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Let $0\rightarrow M_2\overset{f}{\rightarrow}M_1\overset{g}{\rightarrow}M_3\rightarrow0$ be a short exact sequence of modules. Prove that $M_1\cong M_2\times M_3$ if and only if there exists homomorphisms $\phi:M_1\to M_2$ and $\psi:M_3\to M_1$ such that $f\circ\phi+\psi\circ g=\operatorname{Id}_{M_1}$

It looks like splitting lemma, but is a weakest condition. ¿How I can prove this?

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Composing by $g$ on the left, it follows $(g \circ \psi-id) \circ g=0$. As $g$ is onto, we get $g \circ \psi=id$, ie $\psi$ is a section. Similarly, $\phi \circ f=id$.

We have morphisms $\alpha=(\phi,g): M_1 \rightarrow M_2 \times M_3$ and $\beta=f \oplus \psi: M_2 \times M_3 \rightarrow M_1$ with $\beta \circ \alpha=id$. Let $x \in M_2,y \in M_3$ be such that $\beta(x,y)=0$. Then $f(x)=-\psi(y)$. So $-y=-g \circ \psi(y)=g \circ f(x)=0$. So $y=0$ so $f(x)=0$ so $x=0$ and $\beta$ is injective.

Thus $\alpha$ and $\beta$ are inverse isomorphisms.

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  • $\begingroup$ Thanks, I already have proven the converse too. $\endgroup$ – Mephisto Aug 29 '20 at 15:52
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Here's a counterexample with $\Bbb Z$-modules:

Let $M_3=\bigoplus_{n\in\Bbb N}\Bbb Z/2\Bbb Z$ and $M_1=M_2=\Bbb Z\oplus M_3$. Then by a Hilbert hotel argument, one readily sees that $M_2\cong M_1\oplus M_3$. Now let $$f(a_0,a_1+2\Bbb Z, a_2+2\Bbb Z,\ldots) = (2a_0,a_1+2\Bbb Z, a_2+2\Bbb Z,\ldots) $$ and $$g(a_0,a_1+2\Bbb Z, a_2+2\Bbb Z,\ldots)=(a_0+2\Bbb Z,a_1+2\Bbb Z, a_2+2\Bbb Z,\ldots),$$ which gives us the short exact sequence. But there cannot exist $\phi,\psi$ such that $f\circ \phi+\psi\circ g$ is the identity. Indeed, $(1,0+2\Bbb Z,\ldots)$ is impossible to reach.

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  • $\begingroup$ You don't have $g \circ f = 0$ here, as written $g\circ f$ acts as the identity on the not-first factors. One way to fix that is to take e.g. $f(a_0, a_1, a_2, \dotsc) = (2a_0, 0, a_1, 0, a_2, \dotsc)$ and $g(a_0, a_1, a_2, a_3, \dotsc) = ([a_0], a_1, a_3, \dotsc)$. $\endgroup$ – Daniel Fischer Aug 28 '20 at 20:47

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