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It is part of the wedge axiom: $X\vee Y$ is the coproduct in pointed spaces, and therefore comes with distinguished maps ("inclusions") from $X$ and $Y$, giving a coproduct diagram $X\rightarrow X\vee …

Let me start by saying that these ideas are not due to me. I overheard them in a seminar I attended recently (see Footnote). There are many situations in which one is working with a compact Lie group …

Let $E$ be a Landweber exact ring spectrum. That is, we have a map of homotopy ring spectra $MU\rightarrow E$ and an isomorphism of homology theories $E_*X\simeq MU_*X\otimes_{MU_*}E_*$. Is the homoto …

Yes it does! The map in your display is (more or less) the cup product with the Thom class $\lambda_N$. So if you choose a different Thom class you'll get a different map. For example, take the standa …

The answer is no. I guess I owe Fernando Muro 10 dollars. Let $\mathbb{S}\rightarrow\Sigma^{-2}\mathbb{CP}^\infty$ be the inclusion of the bottom cell, and let $f:F\rightarrow\mathbb{S}$ be the fiber. …

Recall that the Bousfield class of a spectrum $E$, written $\langle E\rangle$, is the class of spectra $X$ such that $X\wedge E$ is not contractible. For example the Bousfield class of any of the sphe …

A ring spectrum $E$ is complex oriented if it is equipped with a ring map $MU\rightarrow E$. It is complex orientable if such a ring map exists. An $E_\infty$-ring $E$ is $E_\infty$-complex oriented i …

I would like an example of the following situation, or a proof that no such example exists. $\textbf{Situation}$: Two connective (EDIT: I'm fine with dropping this condition) spectra $X$ and $Y$ such …

Here's a connective example. It is also an example of Maxime's variant question in the comments (regarding $\tau_{\leq m}$ truncations). And thanks to Maxime for looking this argument over before I po …

Suppose $E$ is a complex-oriented ring spectrum whose formal group law is isomorphic to the additive one. As the title suggests, we might as well change the complex orientation so that the formal grou …

In the case that $E$ is trivial, there is a "universal" example of the construction you describe, which is the vector bundle on $S^1\times U(n)$ formed the "canonical" automorphism (each point acts o …

This is an answer to the question in the title, which is what I had meant to ask: is an $E$ as in the question body an $H\mathbb{Z}$-module? (the last sentence of the question body is stronger and lik …

To every complex-oriented ring spectrum $E$ there is associated a formal group law, which is a power series $F_E(x,y)\in E_*[[x,y]]$. Suppose $E$ and $F$ are two complex-oriented ring spectra and supp …

Under the identification of the stable homotopy groups with the (stably) framed bordism groups, it is well known that $\eta\in\pi_1\mathbb{S}$ is represented by $S^1$ with its Lie group framing. Produ …

Let's apply your definition (which I think has typos on the RHS - the two "+" subscripts on the EG should not be there I think). Let's model everything as topological spaces and do the calculation the …