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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?

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    $\begingroup$ 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 $(2)$? $\endgroup$ – dpb492 Jul 1 at 4:05
  • $\begingroup$ @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. $\endgroup$ – Kumar Jul 1 at 4:09
  • $\begingroup$ The homomorphism $i$ is injective due to the assumption that $E$ is short exact, can you use that to define $\gamma$? $\endgroup$ – dpb492 Jul 1 at 4:18
  • $\begingroup$ @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. :( $\endgroup$ – Kumar Jul 1 at 4:24
  • $\begingroup$ 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. $\endgroup$ – Kumar Jul 1 at 5:09
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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.

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