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Jamie Gabe
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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).


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.

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).


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. $B= C^\ast_r(\mathbb F_2)$.

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.

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).


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.

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Jamie Gabe
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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).


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. $B= C^\ast_r(\mathbb F_2)$.

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.

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).

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).


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. $B= C^\ast_r(\mathbb F_2)$.

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.

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Jamie Gabe
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This is false: 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-HaagerupYes, 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 andthe quotient $B$ is separable, exactnuclear, then $A$ is exact (if and) only if the extension has a completely positive splitting.

Inis locally semisplit by the case where $I = \mathcal K(\ell^2(\mathbb N))$, (and $A, B$ are unital), it follows from basic Brown-DouglasChoi-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 counterexampleEffros lifting theorem. Kirchberg showedHence if (which essentially boils down to Theorem 13.4.1 that I mentioned abovein addition) that if $B$$I$ 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(see for instance Exercise 3. $B= C^\ast_r(\mathbb F_2)$9.

A different (also very deep8 in the book of Brown and Ozawa) 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.

This is false: 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. $B= C^\ast_r(\mathbb F_2)$.

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.

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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Jamie Gabe
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