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%opening
\title{Complement of a projective curve}
\author{ Enrique Artal Bartolo \\ Alba Larraya Sancho \\ Miguel A. Marco-Buzunariz}

\begin{document}

\maketitle

\begin{abstract}

\end{abstract}
\section{Definition of the problem}
Given $C$ a proyective curve in $\PP^2$ over $\CC$, we know that the complement of the curve has the homotopy type of a CW-complex of real dimension $2$, our problem is how to find an effective model for said CW-complex.
\section{Smooth curves}
Given a smooth curve of degree $d$ in $\PP^2$, we know it is isomorphic to a curve of the form
\begin{equation}
	C_d=\{[x:y:z]\in\PP^2 | x^d+y^d+z^d=0\}
\end{equation}
Therefore, we will work with $C_d$, from now on.

We will denote:
\begin{eqnarray}
	M_d&=&\{(x,y,z)\in\CC | x^d+y^d+z^d=1\} \\
	S_d&=&\{(x,y,z)\in M_d | x^d, y^d, z^d\in\RR^{\geq0}\} \\
	K &=&\{[x:y:z]\in\PP^2 | x^d, y^d, z^d\in\RR^{\geq0}\} 
\end{eqnarray}

We then have a commutative diagram:
\begin{equation*}
	\begin{array}{ccc}
		S_d & \longrightarrow & M_d \\
		\downarrow & & \downarrow \\
		K & \longrightarrow & \PP^2\setminus C_d \\
	\end{array}
\end{equation*}

The inclusion $i:S_d\longrightarrow M_d$ induces a homotopy equivalence. This follows from the following lemma:
\begin{theorem}
$S_d$ is a strong deformation retract of $M_d$.
\end{theorem}
\begin{proof}
We start by getting a retraction $r:M_d\longrightarrow S_d$ for $d=1$, and then we'll see that we can lift it.

We can construct $r$ in two steps:
\begin{equation*}
	\begin{array}{cccccc}
		r:&M_1 & \longrightarrow & M_1\cap\RR^3&\longrightarrow&S_1 \\
	\end{array}
\end{equation*}
Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow S_1$. Then $r_1$ is given by:
\begin{equation}
	\begin{array}{ccc}
		r_1: M_1\times I & \longrightarrow & M_1\cap\RR^3 \\
		(x_r+ix_i,y_r+iy_i,z_r+iz_i, t) & \longrightarrow & (x_r+(1-t)ix_i,y_r+(1-t)iy_i,z_r+(1-t)iz_i) \\
	\end{array}
\end{equation}
\end{proof}

\end{document}