Question

Define an $S^{1}$-spectrum $E$ to be a sequence of pointed simplicial sets $E_{n},\ n=0,1,2...$ with assembly morphisms $\sigma_{n}:S^{1}\wedge E_{n}\rightarrow E_{n+1}$. An $S^{1}$-spectrum $E$ is now called $\textit{fibrant}$ if all the simplicial sets $E_{n}$ are Kan-fibrant and the adjoint $E_{n}\rightarrow\Omega (E_{n+1})$ of $\sigma_{n}$ is a simplicial weak equivalence. My questions is now if the coproduct $\vee {E_{i}}$ of fibrant spectra $E_{i}$ is fibrant again?

Thank you

Answer 1

No. Here are two reasons:

• The coproduct of two pointed Kan complexes is usually not a Kan complex (say, if the spaces are connected): we can map $\Lambda^2_1$ into "both summands" and the map will not extend to $\Delta^2$.
• The functor $\Omega$ will basically never commute with wedge sums (even up to homotopy). For example, there is a $\Omega$-spectrum $E = H\mathbb{F}_2$ with $E_0 = \mathbb{F}_2$ and $E_1 = K(\mathbb{F}_2, 1)$; so $\pi_0(E_0 \vee E_0)$ is a three-element set but $\pi_0(\Omega(E_1 \vee E_1))$ is the coproduct of $\mathbb{F}_2$ with itself in the category of groups, which is the infinite dihedral group $D_\infty$.