# unique $*$ homomorphism of spatial tensor product [closed]

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 $$A$$.

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

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## 1 Answer

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.

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