diff --git a/lec_4.tex b/lec_4.tex
index 60a9465..7d0771d 100644
--- a/lec_4.tex
+++ b/lec_4.tex
@@ -47,8 +47,8 @@ $K \# m(K) \sim $ the unknot.
 \end{fact}
 \noindent
 \begin{theorem}
-Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $\{0\}$ denote class of all knots concordant to a trivial knot. 
-$\mathscr{C}$ is a group under taking connected sums. The neutral element in the group is $\{0\}$ and the inverse element of an element $\{K\} \in \mathscr{C}$ is $-\{K\} = \{mK\}$.
+Let $\mathscr{C}$ denote a set of all equivalent classes for knots and $[0]$ denote class of all knots concordant to a trivial knot. 
+$\mathscr{C}$ is a group under taking connected sums. The neutral element in the group is $[0]$ and the inverse element of an element $[K]\in \mathscr{C}$ is $-[K] = [mK]$.
 \end{theorem}
 \begin{fact}
 The figure eight knot is a torsion element in $\mathscr{C}$ ($2K \sim $ the unknot).
@@ -70,7 +70,7 @@ Remark: $K \sim K^{\prime} \Leftrightarrow K \# -K^{\prime}$ is slice.
 \end{figure}
 \noindent
 \\
-Pontryagin-Thom construction tells us that there exists a compact three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
+Pontryagin-Thom construction tells us that there exists a compact oriented three - manifold $\Omega \subset B^4$ such that $\partial \Omega = Y$.
 Suppose $\Sigma$ is a Seifert surface and $V$ a Seifert form defined on $\Sigma$: ${(\alpha, \beta) \mapsto \Lk(\alpha, \beta^+)}$. Suppose $\alpha, \beta \in H_1(\Sigma, \mathbb{Z})$, i.e. there are cycles and
 $\alpha, \beta \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$. Then there are two cycles $A, B \in \Omega$ such that $\partial A = \alpha$ and $\partial B = \beta$. 
 Let $B^+$ be a push off of $B$ in the positive normal direction such that 
@@ -78,16 +78,13 @@ $\partial B^+ = \beta^+$.
 Then 
 $\Lk(\alpha, \beta^+) = A \cdot B^+$. But $A$ and $B$ are disjoint, so $\Lk(\alpha, \beta^+) = 0$. Then the Seifert form is zero. 
 \\
-????????????????? 
-\\
+\noindent
 Let us consider following maps:
 \[
 \Sigma \overset{\phi} \longhookrightarrow Y \overset{\psi} \longhookrightarrow \Omega.
 \]
 Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $\gamma \in \ker (H_1(\Sigma, \mathbb{Z}) \longrightarrow H_1(\Omega, \mathbb{Z}))$, then $\gamma \in \ker \phi_*$ or $\gamma \in \ker \psi_*$.
 %
-\\
-????????????\\
 %
 %
 \begin{proposition}
@@ -97,20 +94,76 @@ Let $\phi_*$ and $\psi_*$ be induced maps on the homology group. If an element $
 where $b_1$ is first Betti number.
 \end{proposition}
 \begin{proof}
+Consider the following long exact sequence for a pair $(\Omega, Y)$:
 \begin{align*}
 & 0 \to  H_3(\Omega) \to H_3(\Omega, Y) \to 
 \\ 
 \to & H_2(Y) \to H_2(\Omega) \to H_2(\Omega, Y) \to \\
-\to & H_1(Y) \to \H_1(\Omega) \to H_1(\Omega, Y) \to \\
+\to & H_1(Y) \to H_1(\Omega) \to H_1(\Omega, Y) \to \\
 \to & H_0(Y) \to H_0(\Omega) \to 0
 \end{align*}
-\end{proof}
+By Poincar\'e duality we know that:
+\begin{align*}
+H_3(\Omega, Y) &\cong H^0(\Omega),\\
+H_2(Y) &\cong H^0(Y),\\
+H_2(\Omega) &\cong H^1(\Omega, Y),\\
+H_1(\Omega, Y) &\cong H^1(\Omega).
+\end{align*}
+Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
+= \dim_{\mathbb{Q}} V
+$.\\
+\noindent
+Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on a $K$
+has a subspace of dimension $g_{\Sigma}$ on which it is zero:
+
+\begin{align*}
+\newcommand\coolover[2]%
+    {\mathrlap{\smash{\overbrace{\phantom{%
+            \begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
+\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
+    \begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
+\newcommand\coolleftbrace[2]{%
+    #1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
+\newcommand\coolrightbrace[2]{%
+     \left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
+ \vphantom{% phantom stuff for correct box dimensions
+    \begin{matrix}
+    \overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
+    \underbrace{pqr}_{\mbox{$S$}}
+    \end{matrix}}%
+ V =
+\begin{matrix}% matrix for left braces
+    \coolleftbrace{g_{\Sigma}}{ \\  \\ \\}
+     \\ \\ \\ \\ 
+\end{matrix}%
+\begin{pmatrix}
+    \coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\
+    \sdots &  & \sdots & \sdots &  & \sdots \\
+    0 & \dots & 0 & * & \dots & *\\
+    * & \dots & * & * & \dots & *\\
+   \sdots &  & \sdots & \sdots &  & \sdots \\
+     * & \dots & * & * & \dots & * 
+ \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
+\end{align*}
+\end{proof}
+\noindent
+Let $V = 
+\begin{pmatrix}
+          0 & A\\
+          B & C
+\end{pmatrix}$
+\begin{align*}
+\det (tV - V^T) = \det (tA - B^T) - \det(tB - A^T) 
+\end{align*}
+\begin{corollary}
+If $K$ is a slice knot then there exists $f \in \mathbb{Z}[t^{\pm 1}]$ such that $\Delta_K(t) = f(t) \cdot f(t^{-1})$.
+\end{corollary}
+\begin{example}
+Figure eight knot is not slice.
+\end{example}
+\begin{fact}
+If $K$ is slice, then the signature $\sigma(K) \equiv 0$.
+\end{fact}
+
+
 
