Timeline for Can the Lawvere fixed point theorem be used to prove the Brouwer fixed point theorem?
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37 events
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| May 3, 2020 at 9:51 | answer | added | Rupert | timeline score: 2 | |
| S Apr 20, 2017 at 20:11 | history | bounty ended | Qiaochu Yuan | ||
| S Apr 20, 2017 at 20:11 | history | notice removed | Qiaochu Yuan | ||
| Apr 19, 2017 at 5:19 | answer | added | Todd Trimble | timeline score: 10 | |
| Apr 17, 2017 at 3:02 | comment | added | Todd Trimble | @YemonChoi If that exponential $Y = [0, 1]^X$ exists in the sense of category theory, meaning $\hom(-, Y) \cong \hom(X \times -, [0, 1])$, then necessarily its topology is the topology of continuous convergence. See this post for more information: mathoverflow.net/a/242831/2926 | |
| Apr 14, 2017 at 11:17 | comment | added | François G. Dorais | @BenoîtKloeckner: see mathoverflow.net/questions/131413/… and mathoverflow.net/questions/202811/… | |
| Apr 14, 2017 at 6:13 | comment | added | Benoît Kloeckner | @FrançoisG.Dorais: I am not sure I get what is non-constructive about Brouwer's fixed point. The Sperner's lemma proof proceeds by constructing a convergent sequence of points that are approximate fixed points, in particular the limit of this sequence is a fixed point. Would you ask for more? | |
| Apr 14, 2017 at 5:39 | answer | added | Will Sawin | timeline score: 8 | |
| Apr 14, 2017 at 5:31 | comment | added | Yemon Choi | Apologies if this was clarified elsewhere, but what is the definition of the topology on $[0,1]^X$? For instance, if $X=\beta {\bf N}$ then $C(X,[0,1])$ is the unit ball of $\ell^\infty({\bf N})$ and this is compact in the relative weak-star topology (although the resulting compact space is surely not a continuous image of $[0,1]$) | |
| Apr 14, 2017 at 2:04 | comment | added | Qiaochu Yuan | @David: I'm in a room full of people thinking about this problem and so far we can show that if $X$ has the desired property that there's a surjection $X \to [0, 1]^X$, then $X$ can't be compact Hausdorff, so in particular can't be $[0, 1]$. This is because, if $X$ is compact Hausdorff, then either $X$ is finite or $[0, 1]^X$ fails to be compact (e.g. because it is the unit ball of an infinite-dimensional Banach space). | |
| Apr 14, 2017 at 1:24 | comment | added | David E Speyer | Is it clear that $[0,1]$ doesn't have the stated property? There are only continuum many continuous maps $[0,1] \to [0,1]$, so it isn't clear to me that there couldn't be a surjection $[0,1] \to [0,1]^{[0,1]}$. | |
| S Apr 13, 2017 at 22:50 | history | bounty started | Qiaochu Yuan | ||
| S Apr 13, 2017 at 22:50 | history | notice added | Qiaochu Yuan | Draw attention | |
| Apr 13, 2017 at 22:49 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Jul 15, 2013 at 17:37 | comment | added | Todd Trimble | Well, I'm not sure. All I can say is that there are a lot more wise-cracks about category theory, and a lot of people think they understand it well enough to make pronouncements on (and jokes about) it. Happily, such jokes are becoming more and more pass\'e, as more and more people see that category theory is seriously useful stuff, and that professional category theorists are not in the business of "trying to get something from nothing". (Still, I can't think of any other field where people make such jokes. Can you?) | |
| Jul 15, 2013 at 17:09 | comment | added | user36938 | @Todd Trimble: I have so little understanding of logic that I pay no attention to MO questions about logic; I did not know that a lot of them are naive. Maybe many other people here are in the same boat as me. So ironically, perhaps the greater success of category theory over logic at having fruitful interactions with the rest of mathematics causes more people to have an opinion on categorical questions while passing over in silence anything about logic. | |
| Jul 15, 2013 at 6:22 | comment | added | Todd Trimble | @user36938 : well, I see your point, but I think maybe you're being a little harsh ("does not belong on MO"). Even if we assume the question was a shade offhand (e.g., even if Qiaochu hadn't considered the constructivity aspects before posting), that's true of so many questions here; it's okay to kick it around for a few minutes before deciding that it's naive. It all comes under the heading of exploring. For a comparison: there are a lot of naive questions about logic here. But they usually don't elicit the same kinds of cracks, do they? No "logic corrupting the youth of today" trope. | |
