# Equivalent condition to $M_1\cong M_2\times M_3$ in a short exact sequence of modules.

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?

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.

• Thanks, I already have proven the converse too. – Mephisto Aug 29 '20 at 15:52

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.

• 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 $g(a_0, a_1, a_2, a_3, \dotsc) = ([a_0], a_1, a_3, \dotsc)$. – Daniel Fischer Aug 28 '20 at 20:47