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?



classifying space of linear embeddings
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.


