# Projection of Veronese embedding

Let $$C \subset \mathbb{P}^3$$ be the twisted cubic curve given by the parametrization $$\mathbb{P}^1 \mapsto \mathbb{P}^3$$ $$(s:t)\mapsto (s^3 :s^2t:st^2 :t^3)$$. Let $$P = (0 : 0 : 1 : 0) \in \mathbb{P}^3$$, and let $$H$$ be the hyperplane defined by $$x_2 = 0$$. Let $$\varphi$$ be the projection morphism from $$P$$ to $$H \simeq \mathbb{P}^2$$.

Is $$\varphi: C \mapsto \varphi(C)$$ an isomorphism?

I know that the Veronese embedding (in this case the twisted cubic curve) is isomorphic onto its image, but I don't think this is an isomorphism.

Let $$U_0:=\{(x_0:x_1:x_2) \in \varphi(C) | x_0 \neq 0\}$$ and define $$\varphi_0 : U_0 \mapsto C$$ such that $$\varphi((x_0:x_1:x_2))=(x_0^2:x_0x_1)$$. I would like to define a similar map on $$U_2:=\{(x_0:x_1:x_2) \in \varphi(C) | x_2 \neq 0\}$$ and then see that this map coincide on $$U_0 \cap U_2$$ but I think that without the coordinate we projected ($$st^2$$) is impossible.

The point $$P$$ lies on a tangent line to $$C$$, so the image is a cuspidal cubic curve $$C'$$ (with equaltion $$x_0^2x_3 = x_1^3$$). The map $$C \to C'$$ is the normalization map; it is not an isomorphism.