-\begin{figure}[h]
-\fontsize{10}{10}\selectfont
-\centering{
-\def\svgwidth{\linewidth}
-\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
-}
-%\caption{Sketch for Fact %%\label{fig:concordance_m}
-\end{figure}
diff --git a/lectures_on_knot_theory.tex b/lectures_on_knot_theory.tex
index 038146d..ac1ba7b 100644
--- a/lectures_on_knot_theory.tex
+++ b/lectures_on_knot_theory.tex
@@ -200,59 +200,7 @@ Let $A$ and $B$ be closed, oriented surfaces in $X$.
 \begin{proposition}
 $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes. 
 %$A \cdot B$  gives the pairing as ??
- 
-\end{proposition}
-\begin{proof}
-
-By Poincar\'e duality we know that:
-\begin{align*}
-H_3(\Omega, Y) &\cong H^0(\Omega),\\
-H_2(Y) &\cong H^0(Y),\\
-H_2(\Omega) &\cong H^1(\Omega, Y),\\
-H_2(\Omega, Y) &\cong H^1(\Omega).
-\end{align*}
-Therefore $\dim_{\mathbb{Q}} \quot{H_1(Y)}{V}
-= \dim_{\mathbb{Q}} V
-$.\\
-\noindent
-Suppose $g(K) = 0$ ($K$ is slice). Then $H_1(\Sigma, \mathbb{Z}) \cong H_1(Y, \mathbb{Z})$. Let $g_{\Sigma}$ be the genus of $\Sigma$, $\dim H_1(Y, \mathbb{Z}) = 2g_{\Sigma}$. Then the Seifert form $V$ on a $K$
-has a subspace of dimension $g_{\Sigma}$ on which it is zero:
-
-\begin{align*}
-\newcommand\coolover[2]%
-    {\mathrlap{\smash{\overbrace{\phantom{%
-            \begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
-\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
-    \begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
-\newcommand\coolleftbrace[2]{%
-    #1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
-\newcommand\coolrightbrace[2]{%
-     \left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
- \vphantom{% phantom stuff for correct box dimensions
-    \begin{matrix}
-    \overbrace{XYZ}^{\mbox{$R$}}\\ \\ \\ \\ \\ \\
-    \underbrace{pqr}_{\mbox{$S$}}
-    \end{matrix}}%
- V =
-\begin{matrix}% matrix for left braces
-    \coolleftbrace{g_{\Sigma}}{ \\  \\ \\}
-     \\ \\ \\ \\ 
-\end{matrix}%
-\begin{pmatrix}
-    \coolover{g_{\Sigma}}{0 & \dots & 0 } & * & \dots & *\\
-    \sdots &  & \sdots & \sdots &  & \sdots \\
-    0 & \dots & 0 & * & \dots & *\\
-    * & \dots & * & * & \dots & *\\
-   \sdots &  & \sdots & \sdots &  & \sdots \\
-     * & \dots & * & * & \dots & * 
- \end{pmatrix}_{2g_{\Sigma} \times 2g_{\Sigma}}
-\end{align*}
-
-
-\end{proof}
-
-
-
+ \end{proposition}
 
 
 \section{\hfill\DTMdate{2019-04-15}}
@@ -776,5 +724,15 @@ The proof is the same as over $\mathbb{Z}$.
 
 %Lectures on the Topology of 3-Manifolds
 %An Introduction to the Casson Invariant
+
+\begin{figure}[h]
+\fontsize{10}{10}\selectfont
+\centering{
+\def\svgwidth{\linewidth}
+\resizebox{0.5\textwidth}{!}{\input{images/ball_4_alpha_beta.pdf_tex}}
+}
+%\caption{Sketch for Fact %%\label{fig:concordance_m}
+\end{figure}
+
 \end{document}