mitad de pagina 7

generaldegreeHypersurfaces.tex

@@ -1,5 +1,6 @@
 \documentclass[a4paper,10pt]{amsart}
 \usepackage[utf8]{inputenc}
+\usepackage[T1]{fontenc}
 \usepackage{amsfonts}
 \usepackage{amsthm}
 \usepackage{amsmath}
@@ -28,6 +29,9 @@ decorations.markings, calc}
 \newcommand{\proy}[1]{\pi_{#1}}
 
 
+\addtolength{\hoffset}{-1cm}
+\addtolength{\textwidth}{2cm}
+\linespread{1.2}
 \numberwithin{equation}{section}
 
 %opening
@@ -44,7 +48,7 @@ Alameda de Mazarredo 14,
 48009 Bilbao, Spain}
 %\urladdr{\url{http://riemann.unizar.es/~mmarco}}
 \email{\href{mailto:alarraya@bcamath.org}{alarraya@bcamath.org}}
-\author[M.Á. Marco]{Miguel Á. Marco-Buzunáriz}
+\author[M.Á. Marco]{Miguel \'A. Marco-Buzunáriz}
 \address{Departamento de Matem\'{a}ticas, IUMA \\
 Universidad de Zaragoza \\
 C.~Pedro Cerbuna 12, 50009, Zaragoza, Spain}
@@ -55,6 +59,13 @@ C.~Pedro Cerbuna 12, 50009, Zaragoza, Spain}
 MICIU/AEI/10.13039/501100011033. The second named author is partially supported by  PID2024-156181NB-C31 funded by
 MICIU/AEI/10.13039/501100011033 and CEX2021-001142-S-20-9.}
 
+
+\tikzcdset{%
+   harpoonlurd/.code={\pgfsetarrows{tikzcd right to-tikzcd right to}},
+   harpoonldru/.code={\pgfsetarrows{tikzcd left to-tikzcd left to}},
+   harpoonluru/.code={\pgfsetarrows{tikzcd right to-tikzcd left to}},
+   harpoonldrd/.code={\pgfsetarrows{tikzcd left to-tikzcd right to}},
+}
 \begin{document}
 
 
@@ -134,12 +145,10 @@ and the restriction of $r_{1t}$ to $M_1\cap\RR^{n+1}$ is the inclusion.
 %where $\Re$ and $\Im$ take the real and imaginary part of each coordinate of a point, respectively.
 
 Note that for $(x_0,\dots,x_n)\in M_1$, we already have that
-\begin{equation*}
+\begin{align*}
-\begin{array}{lll}
+	\sum_{j=0}^n x^r_j & =  1 \\
-	\sum x^r_j & = & 1 \\
+	\sum_{j=0}^n  x^i_j & =  0 \\
-	\sum x^i_j & = & 0 \\
+\end{align*}
-\end{array}
-\end{equation*}
 and therefore $r_1(x_0,\dots,x_n,t)\in M_1$ for all $t\in[0,1]$, and in particular, we get that $r_1(x_0,\dots,x_n,1)=(x^r_0,\dots,x^r_n)\in M_1\cap\RR^{n+1}$ for all $(x_0,\dots,x_n)\in M_1$.
 Let us summarize it.
 
@@ -166,13 +175,11 @@ The space $S_1$ is a $n$-simplex.
 \end{lemma}
 \begin{proof}
 Let $e_i=(0,\dots,1,\dots,0)$ be the vectors of the canonical base. Then
-\begin{equation*}
+\begin{align*}
-\begin{array}{lllll}
+S_1&=\left\{(x_0,\dots,x_n)\in\RR^{n+1}\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\} \\
-S_1&=&\{(x_0,\dots,x_n)\in\RR^{n+1}|\sum x_i=1, x_i\geq1\}&& \\
+&=\left\{x_0e_0+\dots+x_ne_n\,\middle|\,\sum_{i=0}^n x_i=1, x_i\geq1\right\}=\Delta_n
-&=&\{x_0e_0+\dots+x_ne_n|\sum x_i=1, x_i\geq1\}&=&\Delta_n\\
-\end{array}
 \qedhere
-\end{equation*}
+\end{align*}
 \end{proof}
 
 \begin{figure}[ht]
@@ -231,12 +238,12 @@ As we can see in Figure \ref{fig:triangulo}, the hyperplanes defined by $\{x_i=0
 Note that there is no region $(-,\dots,-)$.
 
