State and prove the injective and surjective forms of the four lemma for commutative diagrams of modules with exact rows.
Solution
Consider a commutative diagram with exact rows \[ \begin{array}{ccccccccc} A_1&\to&A_2&\to&A_3&\to&A_4\\ \downarrow f_1&&\downarrow f_2&&\downarrow f_3&&\downarrow f_4\\ B_1&\to&B_2&\to&B_3&\to&B_4. \end{array} \]
1. Injective four lemma: if \(f_1\) is surjective and \(f_2,f_4\) are injective, then \(f_3\) is injective.
Let \(x\in A_3\) with \(f_3(x)=0\). Its image in \(A_4\) maps under \(f_4\) to zero by commutativity. Since \(f_4\) is injective, the image of \(x\) in \(A_4\) is zero. Exactness gives \(y\in A_2\) mapping to \(x\).
The element \(f_2(y)\) maps to \(f_3(x)=0\), so exactness in the bottom row gives \(b\in B_1\) mapping to \(f_2(y)\). Since \(f_1\) is surjective, choose \(a\in A_1\) with \(f_1(a)=b\). By commutativity, \[ f_2\bigl(y-\operatorname{im}(a)\bigr)=0. \] Injectivity of \(f_2\) implies \(y=\operatorname{im}(a)\), hence \(x=0\). Therefore \(f_3\) is injective.
For the surjective form, consider exact rows \[ \begin{array}{ccccccccc} A_2&\to&A_3&\to&A_4&\to&A_5\\ \downarrow f_2&&\downarrow f_3&&\downarrow f_4&&\downarrow f_5\\ B_2&\to&B_3&\to&B_4&\to&B_5. \end{array} \]
2. Surjective four lemma: if \(f_2,f_4\) are surjective and \(f_5\) is injective, then \(f_3\) is surjective.
Take \(y\in B_3\) and let \(y_4\) be its image in \(B_4\). Choose \(a_4\in A_4\) with \(f_4(a_4)=y_4\). The image of \(a_4\) in \(A_5\) maps under \(f_5\) to zero, because \(y_4\) maps to zero by exactness. Since \(f_5\) is injective, \(a_4\) maps to zero. Exactness gives \(a_3\in A_3\) mapping to \(a_4\).
Then \(f_3(a_3)-y\) maps to zero in \(B_4\), so exactness gives \(b_2\in B_2\) mapping to \(f_3(a_3)-y\). Choose \(a_2\in A_2\) with \(f_2(a_2)=b_2\), and let \(a_3'\) be the image of \(a_2\) in \(A_3\). By commutativity, \[ f_3(a_3')=f_3(a_3)-y. \] Hence \[ f_3(a_3-a_3')=y. \] Thus \(f_3\) is surjective.