Postnikov complete implies hypercomplete

Let \(f\colon X \to Y\) be an \(\infty\)-connected map in \(T\). Since \(T\) is Postnikov-complete, the map \(f\) is the limit of the maps \(\tau_n X \to \tau_n Y\), so it suffices to show that each of these maps is an isomorphism. The maps \(X \to \tau_n X\) and \(Y \to \tau_n Y\) are \(n\)-connected, and so is \(f\colon X \to Y\). By right cancellation, it follows that \(\tau_n X \to \tau_n Y\) is also \(n\)-connected. On the other hand, since \(\tau_n X\) and \(\tau_n Y\) are \(n\)-truncated, the maps \(\tau_n X \to *\) and \(\tau_n Y \to *\) are \(n\)-truncated. By left cancellation, it follows that \(\tau_n X \to \tau_n Y\) is \(n\)-truncated as well. Thus \(\tau_n X \to \tau_n Y\) is both \(n\)-connected and \(n\)-truncated, hence an isomorphism.