Timeline for Is there a good definition of (topological) K-Theory over arbitrary spaces?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 1, 2011 at 19:45 | comment | added | Alain Valette | @pudin: Here is a very adequate quote from John von Neumann: "Young man, in mathematics you don't understand things. You just get used to them." | |
| Sep 1, 2011 at 18:13 | comment | added | Alain Valette | @pudin: I quit, by lack of interest for the direction taken by the discussion. (We started from K-theory, remember?). I suggest you pass the question to a set-theorist. | |
| Sep 1, 2011 at 12:29 | comment | added | old account | I've just been informed over lunch, that you cannot even form $\{C\}$ as my previous example states it, and neither can one form the collection of isomorphism classes of a non-small category, because set-theory simply does not allow a formulation of this... | |
| Sep 1, 2011 at 9:57 | comment | added | old account | they certainly are all isomorphic, but you still have given no justification of why that collection of isomorphism classes is a set. I mean it certainly is a class with only one element, but does that make it a set? (we're going off topic i fear, is that a problem?) I guess it all boils down to wether $\{C\}$ is a set for a proper class $C$, which nobody here in muenster's topology group seems to know... But it would certainly violate ZFC since the union axiom states that the union of a set is a set and in our case that would be $C$, but ZFC doesn't seem like the correct framework... | |
| Aug 31, 2011 at 19:47 | comment | added | Alain Valette | @ pudin. Remember that for $K(X)$, only finite-dimensional vector bundles are considered. Now, let us stick to the case where $X$ is a point. Let $F$ be a field. You can probably claim that the class of 1-dimensional $F$-vector spaces is proper, since to every set $E$ you may associate the 1-dimensional space $F^{\{E\}}$. But they are all isomorphic, right? Conclusion: the class of all vector bundles over $X$ is proper, but the set of isomorphism classes of such bundles is indeed a set. | |
| Aug 31, 2011 at 15:30 | comment | added | old account | for different choices of $S$ i get bijections between those "sets of isomorphsim classes of vecotrbundles in $S$". The same goes for the "set of isomorphismclasses of finitely generated projective modules over some ring". was this more understandable? i find it very hard conveying these things without a blackboard | |
| Aug 31, 2011 at 15:27 | comment | added | old account | All vectorbundles do not form a set right? (if i am wrong here than just ignore me) Just like all vectorspaces don't or all sets don't. At least there is no a priori reason why they should. Now saying "Set of isomorphism classes of vectorbundles" is without meaning somehow, isn't it? The only way I know out of this is to pick a SET $S$ of vectorbundles with the property that any vectorbundle is isomorphic to one in $S$. (for a compact space X the set $S$ might be the set of finitedim. subbundles of $X \times \mathbb R^\infty$). Here "set of isoclasses" does make sense and... | |
| Aug 31, 2011 at 15:08 | comment | added | Alain Valette | @ pudin. Sorry, I don't understand your comment. What is an isoclass, maybe? Can you explain what you mean when $X$ is a point, i.e. we deal with finite-dimensional vector spaces over a given field? | |
| Aug 31, 2011 at 14:10 | comment | added | old account | hmm doesn't the "up to isomorphism" destroy your argument? one may restrict to equivalence classes of some set of bundles such that any bundle is isomorphic to one in there (or in your case the idempotents) but the set of ALL isoclasses? | |
| Aug 30, 2011 at 15:58 | history | answered | Alain Valette | CC BY-SA 3.0 |