@LeakyNun You can use a spectral sequence argument
In fact, I can tell you the idea. Let $X$ be a space and $C^*$ be the presheaf of singular cochains on $X$, which assigns to each open set $U$ the singular cochain group $C^*(U)$.
The inclusion $U \to V$ gives a natural restriction $C^*(V) \to C^*(U)$
This is a "homotopy sheaf" in the sense that there is a homotopy-equalizer diagram $C^*(U \cup V) \to C^*(U) \oplus C^*(V) \Rightarrow C^*(U \cap V)$, in the sense that the pullback of $C^*(U) \to C^*(U \cap V) \leftarrow C^*(V)$ is chain-homotopy equivalent to $C^*(U \cup V)$. This is exactly what Mayer-Vietoris says.
In particular if $\{U_i\}$ is a cover of $X$, there is a big Mayer-Vietoris diagram $$0 \leftarrow C^*(X) \leftarrow \bigoplus C^*(U_i) \leftarrow \bigoplus C^*(U_i \cap U_i) \leftarrow \bigoplus C^*(U_i \cap U_j \cap U_k) \leftarrow \cdots$$
This is actually an exact sequence. The same thing holds in eg de Rham cohomology but arguing this is much simpler there, because the sheaf of differential forms is a fine sheaf. This is sort of like a replacement for that in this weird sheaf of singular cochains!
Now you can append that cochain complex below with the complex of locally constant functions on $X$, and you have a double complex interpolating between the singular complex of $X$ and the Cech complex of $X$, such that the rows are exact.
This is standard spectral sequence argument now. Filter the double complex from top-to-bottom, so that the $E^1$ page consists of the Cech complex at the last row and nothing else, assuming $\{U_i\}$ is a good cover of $X$
$E^2$ page is just the Cech cohomology groups at the last row, and nothing else, and the whole spectral sequence degenerates at $E^2$. Total cohomology is Cech cohomology
Similarly run the filtered specseq for the filtration from right-to-left, and conclude total cohomology is singular cohomology
@Mathei @ÍgjøgnumMeg and the other algebraists, do you know a good source to learn about totally ordered groups and the semigroups which can be realized as the positive cone of a totally ordered group?
well the bible says that through god all things are possible, so from the contrapositive we infer that "God doesnt exist because infinite regress isnt possible"
well Aristotle was fairly reasonable about his cosmological argument, it's more applicable to people like Aquinas who tried to incorporate it into Christian theology
think of wrapping up an exact sequence on a triangle
if you have an exact sequence $C_\bullet$ just take $A_i = \bigoplus_{n \equiv i \pmod{3}} C_i$ for $i = 0, 1, 2$ and then you have an exact triangle $A_0 \to A_1 \to A_2$
and my prediction is that it will escalate into police-protester conflict (just like it did in the last N protests), and then the govt will condemn the "violence" (just like they did in the last N protests) without doing anything (just like the last N protests)
Do you buy that, though? Why would the police hate the protesters to the extreme that they'd be as violent that the government has to condemn them afterwards?
@LeakyNun You might be amused by Bishop Berkeley's argument for God, which is that events occur only because they are observed by someone - as long as no one is observing an event it isn't happening (if it's happening, you know it did, or at least you know someone who knows it did, so on). So occurrence is a causation of collective conscious, but then if this has to be consistent in a way independent of the intersubjective space of consciousness and objective in any way, there must be God
which is the entity which perceives all that happens :P
If $A$ is torsion-free, then $A$ embeds into $\Bbb Q \otimes_{\Bbb Z} A$. That's just a $\Bbb Q$-vector space, you can choose a basis and order lexicographically
I think there's some conjectures around the fact that if $M$ is a $3$-manifold which admits a taut foliation then $\pi_1 M$ is left-orderable.
If $G$ is left-orderable that's the same thing as saying $G$ acts on $\Bbb R$ by orientation-preserving homeomorphisms
If $M$ admits a taut foliation then you pass to the universal cover. You'd expect the closed transversal to become a copy of $\Bbb R$ upstairs (it may not be; it can be many copies of $\Bbb R$), then $\pi_1 M$ will act on $\Bbb R$ by deck transformations
This will give the purported left-order on $\pi_1 M$
But I think this is wide open
Ah OK here is the conjecture
Let $Y$ be a rational homology 3-sphere. The following are equivalent: (1) $\widehat{HF}(Y)$ is non-minimal, which apparently means $|\widehat{HF}(Y)| > |H_1(F; \Bbb Z)|$, where $\widehat{HF}(Y)$ is some flavour of Heegaard-Floer homology which always has cardinality $\geq |H_1(F; \Bbb Z)|$
(2) $\pi_1 Y$ is left-orderable
(3) $Y$ admits a co-orientable taut foliation (a locally nice decomposition of $Y$ into $2$-manifolds, the leaves of the foliation, such that there's a closed loop transverse to all those leaves).
@MikeMiller should explain (1)
@Alessandro See Theorem 6.8 in Ghys for a proof that $G$ is countable left-orderable iff $G$ injects into $\text{Homeo}^+(\Bbb R)$.
First start moving everything slightly than they had left. Just a few cms. Tilting things at a slight angle etc. Unpairing their socks. Hiding things then "finding" it for them at an obvious place. Nothing too obvious. Install tiny cams preemptively because they inevitably will, then mess with those too.
If they become superstitious, you hit gold. Tell them you spend your portion of the rent to exorcise the apartment every month. Pay someone to recite some nonsense at the start of every month and spray holy water at the place and then cut it all out -- until at the very end of the month.
@BalarkaSen I like the topology, I like how it is basically the cofinite topology. The same topology is given in the case of the affine space I think. I guess thats the relation between algebraic geometry and abstract algebra. Though this is Spec in general but Spec(Z) was the first example where how the topology was cofinite was easy to see.
@RyanUnger This looks so cool. I have a PDE course next semester and it goes into the Einstein field equations and Yang Mills (the physics aspect). I hope I can learn this.
@RyanUnger I was in a discussion today and an algebraic geometer had to clarify at some point in the conversation that by finite group he meant a finite group scheme
and then projects another copy of evenly spaced geodesics of the $S^2$ onto the unit disk (but this one is rotated 90 degrees) such that the geodesics all create a mesh
then
My question is: Does the mesh support a hyperbolic space
You might start by looking up youtube videos on them. People tend to try to phrase problems/proofs in easy to understand ways over there.
(There is also the issue that the more easily understood a statement is, the more likely it is leaving out minutia about the subject it concerns. So keep that in mind.)
user131753
@AlessandroCodenotti, @BalarkaSen: The book is published already by AMS.
