# Talk:Surgery obstruction map I (Ex)

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so it depends only on the bundle over $X$. | so it depends only on the bundle over $X$. | ||

There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$ | There are fiber homotopically trivial bundles on $X$ corresponding to classes in $[X,G/TOP]$ | ||

− | which restrict to any given class in $[S^4,G/Top]$, | + | which restrict to any given class in $[S^4,G/Top]$, as follows from the Puppe sequence with $\pi_7(G/TOP)=0$. |

− | + | ||

From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on $S^4$ we have such vector bundles with first Pontryagin class $48k$ times the generator of $H^4(S^4)$. | From [[Fibre_homotopy_trivial_bundles_(Ex)|another exercise]] we know that on $S^4$ we have such vector bundles with first Pontryagin class $48k$ times the generator of $H^4(S^4)$. | ||

This means that on $X$ we have a vector bundle $\xi$ with $p_1(\xi)=48$ whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence $\phi$. | This means that on $X$ we have a vector bundle $\xi$ with $p_1(\xi)=48$ whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence $\phi$. |

## Revision as of 21:53, 29 May 2012

We take : The normal map gives the base point of . An element of is given by a bundle together with a fiber homotopy trivialization . Under the isomorphism , the pair corresponds to a normal map covered by . The surgery obstruction of a normal map covered by equals

so it depends only on the bundle over . There are fiber homotopically trivial bundles on corresponding to classes in which restrict to any given class in , as follows from the Puppe sequence with . From another exercise we know that on we have such vector bundles with first Pontryagin class times the generator of . This means that on we have a vector bundle with whose sphere bundle is fiber homotopically trivial, by a fiber homotopy equivalence . Now is the sum of and in with respect to the Whitney sum. We compute

and so the surgery obstruction is not a group homomorphism with respect to the Whitney sum.