# Four Lemma

Four Lemma. Let $A_{1} \overset{f_1}{\rightarrow} A_{2} \overset{f_2}{\rightarrow} A_{3} \overset{f_3}{\rightarrow} A_{4}$, $B_{1} \overset{g_1}{\rightarrow} B_{2} \overset{g_2}{\rightarrow} B_{3} \overset{g_3}{\rightarrow} B_{4}$ be exact sequences with $\alpha_{i} : A_{i} \rightarrow B_{i}$ and the diagram commutes. If $\alpha_{1}, \alpha_{3}$ are epic, and $\alpha_{4}$ is mono, then $\alpha_{2}$ is epic.

Proof.

For convenience, denote $\alpha_{i}(a_{i})$ by $\alpha (a_{i})$. Let $b_{2} \in B_{2}$, we want to find $a \in A_{2}$ such that $\alpha(a) = b_{2}$.

Since $g(b_2) \in B_{3}$, there is $a_{3} \in A_{3}$ with $\alpha(a_{3}) = g(b_{2})$ by $\alpha$ is epic.

Since $g \alpha (a_3) = \alpha f (a_{3})$ by commutativity and  $g_{3} \alpha_{3} = g_{3} g_{2} = 0$ by exactness, we have $f (a_{3}) = 0$ by $\alpha_{4}$ is mono and there is $a_{2} \in A_{2}$ with $f(a_{2}) = a_{3}$ by exactness.

Since $g (\alpha (a_{2}) - b_{2}) = g_2 \alpha (a_{2}) - g (b_2) = g_2 \alpha (a_{2}) - \alpha f (a_{2}) = 0$ (the last equation is due to the commutativity), we have $b_1 \in B_{1}$ with $g (b_1) = \alpha (a_{2}) - b_{2}$ by exactness.

Finally, we have $a_{1} \in A_{1}$ with $\alpha (a_{1}) = b_{1}$ by $\alpha_{1}$ is epi and we could easily check  that $b_2 = \alpha (a_{2} - f (a_1))$.

Remark.

• 其對偶論述則是 $\alpha_{2}, \alpha_{4}$ are mono, and $\alpha_{1}$ is epic, then $\alpha_{3}$ is mono.
• 而 Five Lemma 則是根據 isomorphism 即為同時 epi 且 mono，利用最左邊的 $\alpha_{1}$ 是 epimorphism 而最右邊是 $\alpha_{5}$ 是 monomorphism 這兩個條件，分別得到 $\alpha_{3}$ 是 epic 且 mono 得證。
• Short Five Lemma 很明顯當 $A_{1} = B_{1} = A_{5} = B_{5} = 0$ 時, 所需要的條件因為是一樣的 module 而得到 trivial isomorphism。