 Solution

True. Choose a basis $v_1,\dots ,v_r$ of $V$. By surjective function, choose $u_i\in U$ with $\alpha u_i=v_i$. Define the linear map $\beta$ by $\beta v_i=u_i$ and extend by linearity. Then $\alpha \beta v_i=v_i$ on every basis vector, and thereforeThis constructs a right inverse and does not require $\alpha$ to be injective. If $V$ is the zero vector space, the unique zero map is the required right inverse.