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.



Colimits of short exact sequences of C*-algebras
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
$$
 \varepsilon >\|p\_n(b\_n)\| = \text{dist}(b\_n,\text{Im}(i\_n))\geq $$ $$ \geq\text{dist}(\beta \_n(b\_n),\text{Im}(i))=
 \text{dist}(b,\text{Im}(i)) >1-\varepsilon .
 $$


