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We know the fact that $K_0(-)$ and $K_1(-)$ are continuous under inductive sequence of $C^*$-algebras (in fact inductive system), i.e. $K_0(\varinjlim A_n)=\varinjlim K_0(A_n)$ similar for $K_1(-)$. In fact it is also true that $M_k(\lim_{\rightarrow} A_n)=\varinjlim M_k(A_n)$ for $k\in \mathbb N$.

Q1: Does $\widetilde{(\varinjlim A_n)}$ coincide with $\varinjlim\tilde{(A_n)}$? In fact this is a claim in someone's book, but without a proof. If we let $(X,\lambda_n)$ be the inductive limit of $\tilde{A_1}\rightarrow \tilde{A_2}\rightarrow~\cdots$, then by universal property we get a unique morphism $\lambda: X\rightarrow \widetilde{\varinjlim A_n}$. How can we show $\lambda$ is injective? NB morphisms need not be unital, even though $C^*$-algebras are unital.

Q2: Can we find any other continuous functors? What about the universal group $C^{\star}$-algebras, tensor product of $C^{\star}$-algebras, cross product of $C^{\star}$-algebras and so on?

Q3: Do we know any functor which is not continuous?

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2 Answers 2

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The answer to Q1 is a standard fact, which can be found in the standard text books like Bruce Blackadar's book on $C^{\star}$-algebras. The answer to Q2 is "yes" and, again, looking in a text book helps.

For Q3, I think that it is worth noting that the Algebraic K-theory of $C^{\star}$-algebras is very interesting but not continuous, in fact the natural map

$$K^{alg}_i({\mathbb C}) = \lim_n \ K^{alg}_i(M_n{\mathbb C}) \to K^{alg}_i(\lim_n \ M_n{\mathbb C}) = K^{alg}_i(\mathcal K)$$

is only an isomorphism if $i=0$ (in fact, it is the zero-map if $i \neq 0$). The comparison map is much better behaved for commutative algebras if $i \leq 0$, but still not known to be continuous.

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  • $\begingroup$ Thanks for your answer and edition. Q1: I knew that, but no proof in the book. THe inductive limit A can be nonunital even though A_n are unital, e.g. A=K(H). So I worry about in whcih category the statment is true. $\endgroup$
    – m07kl
    Commented Oct 10, 2010 at 20:08
  • $\begingroup$ The unitization is a functor from the category of $C^*$-algebras with homomorphisms (not necessarily preserving the unit) to the category of unital $C^*$-algebras with unital homomorphisms. The statement is that this functor preserves inductive limits. The proof is really straightforward. $\endgroup$
    – Andreas Thom
    Commented Oct 10, 2010 at 20:14
  • $\begingroup$ I talked with my teather yesterday about non-continuous functor. He said that $B\otimes_{min}(-)$ and reduced cross product are not continuous in general. Today he tell me that for $B\otimes_{min}(-)$ I shall look at $B=C^*(G)$, where $G$ is discrete, infinite, hyperbolic, RF, has property T. Q1:What shall $G$ be? $G=SL(n,Z)$ for $n\geq 3$? Are these groups hyperbolic? Q2: Can some ones find reference to this? $\endgroup$
    – m07kl
    Commented Oct 12, 2010 at 21:11
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I have proved following functors are continuous: 1) $B\otimes_{max}(-)$ for any $C^*$-algebra $B$

2) $B\otimes_{min}(-)$ for exact $C^*$-algebra $B$

3) unitization of $C^*$- algebras.

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    $\begingroup$ This should have been added as an edit to your original question, and not left as a separate answer. $\endgroup$
    – Yemon Choi
    Commented Oct 12, 2010 at 20:50
  • $\begingroup$ Moreover, Questions 1 and 2 are things you should be asking your teacher (or Google, for that matter) $\endgroup$
    – Yemon Choi
    Commented Oct 12, 2010 at 20:51
  • $\begingroup$ You are right :) $\endgroup$
    – m07kl
    Commented Oct 12, 2010 at 21:12

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