= Solution
Fix a basis for the <transposition map> $T$. If $M\geq0$, write $M=BB^\dagger$. Then
$$
T(M)=M^T=\overline B B^T=\overline B(\overline B)^\dagger\geq0.
$$
Thus transposition is a <positive linear map> and also preserves the <trace>.
For dimension $D\geq2$, apply it to one subsystem of the <maximally entangled state> from part (a):
$$
(\operatorname{id}\otimes T)(|\Omega\rangle\langle\Omega|)
=\frac1D\sum_{i,j}|i\rangle\langle j|\otimes|j\rangle\langle i|
=\frac FD,
$$
where $F$ is the <swap operator>, $F(|u\rangle\otimes|v\rangle)=|v\rangle\otimes|u\rangle$. It satisfies $F^\dagger=F$ and $F^2=I$, so its <eigenvalues> are $+1$ on symmetric vectors and $-1$ on antisymmetric vectors. In particular,
$$
|\eta\rangle=\frac{|1\rangle|2\rangle-|2\rangle|1\rangle}{\sqrt2},
\qquad
\langle\eta|\frac FD|\eta\rangle=-\frac1D<0.
$$
A positive input has therefore acquired a negative expectation value after adjoining an identity channel. \b[Transposition is positive but not completely positive in dimension at least two.] The one-dimensional case is the identity map and is completely positive; the dimension qualification is necessary.
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