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Sep 18, 2013 at 16:08 comment added magya_bloom Thanks Tobias. I should have said, assuming the embedding: For $a_n \in A_n$, $a_n \mapsto a_n \otimes \mathbb{I}_{n+1}$, where $\mathbb{I}_{n+1}$ is the identity on factor indexed $n+1$.
Sep 18, 2013 at 6:14 comment added Tobias Kildetoft Is there a unique way to embed $A_n$ in $A_{n+1}$? Is it obvious that this union does not depend on how they are embedded? (I guess technically the union is a direct limit, and this is why it does not seem obvious that it should be independent of this choice).
Sep 17, 2013 at 15:23 history asked magya_bloom CC BY-SA 3.0