 Solution

For an incoming ${\mu }^{-}$, a putative $e^{-}{\mu }^{-}\to {\nu }_e{\nu }_{\mu }$ process would change total electric charge from $-2$ to zero. It is forbidden: both negative charged leptons would have to emit a $W^{-}$, and a single exchanged charged boson cannot connect those two emission vertices. Ordinary elastic $e^{-}{\mu }^{-}\to e^{-}{\mu }^{-}$ scattering remains possible through $t$-channel photon or $Z$ exchange, shown in the figure, but this is not the charged-current reaction of part d. The neutral-current weak contribution is distinct from the electromagnetic one.