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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