 The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the rest of the parts, we define a retraction:
-\begin{equation}
+\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^{n+1})\times{[0,1]}\rar["r_2"]&M_1\cap\RR^{n+1}\\
 (Q, t)\rar[mapsto]& t P_Q + (1 - t) Q,
 \end{tikzcd}
-\end{equation}
+\end{equation*}
 %\begin{equation}
 %	\begin{array}{ccc}
 %		r_2: M_1\cap\RR^3\times I & \longrightarrow & S_1 \\
@@ -246,14 +253,15 @@ The retraction in $(+,\dots,+)$ is the identity for any value of $t$. For the re
 where $P_Q$ is a point depending on $Q\in M_1\cap\RR^{n+1}$.
 
 In general, for a point $Q$ in a region $(\pm,\dots,\pm)$, we want the retraction to take the negative coordinates to $0$ and to change the positive ones such that at the end the sum is $1$. Therefore, we can write the point $P_Q$ as:
-\begin{equation}
+\begin{equation*}
-P_Q=\left(\frac{x_0\chi_0}{1-\sum x_j(1-\chi_j)}, \dots,\frac{x_n\chi_n}{1-\sum x_j(1-\chi_j)} \right)
+P_Q=\left(\frac{x_0\chi_0}{\displaystyle 1-\sum_{j=0}^n x_j(1-\chi_j)}, \dots,
-\end{equation}
+\frac{x_n\chi_n}{\displaystyle 1-\sum_{j=0}^n x_j(1-\chi_j)} \right)
-
+\end{equation*}
 where $\chi_j=1$ when $x_j\geq0$ and $\chi_j=0$ otherwise. It is easy to check that the points $tP_Q+(1-t)Q$ are in $M_1$ for all $t$. To see this, we remark that $\sum x_i=1$ implies that $\sum x_i\chi_i+\sum x_i(1-\chi_i)=1$ and therefore $\sum x_i\chi_i=1-\sum x_i(1-\chi_i)$. Then
-\begin{equation}
+\begin{equation*}
-\sum \left((1-t)x_i+\frac{tx_i\chi_i}{1-\sum x_j(1-\chi_j)}\right)=(1-t)\sum x_i+\frac{t}{1-\sum x_j(1-\chi_j)}\sum x_i\chi_i=1
+\sum_{i=0}^n \left((1-t)x_i+\frac{tx_i\chi_i}{\displaystyle 1-\sum_{i=0}^n x_j(1-\chi_j)}\right)=
-\end{equation}
+(1-t)\sum_{i=0}^n x_i+\frac{t}{\displaystyle 1-\sum_{i=0}^n x_j(1-\chi_j)}\sum_{i=0}^n x_i\chi_i=1.
+\end{equation*}
 
 Then, we have proved the following result:
 \begin{lemma}
@@ -265,18 +273,17 @@ The map $r_2$ is a strong deformation retraction from $M_1\cap \RR^{n+1}$ to $S_
 Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we will lift them in order to get a retraction from $M_d$ to $S_d$.
 
