Splitting invariants (Ex)

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	[\mathbb{C}P^k, G/PL] \to L_{2k}(\mathbb{Z})$$ splits the above sequence for $k>2$. Additionally $[\mathbb{C}P^2,G/PL] \equiv \mathbb{Z}$ and the isomorphism is given by the surgery obstruction map.		[\mathbb{C}P^k, G/PL] \to L_{2k}(\mathbb{Z})$$ splits the above sequence for $k>2$. Additionally $[\mathbb{C}P^2,G/PL] \equiv \mathbb{Z}$ and the isomorphism is given by the surgery obstruction map.
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Latest revision as of 15:00, 1 April 2012

Prove that two maps $<wikitex>; Prove that two maps $f_1,f_2 \colon \mathbb{C}P^n \to G/PL$ are homotopic iff their splitting invariants agree for \leq i \leq n$. Use the exact sequence $$L_{2k}(\mathbb{Z}) \to [\mathbb{C}P^k,G/PL] \to [\mathbb{C}P^{k-1},G/PL] \to 0$$ and the fact that the surgery obstruction map $$\theta \colon [\mathbb{C}P^k, G/PL] \to L_{2k}(\mathbb{Z})$$ splits the above sequence for $k>2$. Additionally $[\mathbb{C}P^2,G/PL] \equiv \mathbb{Z}$ and the isomorphism is given by the surgery obstruction map. </wikitex> == References == {{#RefList:}} [[Category:Exercises]]f_1,f_2 \colon \mathbb{C}P^n \to G/PL$ are homotopic iff their splitting invariants agree for $2 \leq i \leq n$.

Use the exact sequence

and the fact that the surgery obstruction map splits the above sequence for $k>2$. Additionally $[\mathbb{C}P^2,G/PL] \equiv \mathbb{Z}$ and the isomorphism is given by the surgery obstruction map.