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Oct 2, 2011 at 1:58 comment added Yemon Choi If this answer is sufficient for your purposes, why not accept it?
Mar 29, 2011 at 11:29 history edited Jino CC BY-SA 2.5
edited title; edited title; edited title
Mar 27, 2011 at 16:41 answer added John Klein timeline score: 4
Mar 27, 2011 at 13:31 history edited Jino CC BY-SA 2.5
edited title
Mar 27, 2011 at 13:29 comment added Jino sorry. I's an error. The fibration is $X \rightarrow E \rightarrow S^n$
Mar 27, 2011 at 11:33 comment added José Figueroa-O'Farrill The sequence you have written is not, however, one with fibre $S^n$, but rather one with fibre $X$ and base $S^n$. If you really mean $S^n \to X \to B$, then the sequence is $$ \cdots \to \pi_{n+1}B \to \pi_n S^n \to \pi_n X \to \pi_n B \cdots$$.
Mar 27, 2011 at 11:18 comment added Jino Sean : any fibration $X \rightarrow B$ with fibre $S^n$
Mar 27, 2011 at 4:18 comment added Sean Tilson What fibration?
Mar 27, 2011 at 2:46 history asked Jino CC BY-SA 2.5