Is the algebra of compact operators flat? 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? 
 A: 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.
A: 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).
