Solution
= Solution
\b[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 $\mathbb R^2$.