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Let $\mathbb{K}$ be the compact operators on a separable infinite dimensional Hilbert space. Denote by $\mathcal{P}(\mathbb{K})$ the space of projections in $\mathbb{K}$. If I am not terribly wrong here, this space has the homotopy type of $\coprod_{n \in \mathbb{N}} BU(n)$.

But $\mathcal{P}(\mathbb{K})$ is also the space of objects in a topological groupoid $\mathcal{G}$ with morphism space given by $\{(p,q,u) \in \mathcal{P}(\mathbb{K}) \times \mathcal{P}(\mathbb{K}) \times U(H)\ |\ upu^* = q \}$, i.e. I add as morphisms all the unitaries connecting the different projections via conjugation. I can now take the classifying space of this groupoid. If I am not mistaken, $\mathcal{G}$ is Morita equivalent to the groupoid that has just one projection per dimension and as morphisms the unitaries commuting with this projection. This way, we get that $B\mathcal{G}$ should also have the homotopy type of $\coprod_{n \in \mathbb{N}} BU(n)$.

There is another (stupid) groupoid $\mathcal{G}'$, which has $\mathcal{P}(\mathbb{K})$ as objects and only identities as morphisms. Its classifying space just repoduces $\mathcal{P}(\mathbb{K})$. My question about this construction is:

Is the map $\mathcal{P}(\mathbb{K}) = B\mathcal{G}' \to B\mathcal{G} \simeq \mathcal{P}(\mathbb{K})$ induced by the inclusion functor a (weak?) homotopy equivalence?

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Hmm, I think I can solve this using the source map as a fibration with fiber $U(H)$. Does this work? – Ulrich Pennig Feb 24 '12 at 13:24
This might be a stupid question, but: How good is Kuiper's theorem actually? Can I contract U(H) through group homomorphisms? – Ulrich Pennig Feb 27 '12 at 12:36

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