The category of $C^{\*}$ algebras is denoted by $\mathcal{A}$.


Is there a functor $\mathcal{F}$ on $\mathcal{A}$ which send each object $A\in \mathcal{A}$ to its center $Z(A)$. In the other words, can we extend the maping $A\mapsto Z(A)$ on objects to a functor on this category?



Is this a functor on the category of $C^{*}$ algebras?
Here is another attempt at proving no such functor exists — I apologize to Chris and to Manny if something like this is already in the papers which they cite.$\newcommand{\Mat}{{\bf M}}\newcommand{\Cplx}{{\bf C}}\newcommand{\Cst}{{\rm C}^\*}$


Let $\Mat\_2$ denote the algebra of $2\times 2$ complex-valued matrices, and let $D\_2$ be the algebra $\Cplx\oplus\Cplx$ regarded as the subalgebra of $\Mat\_2$ consisting of diagonal matrices. Let
$$ A = \{ f\in C([0,1], \Mat\_2) \mid f(1)\in D\_2\}. $$
Since $Z(\Mat\_2)=\Cplx$, one shows (with a small appeal to continuity) that
$$ Z(A)= \{ f\cdot I\_2 \mid f\in C([0,1])\}\cong C([0,1]).$$


Let $\phi: D\_2 \to A$ be the $\*$-homomorphic embedding $\phi(d)(t) = d$ for all $t\in [0,1]$ and all $d\in D\_2$. Let ${\rm ev}\_1 : A\to D\_2$ be the homomorphism "evaluate at $1$".


Now suppose $\mathcal F$ is a functor on the category of $\Cst$-algebras and $\*$-homomorphisms, such that ${\mathcal F}(B)=Z(B)$ for every $\Cst$-algebra $B$. Then ${\mathcal F}({\rm ev}\_1) \circ {\mathcal F}(\phi)$ must be the identity homomorphism on ${\mathcal F}(D\_2)=D\_2$, and thus the identity homomorphism on $D\_2$ must factor somehow through the algebra $Z(A)\cong C([0,1])$. But this is impossible, since there is no injective algebra homomorphism from $D\_2$ into $C([0,1])$. One way to see this last claim is to note that $D\_2$ contains two non-trivial projections $e\_1$ and $e\_2$ which sum to $1$, while the only projections in $C([0,1])$ are $0$ and $1$.




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*Update 7th Oct. 2015*


In response to a comment: we can find examples of a retract $D\to B \to D$ (that is, the composition of these two maps is the identity map on $D$) where $D$ is commutative and infinite-dimensional and $B$ has trivial centre.


Here is one way. Let $A$ be an infinite, countable group, and $H$ a countable amenable group, equipped with an action $\alpha: A\to {\rm Aut}(H)$. Form the semidirect product $G=H\rtimes\_\alpha A$, which can be defined as the set $H\times A$ equipped with the product
$$
(h,a)(k,b) = (h\cdot\alpha\_a(k), ab).
$$
The subset $\{e\_H\} \times A$ is a subgroup of $G$; there is a quotient map of $G$ onto $A$, defined by "projection in the second variable"; and doing the inclusion $A \to G$ followed by the quotient $G\to A$ gives the identity map on $A$. (In other words, we have a retract.)


Note, for later reference, that 
$$(h,a)^{-1} = ({\alpha\_a}^{-1}(h^{-1}), a^{-1} )$$
and so, using the fact that $A$ is abelian, a short calculation gives
$$ 
(h,a) (k,b) (h,a)^{-1} = ( h\alpha\_a(k)\alpha\_b(h^{-1}), b).
$$


I now want to impose conditions on $\alpha$ which will ensure that $G$ is an ICC group, i.e. all conjugacy classes except $\{e\}$ are infinite. It is well known that if $G$ is countable and ICC then its group von Neumann algebra ${\rm VN}(G)$ has trivial centre: since the centre of $\Cst\_r(G)$ is contained in the centre of ${\rm VN}(G)$, we deduce that $\Cst\_r(G)$ has trivial centre.


To find such conditions: note that $(0,a)(k,b)(0,a)^{-1} = (\alpha\_a(k), b)$
and $(h,0)(e,b)(h,0)^{-1} = (h\alpha\_b(h^{-1}), b)$.


So if we assume that $\alpha$ satisfies the following conditions


1) for each $k \in H \setminus\{e\_H\}$, the orbit $\{\alpha\_a(k) : a\in A\}$ is infinite


2) for each $b\in A\setminus\{e\_A\}$, the set $\{h \alpha\_b(h^{-1}): h\in H\}$ is infinite


then $G$ will be ICC, as desired, and so $\Cst\_r(G)$ will have trivial centre.
$\newcommand{\Cst}{{\rm C}^\*}$


Recall that on the category ${\sf Grp}$ of (discrete) groups and homomorphisms, the full group $\Cst$-algebra defines a functor ${\sf Grp}\to {\sf Cstar}$ is functorial. Since $A \to G \to A$ is a retract in ${\sf Grp}$, we get a retract $\Cst(A)\to \Cst(G)\to\Cst(A)$.
Clearly $\Cst(A)$ is commutative and infinite-dimensional. Now since $A$ and $H$ are amenable, so is $G$; hence $\Cst(G)=\Cst\_r(G)$, which has trivial centre by the remarks above.


To get an actual, concrete example: let $A=\{2^n : n \in {\mathbb Z}\}$ be the multiplicative group formed by all integer powers of $2$, and let $H$ be the set of dyadic rationals, regarded as an additive group. The action of $A$ on $H$ is just by scaling, i.e. for $a\in A$ and $h\in H$ we have $\alpha\_a(h)=ah$. It is easily checked that both conditions 1) and 2) are satisfied; note that $e\_H=0$, $e\_A=1$, and $h\alpha\_b(h^{-1})$ is just $h-bh$.


[I learned of the existence of solvable countable ICC groups in a seminar many years ago, which is why I knew there should be some way to manufacture examples that did what you want, once one has the idea of trying to build an example in ${\sf Grp}$ and transport it functorially to ${\sf Cstar}$.]


