Solution
= Solution
Replacing the resource by $(I\otimes U)|\phi^+\rangle$ applies $U$ to Bob's uncorrected output. If $P\in\{I,Z,X,XZ\}$ labels the Bell outcome, the uncorrected state is $UP|\alpha\rangle$, and the unchanged correction produces
$$
\boxed{P^{-1}UP|\alpha\rangle}.
$$
\b[Thus the four outputs are $U|\alpha\rangle$, $ZUZ|\alpha\rangle$, $XUX|\alpha\rangle$, and $(XZ)^{-1}U(XZ)|\alpha\rangle$, up to global phases.]