Splitting Lemma

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 $B = \mathbf{Im}(f) \oplus \mathbf{Ker}(g)$.

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 …