2

I want to prove that the free product $A*B$ of two unital $C^*$-algebras $A$ and $B$ is a coproduct in the sense of category theory. Remember the construction of $A*B$: Take generators $\{a:a\in A\}\cup\{b:b\in B\}$ and let $A*B$ the $C^*$-algebras generated by this set. Then the inclusion maps $\iota_A$ and $\iota_B$ are obvious and are $*$-homomorphisms. Now let $X$ be another $C^*$-algebra with *-homomorphisms $\mu:A\rightarrow C$ and $\nu:B\rightarrow C$, then we can define $u:A*B\rightarrow C$ by $u(a)=\mu(a)$ and $u(b)=\nu(b)$, then this extends to a $*$-homomorphism $u:A*B\rightarrow C$. My question is: why is this map the unique map such that $u\circ\iota_A=\mu$ and $u\circ\iota_B=\nu$? Can someone help me?

Thank you very much.

OLGCT99
  • 21
  • I do no understand the notation ${a:a\in A}\cup{b:b\in B}$. This clearly cannot be a disjoint union, because both $A$ and $B$ are unital and either of the units should become the unit in $AB$ (or am I wrong?). And how do you exactly define the $C^$-algebra generated by this set? My answer given below assumes everything is OK with the definition, but I cannot see the light. Maybe you have a reference to where it is defined? – Vladimir Jun 15 '14 at 16:12
  • 2
    The construction is wrong. Perhaps you should construct coproduct of complex algebras (without norms etc.) first, as an exercise? – Martin Brandenburg Jun 15 '14 at 16:36
  • 1
    OLGCT99 despaired and quited when MB said his construction is wrong... Nevertheless, I'm also interested in the answer and funnily, MB already gave it in http://math.stackexchange.com/questions/345501/is-a-times-b-the-same-as-a-oplus-b/346140#346140 , the part that says that one should identify $x$, $x\otimes 1$ and $1\otimes x$. I guess that the disjoint union really is a disjoint union, but one has to divide by the freely generated algebra by this relation, in particular identify $1_{A*B}$, $1_{A}$ and $1_{B}$. Then about the norm, I have not even started to think. Someone has any reference? – Noix07 Aug 06 '14 at 17:09

3 Answers3

1

If I'm not mistaken to badly, algebraic set-indexed coproducts exist.

Let $*_\lambda A_\lambda$ be the coproduct in the category of *-algebras.

Take the supremum over all C*-seminorms $\delta:*_\lambda A_\lambda\to\mathbb{R}_+$ that agree with the original C*-norms, i.e. $\delta(a_\lambda)=\|a_\lambda\|_\lambda$ for all $a_\lambda\in A_\lambda$.

Note that a priori there may be no such C*-seminorm at all, but which wouldn't be an issue since in that case the supremum would give the trivial C*-seminorm. More importantly, however, is that it is finite for every element, which is luckily the case since $$\delta(\sum_ka_kb_k\ldots)\leq\sum_k\delta(a_k)\delta(b_k)\ldots\leq\sum_k\|a_k\|_{\lambda(a_k)}\|b_k\|_{\lambda(b_k)}\ldots$$ which is bounded independently of the C*-seminorm.

Thus one may take the Hausdorff completion.

The universal property is then easily shown.

freishahiri
  • 17,045
1

Since $u\circ i_A=\mu$, we have $u(i_A(a))=\mu(a)$ for all $a\in A$; hence $u(a)=a$ (if, as you do, you identify $A$ with its image $i(A)$). At the same time, $u(b)=\nu(b)$ for $b\in B$. Thus, $u$ is uniquely determined on $A\cup B\subset A*B$, which uniquely determines $u$ on the entire $A*B$.

Vladimir
  • 5,748
  • 16
  • 21
0

Héhé, I found this on internet, just a small paragraph on page 4 (or 246) talking about which norm one should take.

Noix07
  • 3,839