| Jul 15, 2013 at 5:17 | comment | added | user36938 | @Todd Trimble: I have a lot of respect for the elegance of categorical arguments, and even of topoi, but questions like this one are what give category theory its reputation in some quarters for trying to get something from nothing. Unless one can formulate a convincing argument in the case $n=1$ (deduce the Intermediate Value Theorem from the existence of a topological space about whose existence there is no substantial proposed idea whatsoever), this does not belong on MO. It's akin to conflating flatness for modules and flatness for connections on vector bundles. | |
| Jul 14, 2013 at 20:29 | comment | added | Vidit Nanda | @QiaochuYuan yes, so we have ruled out (more than) $n$-manifolds for all $n$ as candidates for $X$. Fantastic question, by the way. | |
| Jul 13, 2013 at 21:23 | comment | added | Qiaochu Yuan | @Vidit: the space $X$ could itself be "infinite-dimensional." | |
| Jul 13, 2013 at 21:18 | comment | added | Vidit Nanda | @QiaochuYuan I mean that your space filling curve analogues, should they exist, would have to jump from finite to infinite dimensions because in general the function space is infinite dimensional. I'm no differential topologist, but this seems tough to arrange. | |
| Jul 13, 2013 at 20:45 | comment | added | Qiaochu Yuan | @Vidit: I'm not sure what you mean by dimension considerations. Analogues of space-filling curves could exist. | |
| Jul 13, 2013 at 17:36 | comment | added | Vidit Nanda | @QiaochuYuan here's an easier question: consider $n=1$, so $Y = [-1,1]$. For which topological spaces $X$ can one find a surjection from $X$ to the space of all maps $X \to [-1,1]$? It seems tough in general, wouldn't all manifolds get ruled out by dimension considerations alone? | |
| Jul 13, 2013 at 11:20 | comment | added | Todd Trimble | Category theory seems to have the universal property of eliciting snarky remarks from those who don't have much taste for it. | |
| Jul 13, 2013 at 8:03 | answer | added | Andrej Bauer | timeline score: 25 | |
| Jul 12, 2013 at 22:39 | comment | added | Vidit Nanda | @QiaochuYuan thanks! Now it makes more sense. | |
| Jul 12, 2013 at 21:32 | comment | added | Qiaochu Yuan | @Vidit: $D^X$ is the space of continuous functions $X \to D$, not the space of continuous functions $D \to X$. Surjectivity on points in $\text{Top}$ reproduces the usual notion of surjectivity, and these are also precisely the epimorphisms in $\text{Top}$ (although this is not true for, say, the category of Hausdorff spaces), although we don't need to know this. | |
| Jul 12, 2013 at 21:08 | comment | added | Gerald Edgar | Hey! Next, let's prove the measure-theoretic Fubini theorem from the Category theory Fubini theorem! | |
| Jul 12, 2013 at 21:06 | comment | added | Vidit Nanda | Something seems wrong: working in Top, what about the case when $X$ is a single point? For any $D$ (not just a disk), there is a single map $D \to X$, so sending the unique point of $X$ to this single map furnishes a surjection. What's going on?? I suspect surjectivity in Top is different from the Hom-set based definition... | |
| Jul 12, 2013 at 20:54 | comment | added | François G. Dorais | I guess I should mention, in passing, that Lawvere's is much more constructive than Brouwer's. So, if there is an implication that way, the space $X$ will have to contain a lot of non constructive data. (What is needed is essentially a choice of a point from every nonempty compact subspace of $\mathbb{R}$.) | |
| Jul 12, 2013 at 20:24 | answer | added | André Henriques | timeline score: 12 | |
| Jul 12, 2013 at 7:22 | comment | added | Dan Ramras | Qiaochu, thanks for the explanation. One might or might not want to work in a "convenient" category e.g. compactly generated (weak) (Hausdorff) spaces. | |
| Jul 12, 2013 at 7:18 | comment | added | Qiaochu Yuan | (Actually it is unnecessary to talk about exponential objects at all; we can instead work directly with morphisms $X \times X \to Y$, but the surjectivity condition becomes slightly more annoying to state.) | |
| Jul 12, 2013 at 7:16 | comment | added | Qiaochu Yuan | @Dan: it means an object $Y^X$ such that there is a natural isomorphism of functors $\text{Hom}(Z, Y^X) \cong \text{Hom}(Z \times X, Y)$. When this object exists in $\text{Top}$ it is often (but I am not sure if it is always) the space of continuous functions $X \to Y$ with the compact-open topology. Wikipedia says that it always exists and is always the space of continuous functions with the compact-open topology when $X$ is locally compact Hausdorff. | |
| Jul 12, 2013 at 7:11 | comment | added | Dan Ramras | Having just learned about Lawvere's theorem the other day (from some MO post), I've been wondering the same thing! | |
| Jul 12, 2013 at 7:10 | comment | added | Dan Ramras | Does "the exponential" in the last paragraph mean something other than the function space (with the compact-open topology)? | |
| Jul 12, 2013 at 6:53 | history | asked | Qiaochu Yuan | CC BY-SA 3.0 |