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Homotopy theory, homological algebra, algebraic treatments of manifolds.

2 votes

Infinite loop spaces

If I am not mistaken, we get a counter-example to your last question easily. First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from $QA$ to $QB$ that is not a $H$-map to get a …
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4 votes
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The Thom map for the Brown-Peterson cohomology

Here is an "answer" which may be or not be good enough for your purpose, but which is easy to prove. Let's start with Ravenel-Wilson-Yagita Theorem 1.20. Applied to Eilenberg-Maclane spaces, it impli …
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5 votes
Accepted

Cohomology algebra of a fibration whose spectral sequence degenerates in the second term

The statement is false, here is a counterexample. First note that for a Lie Group $G$ and its closed subgroup $H$, we have a fibration $G/H\to BH\to BG$. $BG$ and $BH$ are not finite, but they are a …
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8 votes
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What is $Sq^i(\alpha^j)$ for all $i$ and $j$?

The Cartan formula $$Sq^i(xy)=\Sigma _{j+k=i}Sq^j(x)Sq^k(y)$$ together with the instability condition $$Sq^d(\alpha)=\alpha ^2 \mbox{ if $d=deg(\alpha )$}, Sq^i(\alpha )=0 \mbox{ if $d>deg(\alpha )$ …
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2 votes

Loop space generalization

[Edited after the comment below] Well, I don't know exactly what you mean by "geometric interpretation", but for topological group $G$ the unpointed mapping spaces $Map(BG,X)$ are examples of homotop …
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10 votes
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Representation of finite groups in a compact Lie group

In "Maps from $B\pi$ into $X$" Quart. J. Math. Oxford Ser. (2) 39 (1988), no. 153, 117–127., Wojtkowiak proves that the natural map $Rep(H,G)\rightarrow [BH,BG]$ is not surjective when $H=\Sigma _3$ …
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4 votes

Isomorphism between the Burnside ring $A(G)$ and the zeroth $G$-equivariant stable homotopy ...

This is described in Chapter 5 of Lewis, May, Steinberger "Equivariant Stable homotopy theory" (with contribution by McClure). Basically the idea is that a generator of $A(G)$ is a $G$-set, so 0-dimen …
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5 votes
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Equivalence of homotopy categories and model structure theory

You can find a "direct proof" in the paper by Curtis Simplicial Homotopy Theory (Advances in Mathematics, Volume 6, Issue 2, April 1971, Pages 107–209). The main ingredient is the use of barycent …
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-1 votes

The Eilenberg-MacLane spectrum and retractions

If it were the case, then $K(Z,n)$ would be a retract of $\underline{ku}_n$ where $\underline{ku}_n$ is the $n$-th infinite loop space associated to the spectrum $ku$. However, $ku$ is what is called …
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1 vote

Injectivity on rational homtopy implies surjectivity on rational cohomology for classifiying...

Write $kO$ for the connective $k$-theory, and $X$ for the connective delooping of $BTOP$. Then $H_*(BO;Q)$ and $H_*(BTOP,Q)$ are free commutative (in graded sense) generated by $\pi _*(kO,Q)\cong \p …
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3 votes
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Grading in Eilenberg-Moore spectral sequence

Different people use different notation on gradings, for example I would have called the bigrdading of $e_i$ $E_2^{1,2i}$. Supposing that this is not a typo, Quillen meant by $k$ in $E_s^{j,k}$ the …
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0 votes

Semidirect product of torus with cyclic group: representations/cohomology?

As to the cohomology, here is an answer, which is probably an "over-kill". Probably you can find an answer in much older literature like Borel's. Since $BT$ has torsion-free homology, COROLLARY 4.9 o …
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7 votes
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What are the cohomology groups $H^d(BSO_\infty,Z)$ and $H^d(BO_\infty,Z)$?

The Theorem 1.5 and 1.6 you quote give the answer. More precisely, for $SO$, in the range $d<6$, the only polynomial generators are $p_1$ which has degree 4, $\delta(w_2)$ with degree 3 and $\delta(w …
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5 votes
0 answers
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The behaviour of the suspension homomorphism on $H_*(QX;Z/p)$ for odd $p$ (Reference request)

The mod $p$ homology of $QX=\Omega ^{\infty}\Sigma ^{\infty}X$ for connected $X$ was computed by Dyer-Lashof Homology of Iterated Loop Spaces, Amer. J. of Math., vol.84, No.1 pp 35-88. 1962 It follows …
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4 votes
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spectral sequence with non-trivial action on coefficients

I guess here $\mathbb Z/2$-module means $\mathbb Z[\mathbb Z/2]$-module. So let's take $N=1$, $A=\mathbb Z$. $H_0(\mathbb Z/2,A)$ is the coinvariant $A/\mathbb Z/2$ so it is $A$ in the case of the tr …
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