Let the following be a commutative diagram of \(R\)-modules and \(R\)-module homomorphisms with exact rows:
\[ \begin{array}{ccccccccc} A_1&\xrightarrow{f_1}&A_2&\xrightarrow{f_2}&A_3&\xrightarrow{f_3}&A_4&\xrightarrow{f_4}&A_5\\ \downarrow\alpha_1&&\downarrow\alpha_2&&\downarrow\alpha_3&&\downarrow\alpha_4&&\downarrow\alpha_5\\ B_1&\xrightarrow{g_1}&B_2&\xrightarrow{g_2}&B_3&\xrightarrow{g_3}&B_4&\xrightarrow{g_4}&B_5. \end{array} \]
Prove the following:
If \(\alpha_1\) is an epimorphism and \(\alpha_2, \alpha_4\) are monomorphisms then \(\alpha_3\) is a monomorphism.
If \(\alpha_5\) is a monomorphism and \(\alpha_2, \alpha_4\) are epimorphisms then \(\alpha_3\) is an epimorphism.
Solution
1. If \(\alpha_1\) is surjective and \(\alpha_2,\alpha_4\) are injective, then\(\alpha_3\) is injective.
Let \(x\in A_3\) satisfy \(\alpha_3(x)=0\). Commutativity gives \[ \alpha_4(f_3(x))=g_3(\alpha_3(x))=0. \] Since \(\alpha_4\) is injective, \(f_3(x)=0\). Exactness of the top row at \(A_3\) therefore gives \(y\in A_2\) such that \[ f_2(y)=x. \]
Now \[ g_2(\alpha_2(y)) =\alpha_3(f_2(y)) =\alpha_3(x) =0. \] By exactness of the bottom row at \(B_2\), there is \(b_1\in B_1\) with \[ g_1(b_1)=\alpha_2(y). \] Since \(\alpha_1\) is surjective, choose \(a_1\in A_1\) such that \(\alpha_1(a_1)=b_1\). Commutativity then yields \[ \alpha_2(f_1(a_1)) =g_1(\alpha_1(a_1)) =g_1(b_1) =\alpha_2(y). \] Injectivity of \(\alpha_2\) gives \(f_1(a_1)=y\). Hence \[ x=f_2(y)=f_2(f_1(a_1))=0 \] by exactness of the top row. Thus \(\ker\alpha_3=0\) and \(\alpha_3\) is injective.
2. If \(\alpha_5\) is injective and \(\alpha_2,\alpha_4\) are surjective, then\(\alpha_3\) is surjective.
Let \(b_3\in B_3\). Since \(\alpha_4\) is surjective, choose \(a_4\in A_4\) such that \[ \alpha_4(a_4)=g_3(b_3). \] Then \[ \alpha_5(f_4(a_4)) =g_4(\alpha_4(a_4)) =g_4(g_3(b_3)) =0. \] Because \(\alpha_5\) is injective, \(f_4(a_4)=0\). Exactness of the top row at \(A_4\) gives \(a_3\in A_3\) with \[ f_3(a_3)=a_4. \]
Set \[ \delta=\alpha_3(a_3)-b_3\in B_3. \] Then \[ g_3(\delta) =\alpha_4(f_3(a_3))-g_3(b_3) =\alpha_4(a_4)-g_3(b_3) =0. \] By exactness of the bottom row at \(B_3\), choose \(b_2\in B_2\) such that \[ g_2(b_2)=\delta. \] Since \(\alpha_2\) is surjective, choose \(a_2\in A_2\) with \(\alpha_2(a_2)=b_2\). Therefore \[ \alpha_3(a_3-f_2(a_2)) =\alpha_3(a_3)-g_2(\alpha_2(a_2)) =\alpha_3(a_3)-\delta =b_3. \] Thus every \(b_3\in B_3\) lies in the image of \(\alpha_3\), so \(\alpha_3\) is surjective.