I know that in general exact-by-exact extensions of $C^\*$-algebras need not be exact. Is it true that, if we have a short exact sequence of $C^\*$-algebras


$$0 \to I \to A \to B \to 0$$


such that $I$ is exact and $B$ is nuclear, then $A$ is exact?


If it is true, could you give a reference of the result?
If it is false, could you give a counterexample?


Thanks in advance.



About nuclear-by-exact extensions
Yes, if the quotient $B$ is nuclear, then the extension is locally semisplit by the Choi-Effros lifting theorem. Hence if (in addition) $I$ is exact, then $A$ is exact (see for instance Exercise 3.9.8 in the book of Brown and Ozawa).




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NOTE: I wrote the following when I misread the question, and thought the question was what happens when $I$ is nuclear and $B$ is exact. As people might find it interesting, I've added it back into the answer:


There exist extensions $0 \to \mathcal K(\ell^2(\mathbb N)) \to A \to B \to 0$ with $B$ being (separable and) exact and $A$ non-exact. But the results underpinning this are deep. Examples can be found in the book of Brown and Ozawa, Theorem 13.4.1 (see Remark 13.4.2 for why this provides counter examples).


Here is a slightly different and more detailed explanation: By a theorem of Kirchberg, every exact $C^\ast$-algebra is locally reflexive, and by a theorem of Effros-Haagerup, if $A$ is separable, locally reflexive and $I$ is a nuclear two-sided closed ideal in $A$, then there exists a completely positive splitting $A/I \to A$. So in an extension $0\to I \to A \to B \to 0$ for which $I$ is nuclear and $B$ is separable, exact, then $A$ is exact (if and) only if the extension has a completely positive splitting.


In the case where $I = \mathcal K(\ell^2(\mathbb N))$, (and $A, B$ are unital), it follows from basic Brown-Douglas-Filmore theory that every unital extension $0 \to \mathcal K(\ell^2(\mathbb N)) \to A \to B \to 0$ has a completely positive splitting if and only if $\mathrm{Ext}(B)$ is a group. So any example where $B$ is exact and $\mathrm{Ext}(B)$ is not a group, provides a counterexample. Kirchberg showed (which essentially boils down to Theorem 13.4.1 that I mentioned above) that if $B$ is separable, exact, non-nuclear and QWEP, and $C$ is the unitisation of $C\_0((0,1], B)$, then $\mathrm{Ext}(C)$ is not a group, and therefore there exists a unital extension $0 \to \mathcal K(\ell^2(\mathbb N)) \to A \to C \to 0$ where $A$ is not exact even though $C$ is. So you can take any separable, exact, non-nuclear, QWEP $C^\ast$-algebra $B$ and construct a counterexample, e.g. $B= C^\ast\_r(\mathbb F\_2)$ (this is QWEP by Prop. 13.3.8 in Brown-Ozawa).


A different (also very deep) counterexample comes from Haagerup-Thorbjørnsen who show that $\mathrm{Ext}(C^\ast\_r(\mathbb F\_2))$ is not a group. So there exists a unital extension $0\to \mathcal K(\ell^2(\mathbb N)) \to A \to C^\ast\_r(\mathbb F\_2) \to 0$ where $A$ is not exact.


