@BalarkaSen what's a good example to see how the serre spectral sequence works? not to deduce any new information, but just to, say, verify that it is true and be able to move from one page to the next
and I guess it would be a good exercise to work out differentials?
@LeakyNun In all the examples I know either you don't need to compute differentials too much or you need to use machinery (Leibniz formula) to compute differentials in cohomology.
@LeakyNun there are some exercises in spectral sequences in ch1 of Vakil - but iirc they're all quite trivial (but you might like them anyway as trivial exercises to test your spectral sequence knowledge)
I am identifying $H^p(B, H^q(F)) \cong H^p(B) \otimes H^q(F)$ (of course $B$ and $F$ are special here), so you can write a cohomology class there as a tensor product.
@MikeMiller Oh. That's a lot like how the algebra structure on $H^*(\Omega S^n)$ is.
The weirdest thing is that you have a cup product in homology for a finite group $G$ $H_n(BG,A) \times H_m(BG,A) \to H_{n+m+1}(BG,A)$ coming from the cup product in Tate cohomology (yes, that index is right)
$\Omega \Sigma X$ is called the James construction on $X$. It is the "free topological group on X", in some sense. When X has free abelian homology the homology algebra of J(X) is the tensor algebra (read: free algebra) on the reduced homology of X
can one work out geometrically what the cup-product does for $\Bbb{RP}^\infty$? If we take a generator of $H_1(\Bbb{RP}^\infty,\Bbb Z)$, then taking the cup product with it should give an isomorphism $H_{n}(\Bbb{RP}^\infty,\Bbb Z) \to H_{n+2}(\Bbb{RP}^\infty,\Bbb Z)$, I think
Let $(X,d)$ be a metric space whose metric topology is discrete. Is the reason why that $(X,d)$ is not Geodesic essentially due to the fact that there are no continuous injections from $[0,1]$ to $X$?
apparently, there's a geometric description for the cup-product on homology, using joins of cycles, which explains the dimension shift: arxiv.org/pdf/0911.3014.pdf
For Z/2 with Z/2 coefficients the Tate cup product on homology is an isomorphism in every (pair of) degree(s)
Ok this works with Z coefficients as well. It's an iso (Z/2) otimes (Z/2) -> Z/2 when the starting degrees are both odd and otherwise automatically zero
btw i think it would make life easier if you drop the tensor notation and just identify $H^*(\mathbb{C}\mathbb{P}^{\infty},\mathbb{Z})$ with $\mathbb{Z}[x]$ with $|x|=2$ and $H^*(S^1,\mathbb{Z})$ with $\mathbb{Z}[y]/(y^2)$ with $|y|=1$
(or maybe you really like writing $\otimes$ and using a bunch of $1$'s)
$\newcommand{Ext}{\operatorname{Ext}}\newcommand{Hom}{\operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r \in R$.
Applying snake lemma to the following diagram:
$$\begin{array}{c} 0 & \longrightarrow & A~~ & \longrightarrow & B~~ & \longrightarrow & C~~ & \longrightarrow & 0 \\ &
$\newcommand{Ext}{\operatorname{Ext}}\newcommand{Hom}{\operatorname{Hom}_R}$Let $R$ be a commutative ring with unity and $r \in R$.
Applying snake lemma to the following diagram:
$$\begin{array}{c} 0 & \longrightarrow & A~~ & \longrightarrow & B~~ & \longrightarrow & C~~ & \longrightarrow & 0 \\ &