 \begin{theorem}
-	$S_d$ is a strong deformation retract of $M_d$.
+	The space $S_d$ is a strong deformation retract of $M_d$.
 \end{theorem}
 \begin{proof}
 We want to produce the following diagram:
 \begin{equation*}
-	\begin{array}{ccccc}
+\begin{tikzcd}
-		M_d & \rightleftharpoons & M_d\cap\{x^d_i\in\RR\forall i\}&\rightleftharpoons& S_d \\
+	(x_0,\dots,x_n)\dar[mapsto] & [-30pt]	M_d \rar[shift left=2pt,harpoon] \dar & \lar[shift left=2pt,harpoon] M_d\cap\{x^d_i\in\RR,\,\forall i\}\rar[shift left=2pt,harpoon]\dar & \lar[shift left=2pt,harpoon]S_d \dar\\
-		\downarrow & & \downarrow & & \downarrow\\
+	(x^d_0,\dots,x^d_n)	&M_1 \rar[shift left=2pt,harpoon] & \lar[shift left=2pt,harpoon] M_1\cap\RR^{n + 1}\rar[shift left=2pt,harpoon] & \lar[shift left=2pt,harpoon]S_1
-		M_1 & \rightleftharpoons & M_1\cap\RR^{n+1} & \rightleftharpoons & S_1 \\
+\end{tikzcd}
-	\end{array}
 \end{equation*}
-where the horizontal arrows represent retractions, and the vertical arrows are the maps $(x_0,\dots,x_n)\mapsto (x^d_0,\dots,x^d_n)$. That is, we want to lift the retractions built for $M_1$ and $S_1$ to $M_d$ and $S_d$ via this branched cover.
+where the horizontal arrows represent retractions. That is, we want to lift the retractions built for $M_1$ and $S_1$ to $M_d$ and $S_d$ via this branched cover.
 
 
 We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise, we can write it as $r_1((x_0,\dots,x_n),t)=(\bar{r}(x_0,t),\dots,\bar{r}(x_n,t))$ with $\sum\bar{r}(x_i,t)=1$ whenever $(x_0,\dots,x_n)\in M_1$. Then, it suffices to find $G(x,t)$ such that $G(x,t)^d=\bar{r}(x^d,t)$.
@@ -292,47 +299,55 @@ We start seeing how the lifting of $r_1$ works. Since $r_1$ acts componentwise,
 %For the first component, $r_1(x,t)=x_r+(1-t)ix_i$, and we have a map from $M_d$ to $M_1$ given by $F(x)=x^d$.
 %$r_1$ is the vertical projection over the real axis of the complex plane. In this first component, $M_d$ is a ramified cover over $M_1$, where $0$ is the only point where it ramifies. Therefore, outside of $0$ we have a cover, and then $r_1$ has a unique lifting outside of $0$. Since $r_1$ is constant on $0$, we get a lifting of $r_1$ by defining $\tilde{r}_1(0,t)=0$.
 
-In order to see this, we can write an explicit expression of $\bar{r}$. For a point $x\in\CC$ the action of $\bar{r}$ is the projection over the real axis, as seen in figure \ref{realprojection}.
+In order to see this, we can write an explicit expression of $\bar{r}$. For a point $x\in\CC$ the action of $\bar{r}$ is the projection over the real axis, as seen in Figure \ref{realprojection}.
 \begin{figure}[h]
 \begin{center}
 \begin{tikzpicture}[scale=0.6]
 	\draw (-5,0) to (5,0);
 	\draw (0,5) to (0,-1);
 	\draw[fill] (3,2) circle (2pt) node[above]{$x_r+ix_i$};
-	\draw[dashed] (3,2) to (3,0);
+	\draw[dashed] (3,2) to (3,0) node[below] {$x_r$};
 \end{tikzpicture}
 \end{center}
-\caption{projection over the real axis}
+\caption{Projection over the real axis}
 \label{realprojection}
 \end{figure}
+We can then write the retraction as $\bar{r}(x,t)=m(x,t)e^{2i\pi\theta(x,t)}$ where $m(x,t)$ is the function indicating how the modulus of $\bar{r}(x,t)$ changes and $\theta(x,t)$ portrays the change of the argument. Let us give explicit formulas.
 
