Let $A$ and $B$ be C* algebras and let $\varphi : M(A) \to M(B)$ be a unital $*$-homomorphism of their multiplier algebras. Suppose, in addition, that $\varphi$ is strictly continuous (of course, norm continuity is automatic).

I'm looking for a proof of or counterexample to the following assertion:

If $\varphi$ is injective on $A$, then $\varphi$ is injective.

At first I thought this would be true and easy using that injective $\Leftrightarrow$ isometric for $*$-homomorphisms and that $A$ is strictly dense in $M(A)$. The problem with this approach is that the norm is generally discontinuous for the strict topology. Thanks.

Afterthought: as t.b.'s answer shows, most of the hypotheses here are quite besides the point. A $*$-homorphism $M(A) \to B$ is injective if and only if it is injective on $A$. More generally, a $*$-homomorphism $A \to B$ is injective if and only it is injective on some essential ideal $I$ of $A$.

  • $\begingroup$ I agree with your afterthought, that's exactly the point (can't upvote again). $\endgroup$ – t.b. Aug 9 '12 at 6:09

Yes, this is true.

The point is that $A$ is an essential ideal of $M(A)$ — this is part of the universal property of the multiplier algebra. By definition this means that every ideal $J$ of $M(A)$ intersecting $A$ trivially must be zero. Your hypothesis says that $J = \ker{\varphi}$ is an ideal such that $J \cap A = 0$, so $J = 0$.

See also here for a nice thread on some aspects of the multiplier algebra.

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    $\begingroup$ That's a whole pile easier than I was making it for myself! So the right question to ask was just: if a $*$-homomorphism $M(A) \to B$ is injective on $A$, is it injective? Makes perfect sense, thanks :) $\endgroup$ – Mike F Aug 9 '12 at 6:04
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    $\begingroup$ Yes, exactly (I was tempted to write that this is easy -- but in a different way :)). My preferred way of seeing that $A$ is an essential ideal of $M(A)$ is to use the double centralizer viewpoint: Let $(\ell,r)\colon A \to D(A)$ be the double centralizer representation. If $(L,R) \in J$ then $(L,R)(\ell_a,r_a) = (\ell_{L(a)}, r_{L(a)}) \in J \cap A = 0$, so $L(a) = 0$ for all $a \in A$. Similarly $(\ell_a,r_a) (L,R) = (\ell_{R(a)},r_{R(a)})$ gives that $R(a) = 0$ for all $a \in A$, so $(L,R) = 0$. $\endgroup$ – t.b. Aug 9 '12 at 6:05
  • $\begingroup$ That is a nice argument I will remember that. $\endgroup$ – Mike F Aug 9 '12 at 6:12

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