 Solution

False. A left inverse would force $\alpha$ to be injective: if $\alpha u=0$, then $u=\beta \alpha u=0$. Take the surjective linear map $\alpha :{\mathbb{R}}^2\to \mathbb{R}$ given by $\alpha (x,y)=x$. Since $(0,1)$ belongs to its kernel, every composite $\beta \alpha$ annihilates $(0,1)$ and cannot be the identity map on ${\mathbb{R}}^2$.