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.


1 Answer 1


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 $$ \varepsilon >\|p_n(b_n)\| = \text{dist}(b_n,\text{Im}(i_n))\geq $$ $$ \geq\text{dist}(\beta _n(b_n),\text{Im}(i))= \text{dist}(b,\text{Im}(i)) >1-\varepsilon . $$

  • $\begingroup$ 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

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