$0\to A_1 \xrightarrow{\phi} A_2 \xrightarrow{\psi} A_3 \to 0$, if $A_3$ is free abelian then the sequence splits.

Question

I have a proposition that says that for an exact sequence $0\to A_1 \xrightarrow{\phi} A_2 \xrightarrow{\psi} A_3 \to 0$, if $A_3$ is free abelian then the sequence splits.

So to prove this, we create a right-inverse map for $\psi$, which for each basis element $e\in A_3$ gives an element of $\psi^{-1}(e)$. But couldn't we do this even if $A_3$ is not free abelian? i.e. send any $g\in A_3$ to some element of $\psi^{-1}(g)$?

Answer 1

Consider $0\rightarrow f:\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}/2\rightarrow 0$, where $f:\mathbb{Z}\rightarrow \mathbb{Z}$ is defined by $f(x)=2x$ and $\mathbb{Z}\rightarrow\mathbb{Z}/2$ is the quotient map. This sequence does not splits.

Answer 2

Suppose for example that $A_2={\mathbb Z}$ and A_3=${\mathbb Z}/2{\mathbb Z}$. Let $g$ be the non-zero element of $A_3$. Try mapping $A_3$ to $A_2$ by mapping $g$ to $1$ and $0$ to $0$.

Then your map $\psi^{-1}$ is not a homomorphism:

$$\psi^{-1}(2g)=\psi^{-1}(0)=0$$

$$2\psi^{-1}(g)=2\cdot 1=2$$

and $0\neq 2$.