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I am trying to understand direct limit in category of $C^*$ algebras.

Is it well known that direct limit behaves well with double dual and quotient of $C^*$ algebras?

Any references or ideas?

P.S: This question was first asked on mathatack but there I did not get any answer!

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It definitely does not work for double duals. Indeed, if you view compact operators as an inductive limit of matrix algebras, then the sequence of double duals is exactly the same, but the double dual of compacts is $B(H)$.

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Direct limits do behave well with respect to quotients. Suppose $A$ is the direct limit of a sequence $(A_n)$ with connecting $*$-homomorphisms $\phi_n: A_n \to A_{n+1}$, and let $I$ be a closed ideal of $A$. Then $I$ pulls back to an ideal $I_n$ of $A_n$ for each $n$, and the connecting maps $\phi_n$ are compatible with the quotients, i.e., they lift to connecting maps $\tilde{\phi}_n: A_n/I_n \to A_{n+1}/I_{n+1}$. Moreover, $A/I$ is then the direct limit of the sequence $(A_n/I_n)$. This is easy because the maps $\tilde{\phi}_n$ have no kernel and hence are isometric, and the whole sequence isometrically embeds in $A/I$.

On the other hand, direct limits do not commute with pullbacks; see this question.

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  • $\begingroup$ Nik, aren't you assuming that the connecting maps are injective? I think that the conclusion holds anyway, because the union of the images of the $A_n$'s is dense in $A$ (and the same for $I_n$'s and $I$). $\endgroup$
    – Mateusz Wasilewski
    Commented Oct 3, 2018 at 9:03
  • $\begingroup$ Do you mean the $\phi_n$ or the $\tilde{\phi}_n$? The $\tilde{\phi}_n$ are injective. $\endgroup$
    – Nik Weaver
    Commented Oct 3, 2018 at 11:30
  • $\begingroup$ I confused the two, I'm sorry. Of course you're right. $\endgroup$
    – Mateusz Wasilewski
    Commented Oct 3, 2018 at 12:06
  • $\begingroup$ No problem at all. $\endgroup$
    – Nik Weaver
    Commented Oct 3, 2018 at 12:09
  • $\begingroup$ @NikWeaver: Can you please add few lines why the limit of the sequence is $A/I$? May be this is obvious but since I am beginner so I don’t follow it. Thanks $\endgroup$
    – Math Lover
    Commented Oct 12, 2018 at 7:47

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