# Short Split Exact Sequence Theorem

Let $$E:0\rightarrow A\xrightarrow{i} B\xrightarrow{q} C \rightarrow 0$$ be a short exact sequence of $$R$$-modules. Then the following are equivalent :
$$(1)$$ there is an $$R$$-module homomorphism $$\gamma: B\rightarrow A$$ such that $$\gamma \circ i$$= id.
$$(2)$$ there is a submodule $$D$$ of $$B$$ such that $$B = i(A)\oplus D$$
$$(3)$$ $$E$$ is a split short exact sequence.

I have proved the equivalence of statement $$(2)$$ and $$(3)$$. but unable to prove $$(1)\implies (2)$$ and $$(3)\implies (1)$$.

Can Anyone help me to prove these two statements?

• You have $(2) \Leftrightarrow (3)$ and it looks like you are trying to complete the implication cycle in the order $(1) \Rightarrow (2) \Rightarrow (3) \Rightarrow (1)$, would it help to look at the implications in the other order? Would it be easier to define $\gamma$ assuming the truth of $(2)$? – dpb492 Jul 1 at 4:05
• @dpb492 I am thinking of defining a function $\tau: i(A)\rightarrow A$ and show that for $b\in B$, $\gamma |_{i(A)} (b)=\tau (b)$. but I don't know how will it work? Also, I don't know whether it would help if I take implications in reverse cyclic order. – Kumar Jul 1 at 4:09
• The homomorphism $i$ is injective due to the assumption that $E$ is short exact, can you use that to define $\gamma$? – dpb492 Jul 1 at 4:18
• @dpb492 I already know the fact that $i$ is injective and $q$ is a surjective module homomorphism. But if I were able to use these facts, then maybe I would have only asked for proof-verification and not the proof. :( – Kumar Jul 1 at 4:24
• Note that $i(A)=A$ because $i$ is injective. Therefore, $B=A\oplus D$. Now, every element of $B$ can be expressed as $(a,d)$ where $a\in A$ and $d\in D$. So, Define $\gamma ((a,d))=a$. Notice, that $\gamma$ is a surjective $R$-module homomorphism. Moreover, note that $\gamma \circ i =$id. Hence, we are done. – Kumar Jul 1 at 5:09

For 1 $$\Longrightarrow$$ 2: Since $$\gamma \circ i$$ = id$$_A$$, that implies $$\gamma$$ is surjective. So A $$\cong$$ B\Ker $$\gamma$$. Let D = Ker $$\gamma$$. We show that D $$\cap$$ $$i$$(A) = 0. Suppose b is in the intersection. Then b = $$i$$(a) for some a. So 0 = $$\gamma$$(b) = $$\gamma \circ i$$(a) = a. Since $$i$$ is injective, b must be 0.