-we can then write the retraction as $\bar{r}(x,t)=m(x,t)e^{2i\pi\theta(x,t)}$ where $m(x,t)$ is the function indicating how the modulus of $\bar{r}(x,t)$ changes and $\theta(x,t)$ portrays the change of the argument.
+For a $x=x_r+ix_i\in\CC$, $\bar{r}(x,t)=x_r+(1-t)ix_i$, we can write 
-
+\[
-For a $x=x_r+ix_i\in\CC$, $\bar{r}(x,t)=x_r+(1-t)ix_i$, we can write $m(x,t)=+\sqrt{x_r^2+(1-t)^2x_i^2}$ and $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$.
+m(x,t)=+\sqrt{x_r^2+(1-t)^2x_i^2} \text{ and } \theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}.
-
+\]
-We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}, where each of the sections will be homeomorphic to either the upper half complex plane or to the lower half.
+We can divide $\CC$ in $2d$ sectors of angle $\frac{\pi}{d}$ as in Figure~\ref{divisionC}, where each of the sections will be homeomorphic to either the upper half complex plane or to the lower half one.
 
 \begin{figure}[h]
 \begin{center}
-\begin{tikzpicture}[scale=0.6]
+%\begin{tikzpicture}[scale=0.6]
-	\draw (-5,0) to (5,0);
+%	\draw (-5,0) to (5,0);
-	\draw (5,1.5) to (-5,-1.5);
+%	\draw (5,1.5) to (-5,-1.5);
-	\draw (5,3.2) to (-5, -3.2);
+%	\draw (5,3.2) to (-5, -3.2);
-	\draw (4.7,5) to (-4.7,-5);
+%	\draw (4.7,5) to (-4.7,-5);
-	\draw (2.8,5) to (-2.8,-5);
+%	\draw (2.8,5) to (-2.8,-5);
-	\draw (0.9,5) to (-0.9,-5);
+%	\draw (0.9,5) to (-0.9,-5);
-	\draw (-0.9,5) to (0.9,-5);
+%	\draw (-0.9,5) to (0.9,-5);
-	\draw (-2.8,5) to (2.8,-5);
+%	\draw (-2.8,5) to (2.8,-5);
-	\draw (-4.7,5) to (4.7,-5);
+%	\draw (-4.7,5) to (4.7,-5);
-	\draw (-5,3.2) to (5, -3.2);
+%	\draw (-5,3.2) to (5, -3.2);
-	\draw (-5,1.5) to (5,-1.5);
+%	\draw (-5,1.5) to (5,-1.5);
+%\end{tikzpicture}
+%
+\begin{tikzpicture}
+\foreach \x in {0, ..., 16}{
+	\draw (0,0) to (360/17*\x:3);}
 \end{tikzpicture}
+
+
 \end{center}
 \caption{Division of the complex plane}
 \label{divisionC}
 \end{figure}
 
-Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$, which we will denote $f_k$. The lifting of $r_1$ will then be given by the $G$ previously described. This $G$ will act by taking the point $(x,y,z)\in M_d$ and mapping it to the upper or lower complex plane via the inverse homeomorphisms $f^{-1}_k$, then taking $r_1$ to take it to the real line, and finally bringing it back to its original sector with the homeomorphism $f_k$.
+Then we can get homeomorphisms from the upper complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=1,3,...,2d-1$, and from the lower complex plane to the sectors of angle $[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$ with $k=2,4,...,2d$, which we will denote $f_k$. The lifting of $r_1$ will then be given by the $G$ previously described. This $G$ will act by taking the point $(x_0,\dots,x_n)\in M_d$ and mapping it to the upper or lower complex plane via the inverse homeomorphisms $f^{-1}_k$, then taking $r_1$ to take it to the real line, and finally bringing it back to its original sector with the homeomorphism~$f_k$.
 
