x^2:xy:xz:yz)$$, my problem is to find the closure of the image of $$f$$, my argument was: in the image $$x$$ is not vanishing so we have in affine coordinates $$(1:\frac{y}{x}:\frac{z}{x}:\frac{yz}{x^2})$$ i.e. is given by $$Z(W-YZ)$$ then considering the projective closure I have to homogenize the polynomial. Then $$\overline{Imf}=Z(XW-YZ)$$ is it correct? Thanks!

Yeah, that's pretty much right. Let's say $$[x:y:z]$$ are homogeneous coordinates on $$\mathbb{P}^2$$ and $$[X:Y:Z:W]$$ are homogeneous coordinates on $$\mathbb{P}^3$$. Then the affine chart $$U_x=\{[x:y:z] : x\ne 0\} \cong \mathbb{A}^2$$, which has coordinates $$(\frac{y}{x}, \frac{z}{x})$$, is mapped into the affine chart $$U_X = \{[X:Y:Z:W] : X\ne 0\} \cong \mathbb{A}^3$$, which has coordinates $$(\frac{Y}{X}, \frac{Z}{X},\frac{W}{X})$$, by the affine morphism $$(\frac{y}{x},\frac{z}{x}) \mapsto (\frac{y}{x},\frac{z}{x},\frac{y}{x}\cdot\frac{z}{x})$$, whose image in $$\mathbb{A}^3$$ is given by $$\frac{Y}{X}\cdot\frac{Z}{X} - \frac{W}{X} = 0$$. As you say, the projective closure of this set is $$C = Z(XW-YZ)$$. So we know the closure of the image of $$f$$ is at least as big as $$C$$. Now you can check that on the rest of the domain, e.g. on $$U_y\cap U_z$$ (which is also affine), the image of $$f$$ is contained in $$C$$, so the closure of the image is actually $$C$$.