-

generaldegree.tex

@@ -78,13 +78,16 @@ M_1\rar["r_1"]\ar[dr, "r"]&M_1\cap\RR^3\dar["r_2"]\\
 %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{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.
+M_1\rar["r_1"]&M_1\cap\RR^3\\ 
+P\rar[mapsto]& \Re P.
+\end{tikzcd}
+\end{equation}
+It is a deformation retraction via
+\begin{equation}
+\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["H_1"]&M_1\cap\RR^3\\ 
+(P, t)\rar[mapsto]& t r_1(P) + (1 - t) P.
 \end{tikzcd}
 \end{equation}
 This map is a well-defined deformation retract since $M_1$ has real equations. Note that $M_1\cap\RR^3$
@@ -92,7 +95,18 @@ $M_1\cap\RR^3$ is a plane in $\RR^3$ as in Figure~\ref{fig:triangulo}, and $S_1$
 \[
 S_1=\{(x,y,z)\in\RR_{\geq0}^3|\,x+y+z=1\}=\{(x,y,z)\in[0,1]^3|\,x+y+z=1\},
 \] 
-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:
+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 as follows. 
+The restriction of the retraction to $(+,+,)$ is the identity. For $(+,-,-)$ we sent any point in there
+to $(1,0,0)$ (a similar argument holds when there is only one $+$). 
+For $(-,+,+)$, for a given $(x,y,z)$ we consider intersection of the segment $[(x,y,z),(1,0,0)]$ with $x=0$, i.e., $r_2(x,y,z)=(0,\frac{y}{1-x},\frac{z}{1-x})$ (a similar argument holds when there is only one $-$). Since this definition is compatible in the intersection, it defines
+a retraction $r_2:M_1\cap\RR^3\to S_1$ and we construct the deformation using the segments:
+\begin{equation}
+\begin{tikzcd}[row sep=0pt,/tikz/column 1/.append style={anchor=base east}, /tikz/column 2/.append style={anchor=base west}]
+(M_1\cap\RR^3)\times{[0,1]}\rar["H_1"]&S_1\\ 
+(P, t)\rar[mapsto]& t r_2(P) + (1 - t) P.
+\end{tikzcd}
+\end{equation}
+
 
 \begin{figure}[h]
 \begin{center}
@@ -132,27 +146,28 @@ i.e, $S_1$ is the triangular section defined by the intersection of this plane w
 \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 $S_1$ is $(+,+,+)$.
-
-The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
-\begin{equation}
-	\begin{array}{ccc}
-		r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
-		(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
-	\end{array}
-\end{equation}
-Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
-\begin{equation}
-P_{x,y,z}=\left\{\begin{array}{lr}
-	(x,y,z)&(+,+,+) \\
-	(1,0,0)&(+,-,-) \\
-	(0,1,0)&(-,+,-) \\
-	(0,0,1)&(-,-,+) \\
-	(\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
-	(\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
-	(0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
-\end{array}\right.
-\end{equation}
+%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 $(+,+,+)$.
+%We describe the retraction $r_2$ in two steps. First, we define a map $p:M_1\cap\RR^3\to S_1$ as follows
+%
+%The retraction in $(+,+,+)$ is the identity. For the rest of the parts, we define a retraction:
+%\begin{equation}
+%	\begin{array}{ccc}
+%		r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
+%		(x,y,z, t) & \longrightarrow & (1-t)(x,y,z)+tP_{x,y,z} \\
+%	\end{array}
+%\end{equation}
+%Where $P_{x,y,z}$ is a point depending on $(x,y,z)$. If $(x,y,z)$ is in $(+,-,-)$, $(-,+,-)$, $(-,-,+)$, then $P_{x,y,z}$ is $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. If $(x,y,z)$ is in $(+,+,-)$, $(+,-,+)$, $(-,+,+)$, then in order to choose $P_{x,y,z}$ we take the line to the point that is $0$ everywhere except for the negative coordinate of our point, where it is $1$. Then $P_{x,y,z}$ is the point of this line that intersects with the corresponding edge of the triangle. Then:
+%\begin{equation}
+%P_{x,y,z}=\left\{\begin{array}{lr}
+%	(x,y,z)&(+,+,+) \\
+%	(1,0,0)&(+,-,-) \\
+%	(0,1,0)&(-,+,-) \\
+%	(0,0,1)&(-,-,+) \\
+%	(\frac{x}{1-z},\frac{y}{1-z},0)&(+,+,-) \\
+%	(\frac{x}{1-y},0,\frac{z}{1-y})&(+,-,+) \\
+%	(0,\frac{y}{1-x},\frac{z}{1-x})&(-,+,+) \\
+%\end{array}\right.
+%\end{equation}
 We need to see now that this point is well defined, for this, we'll see that in the borders of the different parts it coincides. We will see this for the part $(+,+,-)$, for the rest of permutations it is equivalent.
 
 We start with the border between $(+,+,-)$ and $(+,+,+)$, this is, the line with $z=0$. In this case