Five Lemma.tex

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\title{Homological Algebra: Five Lemma}
\author{Luis R. Guzman, Jr.}
\date{15 February 2012}
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\frametitle{Overview}
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\begin{itemize}
\item Prerequisites
\item Five Lemma
\item Proof of the Five Lemma
\end{itemize}
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\frametitle{Prerequisites}
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\begin{itemize}
\item Modules
\item Free Modules
\item Exact Sequences
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\frametitle{Five Lemma}
Lemma (Five Lemma) Consider a commutative diagram
\textbf{ }\[
\xymatrix{A_1\ar^{f_1}[r]\ar^{t_1}[d] & A_2\ar^{f_2}[r]\ar^{t_2}[d] & A_3\ar^{f_3}[r]\ar^{t_3}[d] & A_4\ar^{f_4}[r]\ar^{t_4}[d] & A_5\ar^{t_5}[d]\\
B_1\ar_{g_1}[r] & B_2\ar_{g_2}[r] & B_3\ar_{g_3}[r] & B_4\ar_{g_4}[r] & B_5}
\]
of R-Modules and homomorphisms with exact rows. (i) If $t_2$ and $t_4$ are epimorphisms and $t_5$  is a monomorphism then $t_3$ is a epimorphism.
(ii) If $t_2$ and $t_4$ are monomorphisms and $t_1$ is an epimorphism then $t_3$ is a monomorphism.
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\frametitle{Proof of the Five Lemma}
\textbf{Proof}
(i) Suppose that $t_2,t_4$ are epimorphisms and $t_5$ is a monomorphism. Let $b_3\in B_3$. Then $g_3(b_3)\in B_4$ and since $t_4$ is an epimorphism, there exists an element $a_4\in A_4$ such that $g_3(b_3)=t_4(a_4)$. Now $t_5f_4(a_4)=$ $g_4t_4(a_4)=$ $g_4g_3(b_3)=0$ and $t_5$ is a monomorphism. Therefore, $f_4(a_4)=0$. The upper row being exact, there exists an element $a_3\in A_3$ such that $f_3(a_3)=a_4$.  Then $g_3(b_3)= $ $t_4(a_4)=t_4f_3(a_3)=g_3t_3(a_3)$ and so, $g_3(b_3-t_3(a_3))=0$. Therefore, there exists an element $b_2\in B_2$ such that $b_3-t_3(a_3)=g_2(b_2)$ . 
The homomorphism $t_2$ being an epimorphism, there exists an $a_2\in A_2$ such that $b_2=t_2(a_2)$. But then $b_3-t_3(a_3)=g_2(b_2)=g_2(t_2(a_2))=$ $(g_2t_2)(a_2)=t_3f_2(a_2)$  or $b_3=t_3f_2(a_2)+t_3(a_3)=t_3(a_3+f_2(a_2))$. Hence $t_3$ is an epimorphism. \\
(ii) Now suppose that $t_2,t_4$ are monomorphisms and $t_1$ is an epimorphism. Let $a_3\in A_3$ such that $t_3(a_3)=0$. Then
\begin{equation}
0=g_3t_3(a_3)=t_4f_3(a_3).
\end{equation}
$t_4$ being a monomorphism, we have $f_3(a_3)=0$ and, therefore, there exists an element $a_2\in A_2$ such that $a_3=f_2(a_2)$. But then $g_2t_2(a_2)=t_3f_2(a_2)=t_3(a_3)=0$ and, therefore there exists an element $b_1\in B_1$ such that $t_2(a_2)=g_1(b_1)$. Since $t_1$ is an epimorphism, there exists an element $a_1\in A_1$, such that $t_1(a_1)=b_1$. Then $t_2(a_2)=$ $g_1(b_1)=g_1t_1(a_1)=t_2f_1(a_1)$. \\
Now $t_2$ being a monomorphism, we have $a_2=f_1(a_1)$ and, hence, $a_3=f_2(f_1(a_1))=0$ Hence $t_3$ is a monomorphism. $\Box$
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