documentclass{beamer}
usecolortheme{wolverine}
beamertemplatenavigationsymbolsempty
usepackage{tikz}
usetikzlibrary{matrix,arrows,positioning}begin{frame}{Conclusion}

From a fibration we obtain a long exact sequence

begin{center}
begin{tikzpicture}
matrix[matrix of nodes,ampersand replacement=&, column sep=0.5cm, row sep=0.5cm](m)
{
 & $cdots$ & $pi_{n+1}(B)$ \
$pi_{n}(F)$ & $pi_{n}(E)$ & $pi_{n}(B)$ \
$pi_{n-1}(F)$ & $cdots$ &  \
};
draw[->] (m-1-2) edge (m-1-3)
          (m-1-3) edge[out=0, in=180] (m-2-1)
          (m-2-1) edge (m-2-2)
          (m-2-2) edge (m-2-3)
          (m-2-3) edge[out=0, in=180] (m-3-1)
          (m-3-1) edge (m-3-2);
end{tikzpicture}
end{center}

i.e., the image of one homomorphism is equal to the kernel of the next one.
end{frame}