Normal maps and submanifolds (Ex)

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# If $(f, b)$ is normally bordant to a homeomorphism then the splitting obstruction along $Y$ vanishes.
# If $(f, b)$ is normally bordant to a homeomorphism then the splitting obstruction along $Y$ vanishes.
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[[Category:Exercises]]
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Revision as of 14:53, 1 April 2012

Let (f, b) \colon (M, \nu_M) \to (X, \xi) be a degree one normal map. For simplicity, assume that M and X are closed oriented \text{Cat}-manifolds of dimension n. Suppose that Y \subset X is a codimension k oriented submanifold X with normal bundle \nu_{Y \subset X} and that that f is transverse to Y. Prove the following:

  1. There is a canonical degree one normal map (f|_N, b') \colon (N, \nu) \to (Y, \xi|_Y \oplus \nu_{Y \subset X}).
  2. This defines a well-defined map \pitchfork_Y \colon \mathcal{N}(X,\xi)\rightarrow \mathcal{N}(Y,\xi|_Y \oplus \nu_{Y \subset X}).

Of course we have the surgery obstruction map

\displaystyle  \sigma \colon \mathcal{N}(Y, \xi_y \oplus \nu_{Y \subset X}) \to L_{n-k}(\pi_1(Y))

and the composite map

\displaystyle  \sigma \circ \pitchfork_Y \colon \mathcal{N}(X, \xi) \to L_{n-k}(\pi_1(Y))

which is called the splitting obstruction map along Y. In addition prove the following:

  1. If (f, b) is normally bordant to a homeomorphism then the splitting obstruction along Y vanishes.
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