Timeline for No matter how many algebraic invariants we attach to topological spaces, there will always be nonhomeomorphic spaces agreeing on all their invariants
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Mar 24, 2016 at 13:17 | comment | added | Anton Fetisov | @SebastianGoette Yes ncatlab.org/nlab/show/sober+topological+space | |
| Mar 24, 2016 at 13:05 | comment | added | Sebastian Goette | I am sorry I did not come up with a more sensible example. The basic question would be: "what extra conditions does one need to make sure a lattice isomorphism between the collection of open sets of two topological spaces is induced by a bijective set-map of the underlying spaces?" Is sobriety really enough? | |
| Mar 24, 2016 at 12:08 | comment | added | Anton Fetisov | @SebastianGoette Indiscrete spaces are a mock of topology. Of course I consider only sober spaces, which cover all real examples. But if you wish so, you could always label each open subset with its elements, it would be an absolute invariant. | |
| Mar 24, 2016 at 9:45 | comment | added | Sebastian Goette | You write "the lattice of open subsets looks like a perfectly fine algebraic invariant to me, but it certainly discriminates spaces" - is this true even if you don't know which abstract lattice element corresponds to which open set? For example, all indiscrete spaces have the same lattice of open set, but are homoemorphic only if their cardinailties agree. | |
| Mar 23, 2016 at 15:15 | comment | added | Karol Szumiło | I have just realized that the final remarks of the answer mention this issue, so I guess you can ignore my comments. | |
| Mar 23, 2016 at 15:12 | comment | added | Karol Szumiło | Let me clarify: if we interpret invariants as functors, then complete invariants correspond to conservative functors not faithful ones. I suspect that there isn't any conservative functor from either $\mathsf{Top}$ or $\mathsf{HoTop}$ to $\mathsf{Set}$, but this is not what Freyd's theorem is about. | |
| Mar 23, 2016 at 15:05 | comment | added | Karol Szumiło | I don't understand how this counts as an answer to the question as posed. After all, the question was about homeomorphism invariants and the category $\mathsf{Top}$ does admit a faithful functor to $\mathsf{Set}$, namely the standard forgetful functor. How is non-existence of such functor for the homotopy category relevant? | |
| Mar 23, 2016 at 14:46 | comment | added | Tom Goodwillie | Or for a concrete finite-dimensional example, this time one in which both spaces have plenty of nontrivial homotopy groups: the diagonal inclusion $X\to X\wedge X$ when $X=\mathbb CP^2$. | |
| Mar 23, 2016 at 13:25 | comment | added | Sean Tilson | Another example of a map that is 0 in homotopy but not null is given by any cohomology operation in singular cohomology of non-zero degree. | |
| Mar 23, 2016 at 13:09 | comment | added | Tom Goodwillie | Another remark about Whitehead's Theorem: a map $X\to Y$ (of based connected CW complexes) is not necessarily nullhomotopic if it induces the null map on every $\pi_n$. For example, it may be that for every $n$ either $\pi_nX$ or $\pi_nY$ is trivial. One example is a torus mapping to a $2$-sphere. | |
| Mar 23, 2016 at 13:04 | comment | added | Tom Goodwillie | Freyd's result implies that no set of functors from the homotopy category of based spaces to the category of sets can discriminate between all morphisms. If you want to read this as "no matter how many such invariants ..." you should note that "many" means a set as opposed to a proper class. | |
| Mar 23, 2016 at 11:39 | history | edited | Anton Fetisov | CC BY-SA 3.0 |
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| Mar 23, 2016 at 11:33 | history | answered | Anton Fetisov | CC BY-SA 3.0 |