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First of all I claim that asking that $X$ is acyclic for ordinary homology with integral coefficients is the same as asking that it is acyclic with respect to every homology theory. This is because of …

Equivariantly split just means split in the category of spaces with a $C_2$-action, i.e. that there is an equivariant map $r:X→\{x_0\}$ such that $r\circ i$ is equivalent to the identity. This map is …

I think that the answer to your question is no, but not for the reason you may suspect. Let us try to do the "stupid" case first: fix an abelian group $A$ and you want a spectrum $X$ such that $H_0(X) …

This might be a naive answer, but I think it is more than a comment: $E_n$-algebra structures on $E$ give rise to (increasingly commutative) monoidal structures on the category of modules. In general …

In my opinion the construction in "the geometry of iterated loop spaces" by P. May should carry through the simplicial world. If you want a completely simplicial treatment you can find it in theorem 5 …

Let's try to see what a lift $\tilde f : X\to R\textrm{-triv}$ is. Recall that $R\textrm{-triv}$ is the $\infty$-groupoid of $R$-lines with a specified isomorphism with $R$, so a lift $\tilde f$ corre …

A first observation is that homology groups are themselves homotopy groups. Precisely there is a functor $\mathbb{Z}[-]$ from spaces to pointed spaces (it is easier to describe when you think of space …

The easy way for me to think about this things is via the Yoneda lemma. This works better with reduced $K$-theory. This is defined for a pointed space $X$ as $$ \tilde K^0 (X) := ker(K^0(X)\to K^0(*)) …

I cannot answer more precisely without knowing which paper you are referring to. However the point seems to be that you want to characterize the space $Z_f$. In fact there are plenty of spaces $W$ wit …

You can find such a map, but it goes the other way: $\Gamma(Y)\to \Omega X$. It is always wise to keep one's right adjoints on the right hand side.t But first, let me note that every simplicial abeli …

This is not true. Consider the following (split) homotopy fiber sequence $$S^1\to S^1\times S^1\to S^1$$ Then, by a standard argument, we have $$\Sigma(S^1\times S^1)=S^2\vee S^3\vee S^2$$ so for eve …

This is an assemblage of known results, I'll try to put a reference for all of them. By a classical theorem of Serre all stable homotopy groups are finite in positive degree. In particular we have $ …

First of all, note that right before example 3.9 they prove that $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-structu …

In general there is an obstruction living in $H^3(X,\pi_2Y)$. Choose a CW structure on $X$ and $Y$ with only one 0-cell. Then you can use $\varphi$ to define a map at the level of 1-skeleta (just by s …

I think thinking in terms of classifying spaces will help clarify the situation. We know that a rank $n$ complex vector bundle $V$ on $X$ is the same thing as a homotopy class of maps $f:X\to BU_n$. A …