Theorem 35.5:

With notations as in theorem (35.4),

$\displaystyle H_{k-1}(X\sqcup_fE^k) = H_{k-1}(X)/$ im $\displaystyle \;H_{k-1}(f),\quad H_k(X\sqcup_fE^k) = H_k(X)\oplus$ ker $\displaystyle \;H_{k-1}(f)$

nisha 2012-03-20