----- > [!proposition] Proposition. ([[the splitting lemma]]) > > Let $0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0$ > be a [[short exact sequence|short]] [[exact sequence]] of $R$-[[module|modules]]. The following are equivalent: > 1. The sequence [[split short exact sequence|splits]]; > 2. There is an $R$-[[linear map|linear]] map $s:C \to B$ such that $g \circ s=\id_{C}$; > 3. There is an $R$-[[linear map|linear]] map $\ell:B \to A$ such that $\ell \circ f=\id_{A}$. > > Related: [[characterization of left and right inverses in R-mod]]. > > Mnemonic: $s$ is a *s*ection, $\ell$ is a *l*eft inverse. > [!proof]- Proof. ([[the splitting lemma]]) > Notation: $r:=\ell$. > > **$(1) \implies (2)$.** Suppose the sequence splits, so that we may write it as $0 \to A \to A \oplus C \to C \to 0.$ > Define $s: C \to A \oplus C$, $s(c):=(0, c)$. Then $\pi_{C}(s(c))=\pi_{C}(0,c)=c$. > > **$(1) \implies (3)$.** Define $r:A \oplus C \to A$ by $(a,c) \mapsto a$, then $\iota_{A} \circ r=\id_{}$. > > **$(2) \implies (1).$** Define $\varphi:A \oplus C \to B$ via $(a,c) \mapsto f(a)+s(c)$. We want to show there is a commutative diagram > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRGJAF9T1Nd9CKAIzkqtRizYBBLjxAZseAkQBMo6vWatEIAEKzeigUQDM68VrYBhA-L5LByACznNknR26H+ylGSFibtrstgo+jiIBGhLBMl52Rr7IalEW7iBSAAQAOtkQaMxwmTbxYQ6mpKlB1qH2xiguVTFsnmIwUADm8ESgAGYAThAAtkgiIDgQSGrjdFgMbAAWEBAA1iDRljq9tgPDU9QTSGYgCzB0UGw4AO4Qp+cIG+kdO4MjiMeHiE7xu28ArAdJogAGw-V5IADsgKQAA5HsFcmgsAB9GzUBh0ABGMAYAAU6r4QP0sB0FjgXntEHDxkCAJxgyljT7A9FYnH4xKCIkksnrNII7LYEYMt4fIHU6o6XJCilvMg00YipAshWIKH8ti5fA4OjImSs7F4glc4mk8lKtXQxDTSUgXL0fpoBZYWVHK3TbFgC6IAC0Tlp6KwYGCUDocFOF3hbAeIAxho54TYpt5nAonCAA > \begin{tikzcd} > 0 \arrow[r] & A \arrow[r, "f", hook] \arrow[d, "\sim"'] & B \arrow[r, "g", two heads] & C \arrow[r] \arrow[d, "\sim"] \arrow[l, "s"', dashed, bend right=49] & 0 \\ > 0 \arrow[r] & A \arrow[r, "\iota_A"'] & A \oplus C \arrow[r, "\pi_C"'] \arrow[u, "\varphi"] & C \arrow[r] & 0 > \end{tikzcd} > \end{document} > ``` > with $\varphi$ an [[isomorphism]]. Since $\varphi \circ \iota_{A}(a)=\varphi(a,0)=f(a)+0=f(a)$ and $g \circ \varphi(a,c)=g \circ f(a) + g \circ s(c)=0+c=c=\pi_{C}(a,c)$, the diagram certainly commutes. We just have to show that $\varphi$ is injective and surjective. > > **Injective.** Suppose $\varphi(a,c)=0$. We have $\begin{align} > \varphi(a,c)=0 & \\ > & \implies f(a) + s(c) = 0 \\ > & \implies \cancel{g \circ f(a)}^{=0} + \cancel{g \circ s(c)}^{=c} = 0 \\ > & \implies c=0 > \end{align}$ > so $c=0$: $\varphi(a,c)=\varphi(a,0)$. But $\varphi(a,0)=\varphi \circ \iota_{A}(a)=f(a)$, and $f$ is injective. Thus, $a=0$. > > **Surjective.** Let $b \in B$. Put $c:=g(b)$. It is probably not true that $b=s(c)$; let us consider the discrepancy $b-s(c)$. We have $g(b-s(c))=g(b)-c=c-c=0$, and by exactness $\ker g=\im f$, so we can obtain $a \in A$ such that $f(a)=b-s(c)$. Now we have $\varphi(a, c)=f(a)+s(c)=b-s(c)+s(c)=b$ > as desired. Hence $\varphi$ is [[surjection|surjective]]. > > **$(3) \implies (1)$.** Pretty similar (see paper notes). ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```