# Irreducible projective cubic, exists continuous surjection?

Let $C$ be an irreducible projective cubic in $\mathbb{P}_2$ with a singular point $p$. So consider $f: \mathbb{P}_1 \to C$ defined as follows. Identify $\mathbb{P}_1$ with the set of lines in $\mathbb{P}_2$ which pass through $p$, then map a line $L$ through $p$ to $p$ if it is a tangent line to $C$ at $p$, and otherwise to the unique other point of intersection of $L$ with $C$. My question is, does there necessarily exist such a continuous surjection $f$ as I described above?

As the comment below says, perhaps Proposition 1.6.8 of Hartshorne may prove helpful:

Let $X$ be an abstract nonsingular curve, let $P \in X$, let $Y$ be a projective variety, and let $\varphi: X - P \to Y$ be a morphism. Then there exists a unique morphism $\overline{\varphi}: X \to Y$ extending $\varphi$.

I am interested in seeing how the proof goes in general for this question? It's not immediately obvious to me why the general case should be clear looking at the example provided by Georges Elencwajg? I feel as if the proof should follow quite quickly from the result in Hartshorne but I do not see it? Much thanks in advance.

• Yeah, I have fixed it. – user243409 May 26 '15 at 6:27
• Agh, sorry, fixed that as well. – user243409 May 26 '15 at 6:28

Yes, you have described a perfectly legitimate morphism $\mathbb P^1\to C$, not only a continuous map.
So take for $C$ the cuspidal cubic $y^2z=x^3$.
Your map is then $\mathbb P^1\to C: (t:u)\mapsto (x:y:z)=(t^2u:t^3:u^3)$, where $p=(0:0:1)$ and the variable line $L$ through $p$ has equation $tx-uy=0$.