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Let $A,B$ be C*-algebras and let $\varphi: A \to B$ be a $*$-homomorphism. Suppose that $\ker( \varphi) \cap D = \{0\}$ where $D$ is a dense $*$-subalgebra of $A$. Does it follow that $\varphi$ is injective?

I'm pretty sure the answer is "no". At various times I've had to upgrade injectivity on a dense subalgebra to injectivity on the whole algebra, but this has always involved very specific situations or extra hypotheses. I don't think I've ever seen an example which shows this is generally false though.

Added: In hindsight, the following equivalent formulation of this question would have been slightly cleaner.

Find a nonzero, closed ideal $I$ in a C*-algebra $A$ such that $D \cap I = \{0\}$ for some dense $*$-subalgebra $D \subset A$?

Jonas Meyer gives a very simple example. Take $A=C[0,2]$, $I$ to be the ideal of functions which vanish on $[0,1]$, and $D$ to be the polynomial functions in $A$.

A more elaborate example is obtained by taking $A=C^*(G)$, the full group C*-algebra of a discrete, nonamenable group $G$; $D = \mathbb{C} G$, the copy of the group algebra in $A$; and $I$ to be the kernel of projection down to the reduced group C*-algebra $C^*_r(G)$

None's answer appears to be erroneous.

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Well, the analogous statement for Banach spaces is false... – Nate Eldredge Oct 9 '12 at 22:11
Somewhat related:… – Jonas Meyer Dec 14 '12 at 7:42
up vote 4 down vote accepted

If $A=C[0,2]$ and $B=C[0,1]$, define $\phi:A\to B$ to be the restriction map $\phi(f)=f|_{[0,1]}$. Let $D\subset A$ be the algebra of polynomial functions on $[0,2]$.

Then $\ker(\phi)\cap D=\{0\}$ because no nonzero polynomial function vanishes on $[0,1]$. However, $\phi$ is not injective because for example it sends the nonzero continuous function $f(t)=\max\{0,t-1\}$ on $[0,2]$ to the zero function on $[0,1]$. Note that $D$ is dense by the Weierstrass approximation theorem.

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Now that does the trick, thanks! In the meantime, I'd thought of of another example, but it's nowhere close to as simple as yours. Let $G$ be a discrete group. Set $A = C^*(G)$ and $B=C^*_r(G)$, the universal and reduced group C*-algebras. Let $\varphi : A \to B$ be the quotient map. Then $\varphi$ fixes the copy of the group algebra $\mathbb{C} G$ in $A$ and $B$. But, $\varphi$ has a kernel when $G$ is nonamenable. – Mike F Dec 14 '12 at 19:56
Very nice example! – Nate Eldredge Dec 14 '12 at 21:02

A slightly contrived example:

Consider the dense subalgebra $S = S[0,1]$ of simple functions in $L^\infty = L^\infty[0,1]$ and the closed subalgebra $C = C[0,1]$ of $L^\infty$. Notice that $S \cap C = \mathbb{C}$.

Thus, the quotient homomorphism $L^\infty \to L^\infty/ C$ is almost injective on $S$, killing only the constant functions $\mathbb{C}$.

We can remedy this slight defect by factoring out the constant functions.

So: set $A = L^\infty[0,1]/\mathbb{C}$, $D = S/\mathbb{C}$ and $B = (L^\infty/\mathbb{C})\,{\large /}\,(C/\mathbb{C}) \cong L^\infty[0,1]\,/\,C[0,1]$.

Then $D$ is a dense subalgebra of $A$, and the quotient homomorphism $\varphi\colon A \to B$ is injective when restricted to $D$ while $\ker{\varphi} = C[0,1]/\mathbb{C}$.

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Perfect! A nice simple example. Exactly what I was hoping for. Thanks a bundle – Mike F Oct 10 '12 at 4:41
Wait, hang on. Why does the quotient map $L^\infty[0,1]/C[0,1]$ make sense? $C[0,1]$ is not an ideal... – Mike F Oct 11 '12 at 21:35

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