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Take the following small model for the category of finite-dimensional vector spaces and isomorphisms: The set of objects is $\mathbb{N}$ and the set of morphisms $Mor(n,m)$ is empty, if $n \neq m$ and $U(n)$ otherwise. If we take the classifying space of that, we get $\coprod_{n \in \mathbb{N}} BU(n)$.

What happens, if we take $Mor(n,m) = Emb(\mathbb{C}^n, \mathbb{C}^m)$, where the latter denotes the space of isometric linear maps instead, i.e. what is the homotopy type of the corresponding classifying space?

Note that $\mathbb{N}$ does not contain zero, so there is a priori no reason why this should be contractible. Moreover, there is a map from $\coprod_{n \in \mathbb{N}} BU(n)$ into the space described above, induced by the canonical functor. Nevertheless, it is connected, since there is always a canonical embedding $\mathbb{C}^n \to \mathbb{C}^m$ for $m \geq n$.

So, my guess would be that this is a model for $BU$, since it somehow looks like some kind of telescope construction. But I could not prove this, because there is no canonical extension of an embedding $\mathbb{C}^n \to \mathbb{C}^m$ to a unitary. So, am I right or is this just another elaborate description of the point?

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up vote 8 down vote accepted

Let $\mathcal{A}$ be your category. This is a skeleton of the category $\mathcal{B}$ of all nonzero finite-dimensional complex Hilbert spaces and isometric embeddings, so $B\mathcal{A}$ is homotopy equivalent to $B\mathcal{B}$. Now let $\mathcal{B}_0$ be the larger category where we allow the zero space. We have an inclusion $i:\mathcal{B}\to\mathcal{B}_0$ and also a functor $p:\mathcal{B}_0\to\mathcal{B}$ given by $p(V)=V\oplus\mathbb{C}$. There are evident natural maps $1\to pi$ and $1\to ip$, and a natural map between functors gives rise to a homotopy between the induced maps on classifying spaces, so we see that $B\mathcal{B}$ is homotopy equivalent to $B\mathcal{B}_0$. However, $\mathcal{B}_0$ has an initial object, so $B\mathcal{B}_0$ is contractible, so $B\mathcal{A}$ is contractible.

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Ah, I see. So, even though the category itself fails to have a zero object, I can "shift it back", since there is an embedding into $V \oplus \mathbb{C}$. Thank you! – Ulrich Pennig Feb 21 '12 at 22:38

I believe it is (weakly) contractible, for the following amusing reason. Let me write $\mathcal{C}$ for the category described in the question. Direct sum gives a functor $\mu : \mathcal{C} \times \mathcal{C} \to \mathcal{C}$ which is naturally isomorphic to $\mu \circ \tau$ for $\tau$ the twist. In particular, $B\mathcal{C}$ is a connected homotopy associative and homotopy commutative $H$-space.

Observation: the map $x \mapsto x \cdot x : B\mathcal{C} \to B\mathcal{C}$ is homotopic to the identity.

Proof: the collection of morphisms in $\mathcal{C}$ given by $n \to 2n$ using the diagonal $\mathbf{C}^n \to \mathbf{C}^{2n} = \mathbf{C}^n \oplus \mathbf{C}^n$ gives a natural transformation from $\mathrm{Id}_\mathcal{C}$ to $\mathrm{Id}_\mathcal{C} \oplus \mathrm{Id}_\mathcal{C}$, and so a homotopy on classifying spaces.

Then by e.g. Lemma 5.25 of

S. Galatius "Stable homology of automorphism groups of free groups", Ann. of Math. (2) 173 (2011), no. 2, 705–768.

this implies that $B \mathcal{C}$ is weakly contractible.

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This is nice! But I think, I'll accept Neils answer, since it is a little more direct. – Ulrich Pennig Feb 21 '12 at 22:46

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