Four Lemma

前面一篇 Splitting Lemma 中用到 Short Five Lemma, 而 Short Five Lemma (對於 i = 1, 3A_{i} \rightarrow B_{i} 為 isomorphism 時,  0 \rightarrow A_{1} \overset{f_1}{\rightarrow} A_{2} \overset{f_2}{\rightarrow} A_{3} \rightarrow 0 同構於 0 \rightarrow B_{1} \overset{g_1}{\rightarrow} B_{2} \overset{g_2}{\rightarrow} B_{3} \rightarrow 0) 其實是 Five Lemma (左右兩邊為一般的 module, 左上有一 epimorphism 對至左下, 右上有 monomorphism 對至右下, 其餘不變) 的特例。而欲證明 Five Lemma 其實用 Four Lemma (右上有 mono 對到右下) 加上其對偶得證, 也就是說, 只要證明 Four Lemma 就可以得到論述麻煩的 Five 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。
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