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let $p:E\rightarrow B$ be a fibration with fiber $F$, ($E$ and $B$ are cw complexes). $B^k$ denotes the $k$-skeleton of $B$.

1) what does this sentence mean :"we denote the restriction of $E$ to $B^k$ by $E^k$".

2) why there is an exact sequence $$\cdots \rightarrow H_{i+1}(E^{k+1},E^k)\rightarrow H_i(E^k,F)\rightarrow H_i(E^{k+1},F)\rightarrow \cdots$$

3)What is a trivialization of $E$ over the $(k+1)$-cells $D_\alpha^{k+1}$ of $B^{k+1}$

from which we conclude that $$H_{i+1}(E^{k+1},E^k)=H_i(\sqcup_\alpha D_\alpha^{k+1}\times F \, , \, \sqcup_\alpha S_\alpha^{k}\times F)$$

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The definition of $E^k$ just means that it is the elements of $E$ that get sent to $B^k$. – Thomas Andrews May 13 '11 at 12:26

To answer (1), the restriction is defined by $E^k=p^{-1}(B^k)$, which forms a fibration over $B^k$.

To answer (2), this is the long exact sequence for the triple $F\subset E^k\subset E^{k+1}$.

Finally, to answer (3), a trivialization of a fibre bundle means that it is isomorphic to a "trivial" fibre bundle, which is just a product of the base with the fibre. (I'm not sure why we switched to a bundle rather than a fibration.) In any event, the fibre bundle is a product when restricted to the interior of each $k+1$-cell, so that it looks like $D^{k+1}\times F$ on that cell. When we take the relative homology $H_{i+1}(E^{k+1},E^k)$, we are modding out by the boundary of each $k+1$-cell so that the result is the same as looking at each cell separately, noticing that $E^{k+1}$ is a product there, and then modding out by the boundary, which gives the right-hand side of the formula you stated.

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