5
$\begingroup$

Assume I have an inductive system of short exact sequences of $C^{\ast}$-algebras (i.e., short exact sequences $0 \to A_n \to B_n \to C_n \to 0$ together with transformations from the $n$-th to the $(n+1)$-st short exact sequence so that all squares commute). If I form now the colimit of the $C^{\ast}$-algebras, is the resulting sequence $$0 \to \varinjlim A_n \to \varinjlim B_n \to \varinjlim C_n \to 0$$ still exact? Note that I do not want to assume here that the connecting maps in the colimits I form are injective.

$\endgroup$

1 Answer 1

3
$\begingroup$

My notation $$ i_n:A_n\to B_n, $$ $$ p_n:B_n\to C_n, $$ $$ i:\displaystyle \lim_\to A_n\to \displaystyle \lim_\to B_n, $$ $$ p:\displaystyle \lim_\to B_n\to \displaystyle \lim_\to C_n, $$ $$ \beta _n:B_n\to\displaystyle \lim_\to B_n. $$

I suppose the only contentious point is to prove that $\text{Ker}(p) \subseteq \text{Im}(i)$, so suppose that this fails. For each $\varepsilon >0$ we may then choose some $b\in \displaystyle \lim_\to B_n$, such that

  • $\|p(b)\|<\varepsilon $,

  • $\text{dist}(b,\text{Im}(i)) > 1-\varepsilon $.

Since the union of the images of the $B_n$ is dense in $\displaystyle \lim_\to B_n$, we may assume that $b=\beta_n(b_n)$, for some $b_n\in B_n$.

Increasing $n$, if necessary, we may assume that moreover $\|p_n(b_n)\|<\varepsilon $. But this is a contradiction since $$ \varepsilon >\|p_n(b_n)\| = \text{dist}(b_n,\text{Im}(i_n))\geq $$ $$ \geq\text{dist}(\beta _n(b_n),\text{Im}(i))= \text{dist}(b,\text{Im}(i)) >1-\varepsilon . $$

$\endgroup$
2
  • $\begingroup$ Looks good, thanks! I think that we need to use that images of ${}^*$-homomorphisms between $C^*$-algebras are closed, otherwise you might not be able to find the element $b$ in the first step. Actually, that the map $i$ is injective is also not completely trivial: I had to use that each of the maps $i_n$ is isometric since they are injective. $\endgroup$
    – AlexE
    Nov 20, 2020 at 7:00
  • 2
    $\begingroup$ Alex, you are correct on both of your claims. Indeed what makes life so much easier when dealing with $^*$-homomomorphisms on C*-algebras is that every such map is isometric when the kernel is modded out, so, as you say, must have a closed range. I figure Banach space people must be very envious of us for this! $\endgroup$
    – Ruy
    Nov 20, 2020 at 14:23

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.