Timeline for When $C(X)$ is an injective $C(X)$-module? Current answer is erroneous
Current License: CC BY-SA 3.0
14 events
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S Feb 13, 2015 at 0:58 | history | bounty ended | CommunityBot | ||
S Feb 13, 2015 at 0:58 | history | notice removed | CommunityBot | ||
Feb 5, 2015 at 21:53 | history | edited | Norbert | CC BY-SA 3.0 |
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S Feb 4, 2015 at 23:21 | history | bounty started | Narutaka OZAWA | ||
S Feb 4, 2015 at 23:21 | history | notice added | Narutaka OZAWA | Improve details | |
Jan 23, 2015 at 18:12 | vote | accept | Norbert | ||
Feb 2, 2015 at 20:01 | |||||
Dec 7, 2014 at 5:00 | answer | added | Narutaka OZAWA | timeline score: 7 | |
Dec 7, 2014 at 0:08 | comment | added | David Handelman | If you use the absolutely strictest version of injectivity (viewing C(X) as a ring, then saying it is an injective module in the ring-theoretic sense---which is how I interpreted the question initially), then the answer when $X$ is compact is simply that $X$ be finite (since a commutative unital ring with no nilpotents, on being injective as a module, must be von Neumann regular (aka absolutely flat) ...) | |
Dec 6, 2014 at 23:11 | history | edited | Norbert |
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Dec 6, 2014 at 23:02 | comment | added | Yemon Choi | I think this version is usually called "strict injectivity", at least in Helemskii's book. So just to clarify: you require that whenever $M$ is a closed sub-module of a Banach $C(X)$-module $N$, every bounded $C(X)$-module map $M\to C(X)$ has an extension to a bounded $C(X)$-module map $N\to C(X)$. Is that correct? | |
Dec 6, 2014 at 22:58 | comment | added | Norbert | The stricter version, where embeddings have closed range. | |
Dec 6, 2014 at 22:49 | comment | added | Yemon Choi | In other words: are we testing over what Helemskii calls the admissible embeddings, or just the embeddings with closed range? | |
Dec 6, 2014 at 22:47 | comment | added | Yemon Choi | Is this relative injectivity as in Helemskii's theory, or the stricter version? | |
Dec 6, 2014 at 22:45 | history | asked | Norbert | CC BY-SA 3.0 |