A short exact sequence $ 0 \rightarrow A \rightarrow B \rightarrow C\rightarrow 0$ of partially ordered group, where $\alpha : A\rightarrow B$ and $\beta: B\rightarrow C$ are order homomorphism is called lex-extension of $A$ by $C$ if the positive cone of $B$ is $\left\{\alpha(a):a \geq 0 \in A\right\} \cup \left\{b\in B: \beta(b) >0\right\}.$
As an unordered groups the sequence split if there exists an homomorphism $\gamma : B \rightarrow A$ such that $\gamma\circ\alpha$ is an identity on $A.$
In the case of ordered abelian groups do we need this $\gamma$ is order homomorphism or not?
Suppose the sequence splits, then $B = C \oplus A.$ Let $(c, a)\geq (d, e) \in B.$ Then $c > d$ or $c = d$ and $a\geq e.$ Define $\gamma : B \rightarrow A$ by $\gamma(b) = \gamma(x, y) = y,$ where $x\in C$ and $y\in A.$ Clearly $\gamma$ is a retraction on $A.$ If $(x_{1}, y_{1}), (x_{2}, y_{2}) \in B,$ where $x_{1}, x_{2}\in C$ and $y_{1}, y_{2}$ in $A.$ If $x_{1} < x_{2},$ then $(x_{1}, y_{1}) \leq (x_{2}, y_{2})$ even when $y_{1} > y_{2}.$ Then $\gamma((x_{1}, y_{1})) = y_{1}$ and $ \gamma((x_{2}, y_{2})) = y_{2}.$ This implies $\gamma((x_{1}, y_{1})) \geq \gamma((x_{2}, y_{2}))$ if $y_{1} \geq y_{2}.$ This shows $\gamma$ is not order homormorphsim. Could you please explain if we need $\gamma$ to be order homomorphism where is the mistake in this work?