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
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. $\endgroup$– AlexENov 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! $\endgroup$– RuyNov 20, 2020 at 14:23