= Solution
Use the PDF's <Bell state> labels throughout this question: its $\Phi_-$ is the antisymmetric <spin singlet state>, whereas $\Psi_\pm$ are the even-parity states. Write
$$
|a\rangle=a_0|0\rangle+a_1|1\rangle,\qquad
|b\rangle=b_0|0\rangle+b_1|1\rangle.
$$
Expanding the antisymmetric combination cancels its $00$ and $11$ components and gives
$$
\frac{|a\rangle|b\rangle-|b\rangle|a\rangle}{\sqrt2}
=(a_0b_1-a_1b_0)\frac{|01\rangle-|10\rangle}{\sqrt2}.
$$
The two normalized orthogonal vectors are the columns of a <unitary matrix> $U$. Its determinant $a_0b_1-a_1b_0$ has modulus one because $\det(U^\dagger U)=|\det U|^2=1$. Therefore
$$
\boxed{\frac{|a\rangle|b\rangle-|b\rangle|a\rangle}{\sqrt2}
=(\det U)|\Phi_-\rangle=e^{i\alpha}|\Phi_-\rangle.}
$$
This proves <collective-unitary covariance of the two-qubit singlet>: a common basis change alters only its <global phase>, and leaves its density projector unchanged.
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