Proof, [Lemma 4.2.3, AP]. Let $m \in \natp$ and $p \in \text{Proj}(M_{m}(M))$ with $(\text{Tr}_{m} \otimes \tau)(p) = s$, so that $M^{s} = pM_{m}(M)p$, and $\tau_{s} = s^{-1}(\text{Tr}_{m} \otimes \tau)|_{M^s}$ denotes the normalised tracial state on $M^{s}$.

Let $n \in \natp$ and $q \in \text{Proj}(M_{n}(M^{s}))$ with $(\text{Tr}_{n} \otimes \tau_{s})(q) = t$, so that $(M^{s})^{t} = qM_{n}(M^{s})q$. Identify $M_{n}(M^{s}) \subset M_{n}(M_{m}(M))$, so that $q \in M_{n}(M_{m}(M))$, then

Since $q \in M_{n}(M^{s})$, $p$ is the identity on entries of $q$, so

and $(M^{s})^{t} = qM_{n}(M_{m}(M))q = qM_{mn}(M)q$. Finally,

Therefore $(M^{s})^{t} = M^{st}$.$\square$