Direct product implies the splitness of SES.

Question

If $$ 0\rightarrow A \overset{f}{\rightarrow} G\overset{g}{\rightarrow} B\rightarrow 0 $$ is a SES of Abelian groups. Then $G\cong A\oplus B$ if and only if the SES splits on the left, if and only if the SES splits on the right.

In my class, the teacher only shows if the SES splits, then $G= A\oplus B$. I want to show the other direction: If $G\cong A\oplus B$, then the SES splits.

I wonder if $f$ sends $a$ to $(a,0)$ or not. If so, we can construct $\chi:G\rightarrow A$ naturally as $(a,b)\mapsto a$, such that $\chi\circ f=\mathrm{id}_A$. But it may happens that $f(a)\neq (a,0)$, for example:

$$\begin{aligned} 0\rightarrow \mathbb{Z}/2 &\overset{f}{\rightarrow} \mathbb{Z}/2\oplus\mathbb{Z}/2 \overset{g}{\rightarrow} \mathbb{Z}/2\rightarrow 0\\ 1 &\mapsto (1,1)\qquad \ \ \mapsto 0\\ &\quad \ \ (0,1),(1,0) \mapsto 1 \end{aligned}$$

Any hints are welcomed, thanks!

Answer 1

It is unfortunately not uncommon for people to say things frivolously, without paying proper attention to details. This statement:

Then $G\cong A\oplus B$ if and only if the SES splits on the left, if and only if the SES splits on the right.

is false as it is. First a counterexample, consider:

$$A=\mathbb{Z}_2$$ $$G=B=\bigoplus_{i=1}^\infty\mathbb{Z}_4\oplus\bigoplus_{i=1}^\infty\mathbb{Z}_2$$ Where $f$ is embedding of $\mathbb{Z}_2$ into first coordinate $\mathbb{Z}_4$ of $G$, while $g$ is the corresponding quotient (we turn one $\mathbb{Z}_4$ into $\mathbb{Z}_2$, but since we have infinitely many of both, it makes no difference). This sequence does not split, by the same argument why $0\to\mathbb{Z}_2\to\mathbb{Z}_4\to\mathbb{Z}_2\to 0$ does not split. But $G$ is isomorphic to $A\oplus B$.

The reason this happens is that the isomorphism $h:G\to A\oplus B$ is in no way related to the sequence, to $f$ and $g$.

The correct version of the splitting lemma you can find on the wikipedia. The "$G\cong A\oplus B$" condition has to be replaced with the following, stronger condition:

There is an isomorphism $h$ from $G$ to the direct sum of $A$ and $B$, such that $h\circ f$ is the natural injection of $A$ into the direct sum, and $g\circ h^{-1}$ is the natural projection of the direct sum onto $B$.

More precisely, given the following diagram:

$$\require{AMScd} \begin{CD} 0 @>>> A @>{f}>> G @>{g}>> B @>>> 0 \\ & @V{id}VV @V{h}VV @V{id}VV \\ 0 @>>> A @>{(id,0)}>> A\oplus B @>{\pi_2}>> B @>>> 0 \end{CD} $$

The condition above simply says: the diagram commutes.

The hard part is showing that if the sequence splits then such $h$ exists.

The other direction is quite simple: if $i:B\to A\oplus B$ is the obvious embedding $i(x)=(0,x)$, then the right split $r:B\to G$ is simply given by $r=h^{-1}\circ i$. Analogously for the left split. The fact that those mappings work as expected is a direct consequence of the commutativity of the diagram.