Splitting Lemma

嚇死我,差點證不出來。口語解讀是,如果我在 short exact sequence 中找得到一個反向的 morphism 促使正向的 morphism 是 left(or right) split 的話,那麼這串 sequence 就可以寫成前後兩個 module 的 direct sum。

Splittin Lemma. Let 0 \rightarrow A_{1} \overset{f}{\rightarrow} B \overset{g}{\rightarrow} A_{2} \rightarrow 0 be a short exact sequence of R-module homomorphism. The the following are equivalent.

  1. (Right split): There is an R-module homomorphism h : A_{2} \rightarrow B with gh = 1_{A_{2}} ,
  2. (Left split): There is an R-module homomorphism k : B \rightarrow A_{1} with kf = 1_{A_{1}} ,
  3. (Direct Sum): The given sequence is isomorphic to 0 \rightarrow A_{1} \rightarrow A_{1} \oplus A_{2} \rightarrow A_{2} \rightarrow 0, and thus B \cong A_{1} \oplus A_{2}.

Proof.
(1) \Rightarrow (3). By the universal property of coproduct, we have a morphism \varphi : A_{1} \oplus A_{2} \rightarrow B with i_{j} : A_{j} \rightarrow A_{1} \oplus A_{2}, and f = \varphi \circ{}i_{1},~h = \varphi\circ{}i_{2}. Moreover, \varphi{}(a_{1}, a_{2}) = f(a_{1}) + h(a_{2}). Apply it by g, and then we have g\varphi(a_{1}, a_{2}) = gf(a_{1}) + gh(a_{2}). Since gf = 0 by exactness, then 0 + gh(a_{2}) = gh(a_{2}) = 1_{A_{2}}(a_{2}) = a_{2} = \pi_{2}{}(a_{1}, a_{2}). Thus we have proved g\varphi = \pi_{2}. Hence, they all commute between these morphisms, and \varphi is an isomorphism by the short five lemma.

(2) \Rightarrow (3). By the duality of the above statement, we have the dual result. But we could also prove it directly.

By the universal property of product, we have \psi : B \rightarrow A_{1} \oplus A_{2} with \pi_{i} : A_{1} \oplus A_{2} \rightarrow A_{i} and k = \pi_{1}{}\circ\psi,~g = \pi_{2}\psi. Moreover \psi{}(b) = (k(b), g(b)). Hence \psi{}f(a_{1}) = (kf(a_{1}), gf(a_{1})) = (1_{A_{1}}(a_{1}), 0) = i_{1}(a_{1}). Thus we have \psi{}f = i_{1}. It follows that \psi is an isomorphism by the short five lemma.

(3) \Rightarrow (1), (2). Let h = \varphi{}i_{2} and k = \pi_{1}\varphi^{-1}. We have gh = g\varphi{}i_{2} = \pi_{2}i_{2} = 1_{A_{2}} and kf = \pi_{1}\varphi^{-1}f = \pi_{1}i_{1} = 1_{A_{1}}.

Short Five Lemma 比 Five Lemma 簡單多了,用這個會比較好作。另外關於 split 還有

Lemma. If f : A \rightarrow B,~g : B \rightarrow A and gf = 1_{A}, then B = \mathbf{Im}(f) \oplus \mathbf{Ker}(g).
Proof
. Let b \in B and b = b - fg(b) + fg(b) = (b - fg(b)) + fg(b). Sinc fg(b) \in \mathbf{Im} f. and g(b-fg(b)) = g(b) - gfg(b) = g(b) - 1g(b) = g(b) - g(b) = 0. We have proved that B = \mathbf{Im}(f) \oplus \mathbf{Ker}(g).

類似的還有一個,這相當於線代中對投影矩陣的空間拆解。

Lemma. If f^{2} = f for f : A \rightarrow A is a R-homomorphism. Then A = \mathbf{Im}(f) \oplus \mathbf{ker}(f)
Proof.
Let x \in A and x = x - f(a) + f(a). It’s obvious that f(a) \in \mathbf{Im}(f), and f(x - f(a)) = f(x) - f^{2}(x) = f(x) - f(x) = 0 which means x - f(a) \in \mathbf{Ker}(f). Hence A = \mathbf{Im}(f) \oplus \mathbf{Ker}(f).

Five lemma 好像應該也要打出來,不過這實在太囉唆,找找 LaTeX 的繪圖工具 …

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One thought on “Splitting Lemma

  1. Pingback: Four Lemma « XOO’s

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