 Solution

For the odd channel, solve on $x>0$ with $\psi (0)=0$. Inside $0<x<a$, write $\psi =C\sin (kx)$; outside write $\psi =e^{-ikx}-S_{--}{e}^{ikx}$, respecting the outgoing odd-channel minus sign. Continuity at $a$ and the delta-potential derivative jump areEliminating $C$ yieldsThe equations remain valid when $\sin (ka)=0$ by taking the limiting value, rather than dividing blindly. For real $k,U_0$, numerator and denominator inside the fraction are complex conjugates and the prefactor has unit modulus. Thus $|S_{--}{|}^2=1$. This is conservation of flux in the odd parity channel: the real potential causes only a phase shift there, not absorption. It does not assert unit left-to-right transmission in the usual incident-from-one-side basis.