what is the image of $\partial 1_{S^n}$ where $\cdots \pi_n(S^n)\rightarrow \pi_{n-1}(X) \rightarrow \pi_{n-1}(B)\rightarrow\cdots$
what is the image of $\partial( 1_{S^n})$ for the exact sequence for the fibration $X \to E \to S^n$
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$\begingroup$ What fibration? $\endgroup$– Sean TilsonCommented Mar 27, 2011 at 4:18
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$\begingroup$ Sean : any fibration $X \rightarrow B$ with fibre $S^n$ $\endgroup$– JinoCommented Mar 27, 2011 at 11:18
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1$\begingroup$ 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 $$ \cdots \to \pi_{n+1}B \to \pi_n S^n \to \pi_n X \to \pi_n B \cdots$$. $\endgroup$– José Figueroa-O'FarrillCommented Mar 27, 2011 at 11:33
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$\begingroup$ sorry. I's an error. The fibration is $X \rightarrow E \rightarrow S^n$ $\endgroup$– JinoCommented Mar 27, 2011 at 13:29
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$\begingroup$ If this answer is sufficient for your purposes, why not accept it? $\endgroup$– Yemon ChoiCommented Oct 2, 2011 at 1:58
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Any fibration $X \to E \to S^n$ up to fiber homotopy equivalence is given by the "clutching construction" applied to a map $f: S^{n-1} \to G(X)$, where $G(X)$ denotes the topological monoid of self homotopy equivalences of $X$. Let $e: G(X) \to X$ be the evaluation map (evaluate an equivalence at the basepoint of $X$). The answer to your question is given by taking the homotopy class of the composition $$ S^{n-1} \overset{f}\to G(X) \overset{e} \to X . $$ Note on the clutching construction: Given $f$ as above, form $E = (D^n_- \times X) \cup (D^n_+ \times X)$ where the gluing is done along the equivalence $S^{n-1} \times X \to S^{n-1}\times X$ given by $(x,v) \mapsto (x,g(x)(v))$. The first factor projection $E \to S^n$ gives your fibration up to fiberwise equivalence. Small technical point: clutching doesn't give a fibration necessarily (it is what one calls a quasifibration), so we have to convert the map to a fibration in the usual way. However, if we had started with a map $S^{n-1} \to \text{homeo}(X)$ it would clutch to a fiber bundle.
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$\begingroup$ John: Thanks John. I have some questions for your answer. First, two fibrations $F \to E \to \B$ and $F^prime \to E^prime \to \B^prime$ are equivalent means that $F=F^prime$, $B=B^prime$ and there is a homotopy equivalence $f:E \to E^prime$? Second, how prove the fact that any fibration is fiber homotopy equivalent by the "clutching construction". If a good referece, please recommend. $\endgroup$– JinoCommented Mar 28, 2011 at 6:05
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$\begingroup$ See page 22 Hatcher's online $K$-book for a description of the clutching construction in the vector bundle case. The case of a fibration is basically the same. For the fibration case, see Brayton Gray's paper, On the iterated suspension, Topology ** 27 ** (1988), no. 3, 301-310. $\endgroup$ Commented Mar 28, 2011 at 17:03
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$\begingroup$ John: Tkanks John. I will study them. $\endgroup$– JinoCommented Mar 29, 2011 at 2:35