2
$\begingroup$

In the category of C*-algebras and $*$-homomorphisms, a morphism is monic precisely when it is injective, and epic precisely when it is surjective (see Mono- and epi-morphisms for C*-algebras). Is this still true in the category of C*-algebras and completely positive maps?

Suppose $f \colon A \to B$ is a completely positive monomorphism. If $f(a)=0$ for some positive $a \in A$, then $fg=fh$ for the completely positive maps $g,h \colon \mathbb{C} \to A$ defined by $g(z)=0$ and $h(z)=za$, whence $a=0$. What if $f(a)=0$ for an arbitrary $a \in A$? If we could still conclude $a=0$ then $\ker(f)=0$ and $f$ would be injective.

$\endgroup$

1 Answer 1

7
$\begingroup$

Yes. Every positive map $f$ is self-adjoint: $f(x^*)=f(x)^*$ for every $x$. Hence, if $f(a)$=0 and $a=b+ic$ with $b,c$ self-adjoint, then $f(b)=0=f(c)$. Suppose for a contradiction that $b\neq0$. Then, for $b=b_+-b_-$ with positive $b_+,b_-$, one has $f(b_+)=f(b_-)$. The cp maps $g_{\pm}\colon{\mathbb C}\to A$, defined by $g_\pm(1)=b_\pm$, satisfy $f\circ g_+=f\circ g_-$.

For the epimorphism problem, let $f\colon A\to B$ be an epi cp contraction. Then, the unitization $f^{1}\colon A^1\to B^1$ is a unital cp map which is epi for ucp maps. Let $S$ be the norm closure of $f^{1}(A^{1})$ in $T:=B^{1}$. We will prove that if $S\subset T$ is proper inclusion of operator systems (those are unital self-adjoint closed subspaces of a $\mathrm{C}^*$-algebra), there is a states $g,h$ on $T$ such that $g|_S=h|_S$ but $g\neq h$. By the Hahn--Banach theorem, there is a non-zero bounded linear functional $\phi$ on $T$ such that $\phi|_S=0$. By the GNS construction for a bounded linear functional (of an ambient $\mathrm{C}^*$-algebra), we may assume that $\phi(x)=\langle x\eta,\xi\rangle$ for some realization $T\subset B(H)$ and $\xi,\eta\in H$. Notice that $\xi\perp\eta$ because $1\in S$. Now we consider the states $g'$ and $h'$ on $T$ defined by $2g'(x)=\langle x\xi,\xi\rangle+\langle x\eta,\eta\rangle$ and by $2h'(x)=\langle x(\xi+\eta),\xi+\eta\rangle$. They coincide on $\Phi(S)$ but not on $\Phi(T)$. Hence, the states $g=g'\circ\Phi$ and $h=h'\circ\Phi$ on $T$ will do.

$\endgroup$
3
  • $\begingroup$ Neat answer, thanks! What about epimorphisms? $\endgroup$ Apr 11, 2014 at 21:30
  • $\begingroup$ That's much more difficult! It asks if every epimorphism has dense range. I think I know it's true for finite dimensional $\mathrm{C}^*$-algebras. $\endgroup$ Apr 11, 2014 at 22:07
  • 3
    $\begingroup$ In fact, it wasn't too difficult. I add it to the answer. $\endgroup$ Apr 11, 2014 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.