Timeline for Is there a map of spectra implementing the Thom isomorphism?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2010 at 13:10 | comment | added | Johannes Ebert | That was a typo, $H \sigma^n\bZ=\Sigma^n H\bZ$. The composition is a weak equivalence iff $X^{\mu} \to \Sigma^n H \bZ$ is a Thom class. | |
| Nov 23, 2010 at 13:05 | history | edited | Johannes Ebert | CC BY-SA 2.5 |
added 1 characters in body
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| Nov 23, 2010 at 12:27 | comment | added | roger123 | This is interesting, Johannes. May I ask what $H\sigma^n\mathbb{Z}$ is and what the map $\Sigma^n H\mathbb{Z}\wedge H\mathbb{Z}\to H\sigma^n\mathbb{Z}$ does? One likes to have a map $f:X^\mu\wedge H\mathbb{Z}\to X_+\wedge \Sigma^n H\mathbb{Z}$ in the end to get the homology isomorphism, right? Do you say that $X^\mu\to \Sigma^n H\mathbb{Z}$ can be defined as a Thom class iff the map $f$ is a weak homotopy equivalence? Thank you. | |
| Nov 21, 2010 at 19:33 | vote | accept | skupers | ||
| Nov 21, 2010 at 16:32 | history | edited | Charles Rezk | CC BY-SA 2.5 |
Tex fix
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| Nov 21, 2010 at 10:50 | history | answered | Johannes Ebert | CC BY-SA 2.5 |