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John Rognes
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Looping the fiber sequence $S^1 \to S^3 \to S^2$ gives $\Omega^2 S^2 \simeq \mathbb{Z} \times \Omega^2 S^3$. This is the group completion $\mathbb{Z} \times BB_\infty^+$ of $\coprod_{n\ge0} BB_n$, so by Segal's theorem $\bigoplus_{n \in \mathbb{Z}} H_*(B_\infty)$ is isomorphic to $H_*(\Omega^2 S^2) \cong \bigoplus_{n \in \mathbb{Z}} H_*(\Omega^2 S^3)$.

Looping the fiber sequence $S^1 \to S^3 \to S^2$ gives $\Omega^2 S^2 \simeq \mathbb{Z} \times \Omega^2 S^3$. This is the group completion $\mathbb{Z} \times BB_\infty^+$ of $\coprod_{n\ge0} BB_n$, so by Segal's theorem $\bigoplus_{n \in \mathbb{Z}} H_*(B_\infty)$ is $H_*(\Omega^2 S^2) \cong \bigoplus_{n \in \mathbb{Z}} H_*(\Omega^2 S^3)$.

Looping the fiber sequence $S^1 \to S^3 \to S^2$ gives $\Omega^2 S^2 \simeq \mathbb{Z} \times \Omega^2 S^3$. This is the group completion $\mathbb{Z} \times BB_\infty^+$ of $\coprod_{n\ge0} BB_n$, so by Segal's theorem $\bigoplus_{n \in \mathbb{Z}} H_*(B_\infty)$ is isomorphic to $H_*(\Omega^2 S^2) \cong \bigoplus_{n \in \mathbb{Z}} H_*(\Omega^2 S^3)$.

Source Link
John Rognes
  • 9.3k
  • 45
  • 52

Looping the fiber sequence $S^1 \to S^3 \to S^2$ gives $\Omega^2 S^2 \simeq \mathbb{Z} \times \Omega^2 S^3$. This is the group completion $\mathbb{Z} \times BB_\infty^+$ of $\coprod_{n\ge0} BB_n$, so by Segal's theorem $\bigoplus_{n \in \mathbb{Z}} H_*(B_\infty)$ is $H_*(\Omega^2 S^2) \cong \bigoplus_{n \in \mathbb{Z}} H_*(\Omega^2 S^3)$.