 
 
@@ -344,40 +359,34 @@ Then we can get homeomorphisms from the upper complex plane to the sectors of an
 
 
 The homeomorphisms are given by
-\begin{equation}
+\begin{align*}
-	f_k(|z|\exp^{i\theta})=|z|^\frac{1}{d}\exp^{i\frac{\theta}{d}+(k-1)i\frac{\pi}{d}}\text{ for }k=1,3,5,\dots,2d-1,
+	f_k(|z|\exp^{i\theta})&=|z|^\frac{1}{d}\exp{\left(i\frac{\theta}{d}+(k-1)i\frac{\pi}{d}\right)}\text{ for }k=1,3,5,\dots,2d-1,\\
-\end{equation}
+	f_k(|z|\exp^{i\theta})&=|z|^\frac{1}{d}\exp{\left(i\frac{\theta-\pi}{d}+(k-1)i\frac{\pi}{d}\right)}\text{ for }k=2,4,6,\dots,2d.
-\begin{equation}
+\end{align*}
-	f_k(|z|\exp^{i\theta})=|z|^\frac{1}{d}\exp^{i\frac{\theta-\pi}{d}+(k-1)i\frac{\pi}{d}}\text{ for }k=2,4,6,\dots,2d
-\end{equation}
 We will also need the inverses of these homeomorphisms, which are:
-\begin{equation}
+\begin{align*}
-	f_k^{-1}(|z|\exp^{i\theta})=|z|^d\exp^{id(\theta-(k-1)\frac{\pi}{d})}\text{ for }k=1,3,5,\dots,2d-1,
+	f_k^{-1}(|z|\exp^{i\theta})&=|z|^d\exp{\left(id\left(\theta-(k-1)\frac{\pi}{d}\right)\right)}\text{ for }k=1,3,5,\dots,2d-1,\\
-\end{equation}
+	f_k^{-1}(|z|\exp^{i\theta})&=|z|^d\exp{\left(id\left(\theta-\pi-(k-1)i\frac{\pi}{d}\right)\right)}\text{ for }k=2,4,6,\dots,2d.
-\begin{equation}
+\end{align*}
-	f_k^{-1}(|z|\exp^{i\theta})=|z|^d\exp^{id(\theta-\pi-(k-1)i\frac{\pi}{d})}\text{ for }k=2,4,6,\dots,2d
-\end{equation}
-
 Then, we define $G=f_k\circ\bar{r}\circ f_k^{-1}$, where $k$ depends on the point $z$ to which we will apply it. For $z\in\CC$, given by $z=|z|\exp^{i\theta}$ for $\theta\in[(k-1)\frac{\pi}{d}, k\frac{\pi}{d}]$, we have:
 \begin{equation}
-\begin{array}{ccc}
+\begin{aligned}
-	G(z,t)&=&f_k(\bar{r}(f_k^{-1}(z),t))\\
+	G(z,t)&=f_k\left(\bar{r}(f_k^{-1}(z),t)\right)\\
-	&=&f_k(\bar{r}(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},t))\\
+	&=f_k\left(\bar{r}(|z^d|\exp^{id(\theta-(k-1)\frac{\pi}{d})},t)\right)\\
-	&=&f_k(+\sqrt{(z^d)_r^2+(1-t)^2(z^d)_i^2}\exp^{i\arctan((1-t)\tan(d(\theta-(k-1)\frac{\pi}{d})))})\\
+	&=f_k\left(\sqrt{(z^d)_r^2+(1-t)^2(z^d)_i^2}\, \exp{\left(i\arctan\left((1-t)\tan\left(d\left(\theta-(k-1)\frac{\pi}{d}\right)\right)\right)\right)}\right)\\
-	&=&+\sqrt[2d]{(z^d)_r^2+(1-t)^2(z^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))}
+	&=\sqrt[2d]{(z^d)_r^2+(1-t)^2(z^d)_i^2}\,
-\end{array}
+	\exp{\left(\frac{i}{d}\arctan\left((1-t)\tan\left(d\theta\right)\right)\right)}
+\end{aligned}
 \label{Gliftedr1}
 \end{equation}
-
 Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since:
-\begin{equation}
+\begin{equation*}
-\begin{array}{ccc}
+\begin{tikzcd}[column sep=0pt, row sep=5pt]
-	G(z,t)^d&=&m(z,t)^d\exp^{i\arctan((1-t)\tan d\theta)}\\
+G(z,t)^d\dar[equal, equal]&=&m(z,t)^d\exp{\left(i\arctan((1-t)\tan d\theta)\right)}\\
-	||& & \\
+\bar{r}(z^d,t)&=&m(z^d,t)\exp{(i\arctan((1-t)\tan d\theta))}\
-	\bar{r}(z^d,t)&=&m(z^d,t)\exp^{i\arctan((1-t)\tan d\theta)}\\
+\end{tikzcd}
-\end{array}
+\end{equation*}
-\end{equation}
+We can then write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
-We can the write the lifting of $r_1$ as $\tilde{r}_1((x_0,\dots,x_n),t)=(G(x,t), \dots, G(x_n,t))$.
 
