= Twin paradox simultaneity change
During either inertial leg, the traveller can correctly say that the moving home clock runs slowly. The missing step is deciding which distant home-clock reading is simultaneous with turnaround. By the <Lorentz transformation>, the outward <inertial frame>'s constant-time line through $P$ obeys $t-vx/c^2=\text{constant}$. At $x=0$ it meets the home <world line> at
$$
t_H^-=\frac T2-\frac{v^2T}{2c^2}.
$$
For the return <inertial frame>, use $t+vx/c^2=\text{constant}$ through the same <spacetime event> $P$; its home-clock assignment is
$$
t_H^+=\frac T2+\frac{v^2T}{2c^2},\qquad
t_H^+-t_H^-=\frac{v^2T}{c^2}.
$$
The dashed lines in the <spacetime diagram> illustrate this <relativity of simultaneity>. For $v=0.6c$ and $T=10\,\mathrm{yr}$, they assign $3.2\,\mathrm{yr}$ and $6.8\,\mathrm{yr}$ to the home clock. The difference of $3.6\,\mathrm{yr}$ is a change in the traveller's distant-time assignment, not a sudden physical advance of the home clock.
On each leg, the traveller's $4\,\mathrm{yr}$ of <proper time> corresponds to $4/\gamma=3.2\,\mathrm{yr}$ of home-clock advance according to that leg's <inertial frame>. Including the <twin paradox simultaneity change> gives $3.2+3.6+3.2=10\,\mathrm{yr}$, exactly the reading at reunion. Actual received <light> signals remain continuous across a physical turnaround; their arrival times are distinct from the distant <spacetime events> assigned as simultaneous by <Einstein synchronisation>.
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