Prove that in the diagram $A_1 \rightleftarrows_{i_{1}}^{\pi_1} A \leftrightarrows_{i_{2}}^{\pi_2}$ we have $A \cong A_1 \oplus A_2$

Question

Let $A_i , A_2, A$ be left $\mathrm{R} -$ modules. If in the diagram below $$A_1 \rightleftarrows_{\pi_{1}}^{i_1} A \leftrightarrows_{\pi_{2}}^{i_2} A_2$$ we have that $\pi_1 i_1 = 1_{A_1}$ and $\pi_2 i_2 = 1_{A_2}$ and $(i_1, \pi_2)$ is exact (which means $im i_1 = ker \pi_2$) prove that $A \cong A_1 \oplus A_2$.

Does anyone have any idea how to solve this? I tried using the fact that $A$ satisfies the universal property of direct sums but didn't get anywhere.

Not necessarily. I have solved that case. this is a different one. — user53970, Aug 19 '18 at 20:33

score 2 · Answer 1 · answered Aug 19 '18 at 20:59

2

Since $R$-mod is an abelian category, $i_1$ is injective, and $\pi_2$ is surjective, you can use the splitting lemma on the short exact sequence $$ 0\rightarrow A_1\xrightarrow{i_1}A\xrightarrow{\pi_2} A_2\rightarrow 0. $$ The arrow $i_2$ is gives a right splitting (and $\pi_1$ gives a left splitting, and having either one of them is enough.)

answered Aug 19 '18 at 20:59

Ben West

12,366

There's even an element-free proof on this site if you prefer: https://math.stackexchange.com/questions/748699/abstract-nonsense-proof-of-the-splitting-lemma/753182#753182 – Ben West Aug 19 '18 at 21:04

score 1 · Accepted Answer · answered Aug 19 '18 at 21:04

$\DeclareMathOperator\Ker{Ker}\DeclareMathOperator\Im{Im}$From $\pi_2\circ i_2=1_{A_2}$ we get an isomorphism $A\cong\Ker\pi_2\oplus A_2$ given by \begin{align} &A\to\Ker\pi_2\oplus A_2& &x\mapsto(x-(i_2\circ\pi_2)(x),\pi_2(x)) \end{align} By assumption, $\Ker\pi_2\cong\Im i_1\cong A_1$ where the last isomorphism holds because $i_1$ is injective, hence $A\cong A_1\oplus A_2$.

Prove that in the diagram $A_1 \rightleftarrows_{i_{1}}^{\pi_1} A \leftrightarrows_{i_{2}}^{\pi_2}$ we have $A \cong A_1 \oplus A_2$

2 Answers2