 
 Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d_0, \ldots , x^d_n\in\RR\}$ into $S_d$.
@@ -385,11 +394,11 @@ For a point $P=(x_0\xi_0,\dots,x_n\xi_n)\in M_d\cap\{x^d_0, \ldots , x^d_n\in\RR
 \begin{equation}
 P^d_Q=\left(\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}, \dots,\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)} \right)
 \end{equation}
-In the $(+,+,+)$ case, $r_2$ is the constant path, and therefore the lift $\tilde{r}_2$ will also be the constant path.
+In the $(+,\dots,+)$ case, $r_2$ is the constant path, and therefore the lift $\tilde{r}_2$ will also be the constant path.
 
-In the rest of the cases, the lift will be given by:
+In the rest of the cases, the lift will be given by $\tilde{r}_2((x_0\xi_0,\dots,x_n\xi_n), t)$ equals
 \[
-\tilde{r}_2((x_0\xi_0,\dots,x_n\xi_n), t)=\left(\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0,\dots,\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right)
+\left(\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0,\dots,\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right).
 \]
 
 
@@ -399,7 +408,7 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
 \end{proof}
 
 \begin{theorem}
- The strong retraction from $M_d$ to $S_d$ induces a strong retraction, and therefore an homotopy equivalence, between $\CC\PP^n\setminus C_d$ and $K$.
+ The strong retraction from $M_d$ to $S_d$ induces a strong retraction, and therefore, a homotopy equivalence between $\CC\PP^n\setminus C_d$ and $K$.
 \end{theorem}
 \begin{proof}
 	The path followed by each point in $\CC\PP^n\setminus C_d$ in the retraction will be given by the path followed by a representative in $M_d$ as it is retracted to $S_d$. The class representatives we choose in $\CC\PP^n\setminus C_d$ will be such that the coordinates to the $d$ power add up to $1$. Since this choice is not unique, we are now going to see that if we apply the retraction $r=r_2\circ r_1$ to two such representatives, they are sent to the same class by $r$.
@@ -407,43 +416,36 @@ We have then proved that $S_d$ is a strong deformation retract of $M_d$.
 	Let $p=[x_0:\dots:x_n]\in\CC\PP^n\setminus C_d$ and $\lambda p=[\lambda x_0:\dots: \lambda x_n]$ be two representatives of some point $p$, since both representatives must be such that $\sum x^d_i=1$, necessarily $\lambda$ is a $d-$th root of unity.
 
