i went to an israeli restaurant a couple days ago that was expensive so i'd avoided it for a while and had fresh mushroom hummus for lunch and it was damn good
So we were talking about how $H_n(X)=H_n(Y)$ for all $n$ does not imply that $X$ and $Y$ are homotopy equivalent, for example using singular homology with coefficients in $\Bbb Q$ we have that $\Bbb R\Bbb P^2$ and a point have the same homology groups in all dimensions
But things can get worse than that, $T^2$ and $S^1\vee S^1\vee S^2$ have the same homology in all dimensions for all homology theories
So the question is: Is there a noncontractible space which has the same homology as a point in all dimensions for all homology theories?
btw you can show that $T^2$ is not homotopy equivalent to $S^1\vee S^1 \vee S^2$ by showing that he cup product doesn't vanish on $T^2$, but is trivial on the latter
@AlessandroCodenotti Recall that for a discrete group $G$, there is an invariant called the "group homology" of $G$, written $H_*(G;\Bbb Z)$, defined in terms of an acyclic resolution of $\Bbb Z$ as a $\Bbb ZG$-module. This invariant is in fact isomorphic to $H_*(K(G,1);\Bbb Z)$.
So first of all, there are no finite groups that are acyclic in the sense that their group homology vanishes, but there are infinite finitely presented groups: encyclopediaofmath.org/index.php/Acyclic_group
So this space, $K(G,1)$, has no homology; this implies the same of $\Sigma K(G,1)$
Now the suspension of any connected space is simply connected
So $\Sigma K(G,1)$ is simply connected and acyclic. It is a corollary of Whitehead's theorem and the Hurewicz theorem that $\Sigma K(G,1)$ is then contractible
So this space $K(G,1)$ is stably trivial but unstably quite interesting, as it has (by definition) $\pi_1 K(G,1) = G$
Thus every (even extraordinary) homology theory on $K(G,1)$ outputs the homology of a point, but it is not contractible
generally to see "all of the information" of a non-simply connected space you want to consider also the homology with respect to local coefficient systems, or equivalently, the homology of all covering spaces
but there are finite groups $H$ and $G$ such that $\Bbb Z[H] \cong \Bbb Z[G]$ (stumbled upon a paper once), so you can't recover $\pi_1$ from local-coefficient systems homology, since we're still "linearizing" at throwing non-abelian information out
You were successful then, thanks! I still have quite a few courses in algebraic topology that I can take so hopefully at some point I'll be able to follow the proofs as well
So really you should be writing down some category of subgroups (where automorphisms are given as G/N_G(H)), and then a functor from that to abelian groups by taking homology
so if you have a covering space $E \to X$, you can take the cohomology of $E$ and that's a $\pi_1(X)$-module, then you can take the group cohomology of that, I hadn't thought about that
is the equivalence of local coefficient systems and homology of covering spaces related to the fact that local coefficient systems are $\mathbf{Ab}$-valued sheaves and covering spaces "are" $\mathbf{Set}$-valued sheaves?
@MatheinBoulomenos It's a $\text{Deck}(E)$-module, right? I hadn't thought to take the group cohomology of it, just keep track of $H_* E$ as a $\text{Deck}(E)$-module
@MatheinBoulomenos Hmm, perhaps there's a proof there. For me it is a calculation
I just realized you can put some hot water in a ramen noddle bag , dump the water , open the ramen noodle block , put some cheese and eat it like a sandwich
@TedShifrin Maybe you can fix $A$ because it doesn't have much of effect on anything. $B$'s placement determines the measure of $AB$, which in turn determines where $C$ can go.
let $Ln$ be the principal branch of the complex logarithm (defined on $(-\pi, \pi]$). So now I am asked to evaluate $Ln((-1+i)^2) = Ln(-2i) = \ln(2) -i \frac{\pi}{2}$. How do I know its not $...= \ln(2) +i \frac{3\pi}{2}$? because its outside the domain of definition?
Right, @CaptainAmerica. So did you see what I said about sampling with $B$ uniformly distributed around the circle and using that to estimate the answer?
@JoeShmo Pick three points at random on a circle. Draw the triangle with them as vertices. What's the likelihood that the center of the circle is inside the triangle?
It's effectively the same as Milnor's but one lemma is speedily proven with symplectic geometry. But sure! I will have to write out the details again, so another time indeed.
