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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.

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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.

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  • $\begingroup$ How do you prove that this normalization map is not an isomorphism? $\endgroup$ – user289143 May 29 '19 at 13:42
  • $\begingroup$ Because the target curve is singular. $\endgroup$ – Sasha May 30 '19 at 14:18

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