# Colimits of short exact sequences of C*-algebras

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.

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

• 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. Nov 20, 2020 at 7:00
• 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!
– Ruy
Nov 20, 2020 at 14:23