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Suppose that $A\hookrightarrow B$ is an inclusion of $C^*$-algebras and let $K$ be the algebra of compact operators on a separable Hilbert space. Is it true that the map $A\otimes K\hookrightarrow B\otimes K$ is injective?

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    $\begingroup$ Firstly, there are some typos in your question to fix. Secondly, can you clarify whether $\otimes$ denotes ${\rm C}^*$-tensor product, or just the uncompleted algebraic tensor product? $\endgroup$ – Yemon Choi Oct 7 '19 at 2:53
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    $\begingroup$ Thirdly, if you want to talk about objects of a category being flat, you need to define your terms rather more precisely, especially when your category is not additive. In particular, "tensoring with this object preserves monomorphisms" might not be enough to categorize "flatness" - if one wants to use a definition based on short exact sequences, then tensoring $0\to J \to A \to A/J$ with a fixed ${\rm C}^*$-algebra need not preserve "exactness in the middle" $\endgroup$ – Yemon Choi Oct 7 '19 at 2:57
  • $\begingroup$ If $A\hookrightarrow B$ is an inclusion of C*-algebras, and $C$ is any other C*-algebra, then $A⊗_{min}C\hookrightarrow B⊗_{min}C$ is injective. The same is also true for $⊗_{max}$ in the following special cases: (1) $C$ is nuclear, (2) $A$ is nuclear, (3) $A$ is a hereditary subalgebra of $B$, or (4) there exists a conditional expectation from $B$ to $A$. Beyond these cases the statement for $⊗_{max}$ may fail. (Ref: Section (3.6) in Brown-Ozawa book). $\endgroup$ – Ruy Nov 3 '19 at 2:29
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Assuming that your $\otimes$ denotes min-tensor product of ${\rm C}^\ast$-algebras, then the answer to the question in the body of your post -- which is NOT the same as the question in the title of your post, at present! -- is "Yes, but this says nothing special about $K$."

In more detail: min-tensoring an inclusion of ${\rm C}^\ast$-algebras with any fixed ${\rm C}^\ast$-algebra $M$ will preserve injectivity, i.e. $A\hookrightarrow B$ always gives $A\otimes M \hookrightarrow B\otimes M$. The interesting question is whether the canonical map of operator spaces

$$ \frac{B\otimes M}{A\otimes M} \to (B/A) \otimes M \qquad\qquad(1)$$

is injective. If this is true for all inclusions $A\hookrightarrow B$ then I think this is equivalent to saying $M$ satisfies Simon Wassermann's "Slice Property (S)". Property (S) forces $M$ to be an exact ${\rm C}^*$-algebra (in the sense of Kirchberg--Wassermann), and it is known that nuclearity implies property (S). So in particular, this does work for the algebra $K$.

By the way, although I'm not sure if it has been noticed explicitly in the literature, the statement (1) above should be equivalent to saying that $(\underline{\quad})\otimes M$ preserves equalizer pairs in the category of ${\rm C}^*$-algebras. I think this can be taken as one notion of being a left exact functor, but I admit I am not so familiar with the general categorical language/results here.

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    $\begingroup$ Nice, I didn't know this. $\endgroup$ – Nik Weaver Oct 7 '19 at 10:51
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    $\begingroup$ @Nik I did write this answer around 4am local time so I should probably check some of the details :) $\endgroup$ – Yemon Choi Oct 7 '19 at 13:03
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The answer is yes. An easy way to see this is to first check that the inclusion of $M_n(A)$ into $M_n(B)$ is isometric. Then taking the union over $n$ yields the result (whether you complete or not, since it's isometric on the uncompleted union).

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