 	First, we see how $r_1$ acts on these representatives.
-	\begin{equation}
+	\begin{align*}
-		\begin{array}{lll}
+			\tilde{r}_1(p,t) & =  [G(x_0,t): \dots: G(x_n,t)], \\
-			\tilde{r}_1(p,t) & = & [G(x_0,t): \dots: G(x_n,t)] \\
+			\tilde{r}_1(\lambda p,t) & =  [G(\lambda x_0,t): \dots: G(\lambda x_n,t)].
-			\tilde{r}_1(\lambda p,t) & = & [G(\lambda x_0,t): \dots: G(\lambda x_n,t)] \\
+	\end{align*}
-		\end{array}
-	\end{equation}
 	If we define $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$ for $x\in\CC$, we get that
-	\begin{equation}
+	\begin{equation*}
 		\theta(\lambda x,t)=\arctan\frac{(1-t)\lambda x_i}{\lambda x_r}=\arctan\frac{(1-t)x_i}{x_r}=\theta(x,t)
-	\end{equation}
+	\end{equation*}
-	By \ref{Gliftedr1}:
+	By \eqref{Gliftedr1}:
-	\begin{equation}
+	\begin{align*}
-		\begin{array}{lll}
+			G(x,t) & =  \sqrt[2d]{(x^d)_r^2+(1-t)^2(x^d)_i^2}\exp{\left(\frac{i}{d}\arctan((1-t)\tan(d\theta)))\right)}  \\
-			G(x,t) & = & +\sqrt[2d]{(x^d)_r^2+(1-t)^2(x^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))}  \\
+			G(\lambda x,t) & =  \sqrt[2d]{(\lambda^d x^d)_r^2+(1-t)^2(\lambda^d x^d)_i^2}\exp{\left(\frac{i}{d}\arctan((1-t)\tan(d\theta)))\right)} \\
-			G(\lambda x,t) & = & +\sqrt[2d]{(\lambda^d x^d)_r^2+(1-t)^2(\lambda^d x^d)_i^2}\exp^{\frac{i}{d}\arctan((1-t)\tan(d\theta)))} \\
+	\end{align*}
-		\end{array}
-	\end{equation}
 	Since $\lambda$ is a $d-$th root of unity, we get that $G(x,t)=G(\lambda x,t)$. Therefore, we get that $\tilde{r}_1(p,t)=\tilde{r}_1(\lambda p,t)$. It's just left to check that $r_2$ also sends them to the same class.
 
 	If $[x^d_0: \dots: x^d_n]$ is in $(+,\dots,+)$, the retraction is the identity, and therefore both representatives will stay the same and therefore go to the same class.
 
 	For the rest of the  regions, we have
-\begin{equation}
+\begin{equation*}
-\begin{array}{lll}
+\tilde{r}_2([x_0\xi_0:\dots:x_n\xi_n], t) =\left[H(x_0,\chi_0)\xi_0:\dots:H(x_n,\chi_n) \xi_n\right]
-\tilde{r}_2([x_0\xi_0:\dots:x_n\xi_n], t)& =& \left[\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right]
+\end{equation*}
-\end{array}
+where
-\end{equation}
+\[
-
+H(x_i, \chi_i):=\sqrt[d]{(1-t)x^d_i+t\frac{x^d_i\chi_i}{1-\sum x^d_j(1-\chi_j)}},\quad i=0,\dots,n.
-
+\]
 Then, for the point $\lambda p$ we get
-\begin{equation}
+\begin{align*}
-\begin{array}{lll}
+\tilde{r}_2([\lambda x_0\xi_0:\dots:\lambda x_n\xi_n], t)&=\left[H(\lambda x_0, \chi_0)\xi_0:\dots:H(\lambda x_n, \chi_n)\xi_n\right]\\ 
-\tilde{r}_2([\lambda x_0\xi_0:\dots:\lambda x_n\xi_n], t)&=&\left[\sqrt[d]{(1-t)\lambda^d x^d_0+t\frac{\lambda^d x^d_0\chi_0}{1-\sum \lambda^d x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)\lambda^d x^d_n+t\frac{\lambda^d x^d_n\chi_n}{1-\sum \lambda^d x^d_j(1-\chi_j)}}\xi_n\right]\\
+&=\left[H( x_0, \chi_0)\xi_0:\dots:H(x_n, \chi_n)\xi_n\right].
-&=&\left[\sqrt[d]{(1-t)x^d_0+t\frac{x^d_0\chi_0}{1-\sum x^d_j(1-\chi_j)}}\xi_0:\dots:\sqrt[d]{(1-t)x^d_n+t\frac{x^d_n\chi_n}{1-\sum x^d_j(1-\chi_j)}}\xi_n\right]\\
+\end{align*}
-\end{array}
-\end{equation}
-
 So we see that $\tilde{r}_2$ also takes both representatives will end to the same point.
 \end{proof}