New document with the retractions and computations for general degree

generaldegree.tex

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+\documentclass[a4paper,10pt]{amsart}
+\usepackage[utf8]{inputenc}
+\usepackage{amsfonts}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage{tikz}
+\usepackage{tikz-cd}
+\usetikzlibrary{positioning}
+\usetikzlibrary{arrows.meta}
+\usetikzlibrary{decorations.pathmorphing}
+
+\usepackage{hyperref}
+
+\newtheorem{theorem}{Theorem}
+\newtheorem{lemma}{Lemma}
+\newtheorem{definition}{Definition}
+\newcommand{\CC}{\mathbb{C}}
+\newcommand{\RR}{\mathbb{R}}
+\newcommand{\DD}{\mathbb{D}}
+\newcommand{\BB}{\mathbb{B}}
+\newcommand{\PP}{\mathbb{P}}
+\newcommand{\s}{\mathbb{S}}
+\newcommand{\ZZ}{\mathbb{Z}}
+
+
+%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 $r_2:M_1\cap\RR^3\longrightarrow S_1$
+\end{proof}
+
+\end{document}