diff --git a/generaldegree.tex b/generaldegree.tex
index cd573da..7c05c59 100644
--- a/generaldegree.tex
+++ b/generaldegree.tex
@@ -4,11 +4,9 @@
 \usepackage{amsthm}
 \usepackage{amsmath}
 \usepackage{tikz}
-\usepackage{tikz-cd}
-\usetikzlibrary{positioning}
-\usetikzlibrary{arrows.meta}
-\usetikzlibrary{decorations.pathmorphing}
-\usetikzlibrary{shapes.multipart}
+
+\usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart,
+decorations.markings, calc}
 
 \usepackage{hyperref}
 
@@ -26,6 +24,7 @@
 \newcommand{\ZZ}{\mathbb{Z}}
 
 \numberwithin{equation}{section}
+\tikzset{->-/.style={decoration={markings, mark=at position #1 with {\arrow{>}}},postaction={decorate}}}
 
 %opening
 \title{Complement of a projective curve}
@@ -67,6 +66,8 @@ The space $S_d$ is a strong deformation retract of $M_d$.
 We start by getting a retraction $r:M_d\longrightarrow S_d$ for $d=1$. Note that for $d=1$ the result is trivial since $M_d$ is an affine complex plane (contractible) and $S_d$ is a closed triangle in an affine real plane. The key point is that in this case the retraction is easily constructed and we will lift it
 for $d>1$.
 
+
+
 We can construct $r$ in two steps:
 \begin{equation*}
 \begin{tikzcd}
@@ -74,34 +75,64 @@ M_1\rar["r_1"]\ar[dr, "r"]&M_1\cap\RR^3\dar["r_2"]\\
 &S_1.
 \end{tikzcd}
 \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:
+%Let $r_1:M_1 \longrightarrow M_1\cap\RR^3$ and $r_2:M_1\cap\RR^3\longrightarrow S_1$. 
+The map $r_1$ 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}
+%	\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}
+\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
+M_1\times{[0,1]}\rar["r_1"]&M_1\cap\RR^3\\ 
+(P, t)\rar[mapsto]& t \Re P + (1 - t) P.
+\end{tikzcd}
 \end{equation}
-$M_1\cap\RR^3$ is a plane in $\RR^3$, and $S_1$ can be written as $S_1=\{(x,y,z)\in\RR_{\geq0}^3|x+y+z=1\}=\{(x,y,z)\in[0,1]^3|x+y+z=1\}$, this is, $S_1$ is a triangular section of the plane defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described below, and we will retract each of them separately:
+This map is a well-defined deformation retract since $M_1$ has real equations. Note that $M_1\cap\RR^3$
+$M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$ is 
+\[
+S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
+\] 
+i.e, $S_1$ is the triangular section defined by the intersection of this plane with the coordinated planes $\{x=0\}$, $\{y=0\}$, $\{z=0\}$. The intersection with these planes divides $M_1\cap\RR^3$ into seven parts, described in Figure~\ref{fig:triangulo}. We will retract each of them separately:
+
 \begin{figure}[h]
 \begin{center}
 \begin{tikzpicture}[scale=0.7,every text node part/.style={align=center}]
-	\draw (3,4.5) to (-6,-9);
-	\draw (-3,4.5) to (6,-9);
-	\draw (-7,-4.5) to (7,-4.5);
-	\node at (0,-2) {$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
-	\node at (4,-2) {$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
-	\node at (5.5,-6) {$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
-	\node at (0,-6) {$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
-	\node at (-5.5,-6) {$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
-	\node at (-4,-2) {$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
-	\node at (0,2) {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
-	\node[right] at (0,0) {(0,0,1)};
-	\node[right,above] at (3,-4.5) {(0,1,0)};
-	\node[left,above] at (-3,-4.5) {(1,0,0)};
+\coordinate (A) at (-3,-4.5);
+\coordinate (B) at (3,-4.5);
+\coordinate (C) at (0,0);
+\fill[green!25!white] (C) to (B) to (A);
+	\draw ($2*(C)-(A)$) to ($2*(A)-(C)$);
+	\draw ($2*(B)-(A)$) to ($2*(A)-(B)$);
+	\draw ($2*(B)-(C)$) to ($2*(C)-(B)$);
+	\node at (0,-2.5) {$(+,+,+)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\geq0$};
+	\node at (5.5,0) {$(-,+,+)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\geq0$};
+	\node at (7.5,-6.5) {$(-,+,-)$};%{$x\leq0$,\\ $y\geq0$,\\ $z\leq0$};
+	\node at (0,-8.5) {$(+.+,-)$};%{$x\geq0$,\\ $y\geq0$,\\ $z\leq0$};
+	\node at (-8,-6.5)  {$(+.-,-)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\leq0$};
+	\node at (-6,0) {$(+.-,+)$};%{$x\geq0$,\\ $y\leq0$,\\ $z\geq0$};
+	\node at (0,4) {$(-,-,+)$};% {$x\leq0$,\\ $y\leq0$,\\ $z\geq0$};
+	\foreach \x[evaluate={\y=1-\x}] in {.125, .25, ..., 1}
+		{
+\draw[->-=.5,green!75!black] ($1.75*(C) -.75*\x*(A) -.75*\y*(B)$) -- (C);
+\draw[->-=.5,green!75!black] ($1.75*(B) -.75*\x*(A) -.75*\y*(C)$) -- (B);
+\draw[->-=.5,green!75!black] ($1.75*(A) -.75*\x*(C) -.75*\y*(B)$) -- (A);
+}
+	\foreach \x[evaluate={\y=1-\x}] in {0.25, .3125, ..., .75}
+		{
+\draw[->-=.5,green!75!black] ($1.75*\x*(C) -.75*(A) + 1.75*\y*(B)$) -- ($\x*(C)+ \y*(B)$);
+\draw[->-=.5,green!75!black] ($1.75*\x*(A) -.75*(B) + 1.75*\y*(C)$) -- ($\x*(A)+ \y*(C)$);
+\draw[->-=.5,green!75!black] ($1.75*\x*(A) -.75*(C) + 1.75*\y*(B)$) -- ($\x*(A)+ \y*(B)$);
+}
+%\fill (6, -4.5) circle[radius=.1cm];
+	\node[right=2pt] at (C) {(0,0,1)};
+	\node[above right] at (B) {(0,1,0)};
+	\node[above left] at (A) {(1,0,0)};
 \end{tikzpicture}
+\caption{Triangle $K$}
+\label{fig:triangulo}
 \end{center}
 \end{figure}
-We denote each section as $(\pm,\pm,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ would be $(+,+,+)$.
+We denote each section as $(\pm,\pm,\pm)$ indicating which coordinates are greater or equal to $0$ and which are lower or equal to $0$. For example, the triangle corresponding to $S_1$ is $(+,+,+)$.
 
 The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
 \begin{equation}