diff --git a/generaldegree.tex b/generaldegree.tex
index e612788..cba54be 100644
--- a/generaldegree.tex
+++ b/generaldegree.tex
@@ -4,6 +4,7 @@
 \usepackage{amsthm}
 \usepackage{amsmath}
 \usepackage{tikz}
+\usepackage[new]{old-arrows}
 
 \usetikzlibrary{cd, positioning, arrows.meta, decorations.pathmorphing, shapes.multipart,
 decorations.markings, calc}
@@ -53,9 +54,9 @@ We will denote:
 We then have a commutative diagram:
 \begin{equation*}
 	\begin{array}{ccc}
-		S_d & \longrightarrow & M_d \\
+		S_d & \longhookrightarrow & M_d \\
 		\downarrow & & \downarrow \\
-		K & \longrightarrow & \PP^2\setminus C_d \\
+		K & \longhookrightarrow & \PP^2\setminus C_d \\
 	\end{array}
 \end{equation*}
 
@@ -205,7 +206,7 @@ similarly to the previous case, in this line $x=1-z$.
 \end{proof}
 
 
-Now that we have the retractions that allow us to retract $M_1$ into $S_1$, we will try to lift them in order to get a retraction from $M_d$ to $S_d$.
+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 $M_d$.
@@ -250,7 +251,7 @@ we can then write the retraction as $\bar{r}(x,t)=m(x,t)e^{2i\pi\theta(x,t)}$ wh
 
 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 $\theta(x,t)=\arctan\frac{(1-t)x_i}{x_r}$.
 
-We can divide $\CC$ in $2d$ sectors of angle $\pi/d$ as in figure \ref{divisionC}.
+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.
 
 \begin{figure}[h]
 \begin{center}
@@ -317,7 +318,7 @@ Then, we can see that $G(z,t)^d=\bar{r}(z^d,t)$, since:
 	\bar{r}(z^d,t)&=&m(z^d,t)\exp^{i\arctan((1-t)\tan d\theta)}\\
 \end{array}
 \end{equation}
-
+We can the write the lifting of $r_1$ as $\tilde{r}_1((x,y,z),t)=(G(x,t), G(y,t), G(z,t))$.
 
 
 Lastly, we are going to see how to lift $r_2$ in order to get a retraction from $M_d\cap\{x^d, y^d, z^d\in\RR\}$ into $S_d$.
@@ -417,9 +418,9 @@ We denote as $\Delta$ the $\Delta-$complex $[v_0, v_1, v_2]$ that has $3$ vertic
 	\begin{array}{ccccc}
 		\Delta & \longleftrightarrow & S_1 & & \\
 		\uparrow& & \uparrow & & \\
-		\Delta^d & \longleftrightarrow & S_d & \longrightarrow & M_d \\
+		\Delta^d & \longleftrightarrow & S_d & \hookrightarrow & M_d \\
 		& & \downarrow & & \downarrow \\
-		& & K & \longrightarrow & \PP^2\setminus C_d \\
+		& & K & \hookrightarrow & \PP^2\setminus C_d \\
 	\end{array}
 \end{equation*}
 where $\Delta^d$ is a $\Delta-$complex that we are trying to find and identify with $S_d$, equivalently as what we have done for $d=1$. Since $S_d$ is a ramified cover of $S_1$ by the power to the $d$ map, we want $\Delta^d$ to be a ramified cover of $\Delta$ by the same map.