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This seems to be an answer, based on discussion with Maxime Ramzi in the comments. The space of $E_\infty$ $KU$-algebra maps from $KU\otimes S^1_+\to F(S^1_+,KU)$ is the same as the space of $E_\infty …

For a more self-contained answer, let $L$ be the tautological line bundle over $\mathbb{C}P^\infty$. This gives a class $x=[L]-[1]\in\widetilde{K}^0(\mathbb{C}P^\infty)$. It is standard that $K^0(\m …

For any finite abelian $G$ and $H\leq G$ we have a geometric fixed-point functor $\phi^H\colon\text{Sp}_G\to\text{Sp}$ which preserves smash products and sends the equivariant sphere $S^0_G$ to $S^0$. …

Depending on exactly what you are trying to do, you may find it more useful to consider $E^0(BV)$, where $E$ is Morava $E$-theory and $V\simeq(\mathbb{Z}/p)^d$. (Here everything is $2$-periodic and co …

There is a paper by Milnor called On the homology of Lie groups made discrete which contains many results on this kind of problem, as well as references to related literature. However, I don't think …

Using the Hopf fibration you can show that $\Omega^2S^2\simeq\mathbb{Z}\times\Omega^2S^3$. Also $\pi_1(\Omega^2S^3)=\pi_3(S^3)=\mathbb{Z}$, so the Universal Coefficient Theorem gives $[\Omega^2S^3,S^ …

There is an Atiyah-Hirzebruch spectral sequence $H^*(X;\mathbb{Z}[u,u^{-1}])\Longrightarrow K^*(X)$, in which the differentials are always torsion-valued. Thus, if $H^*(X;\mathbb{Z})$ is torsion-free …

First, you just have $K^n(\mathbb{Z}\times BU)=\text{Map}(\mathbb{Z},K^n(BU))$, so you can work with $K^*(BU)$, which is technically more convenient. In particular, this is the inverse limit of the r …

The general point is just that if $\mathcal{U}$ is equivalent to the free symmetric monoidal category $F\mathcal{C}$ generated by $\mathcal{C}$ then $K(\mathcal{U})\simeq\Sigma^\infty_+\mathcal{C}$. …

Adams operations exist in quite wide generality. For any even periodic ring spectra $E$ and $F$, we have associated formal groups $G_E$ and $G_F$ over base schemes $S_E$ and $S_F$. There is a moduli …

Usually you are computing $H^*(X)$ or $K^*(X)$ for a reason; for example if $H^*(X)\not\simeq H^*(Y)$ then you know that $X$ and $Y$ are not homotopy equivalent, but if $H^*(X)\simeq H^*(Y)$ and this …

There are cofibrations $$\Sigma^2 kU\xrightarrow{v} kU\to H\xrightarrow{\alpha}\Sigma^3 kU$$ and $$ H\xrightarrow{p}H\to H/p\xrightarrow{\beta} \Sigma H. $$ The composite $\alpha\beta\colon H/p\to\ …

I'll just give an answer for finite complexes $X$. The condition $\widetilde{K}^*(X)=0$ only depends on the suspension spectrum of $X$ so this is naturally regarded as a question in stable homotopy th …

Consider a map $f\colon X\to Y$ of spaces (such as $B(G^{\text{disc}})\to B(G)$). Say that $f$ is a $K(n)$-equivalence if $K(n)^*(f)\colon K(n)^*(Y)\to K(n)^*(X)$ is an isomorphism. We will allow th …

You might like to look at the paper "Algebraic theories, span diagrams and commutative monoids in homotopy theory" (https://arxiv.org/abs/1109.1598) by James Cranch. I think that it does not directly …