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If $A$ is a nuclear $C^*$ algebra, $A^{op}$ is the opposite $C^*$ algebra, is the conclusion: "there is a unique injective $*$-homomphism $\pi \colon A\otimes A^{op}\rightarrow M(A)\otimes M(A)^{op}$" correct?

$M(A)$ is the multiplier algebra of $A$.

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closed as off-topic by math112358, Saad, KReiser, max_zorn, Brahadeesh Nov 23 '18 at 6:32

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You always have the canonical inclusion, given by $\mathrm{id} \otimes \mathrm{id}$. However, you can also consider $\alpha \otimes \beta$, where $\alpha$ and $\beta$ are automorphisms of $A$ resp. $A^{\mathrm{op}}$. In general these $*$-hom. will not be equal.

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  • $\begingroup$ I have edited the post. $\endgroup$ – user42761 Oct 8 '18 at 19:51
  • $\begingroup$ No, why should it be true ? The one I wrote in my answer does not extend the identity. $\endgroup$ – user42761 Oct 11 '18 at 14:37

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