Universal bundles: construction of the map associated to a group homomorphism

For a Lie group $G$ let $EG \to BG$ denote the universal bundle. A Lie group homomorphism $\rho: G \to H$ determines a map $B \rho: BG \to BH$ as the classifying map for the principal $H$-bundle $EG \times_\rho H \to BG$. While this argument yields existence of $B \rho$ (and uniqueness up to homotopy) it is far from being explicit. Does there exist a true construction of $B \rho$? If $\rho$ is a embedding of a subgroup, then on gets the classifying map by a quotient procedure. I was hoping for a similar result for arbitrary group homomorphism. Assume anything you want on $H$ (compactness ect).

Some background motivation: It is well-known that universal bundles characterize equivalence classes of principal bundles. A further, not so common characterization is given by 'homomorphisms $\rho$' between the loop space of $M$ and $G$. The principal bundle $P$ is then the to the path bundle associated bundle $PM \times_\rho G$. Hence there are two classification of principal bundles with connections and I want to better understand the interplay between them.

• Constructions of classifying spaces tend to be functorial (e.g., Milnor or Milgram bar construction), so take a homomorphism $\rho$ of topological groups to continuous maps $B\rho$ explicitly. – Todd Trimble Jul 1 '14 at 14:55
• Following on Todd's comment, note that the simplicial bar construction gives a model for $BG$ when $G$ is a Lie group (see Segal's paper "Classifying spaces and spectral sequences," available here: maths.ed.ac.uk/~aar/papers/segalclass.pdf). – Dan Ramras Jul 1 '14 at 17:42
• Alternatively, if you want to stick with the model for $BG$ which is the quotient of a contractible space ($EG$) by a free $G$ action, note that if $EG$ is one such, then so too is $EG \times EH$, where $G$ acts on $EH$ through the homomorphism $\rho$. Then $BG = (EG \times EH) / G$, and $BH = EH / H$, so $B\rho$ can be given by the projection onto the second factor $(EG \times EH) / G \to EH / \rho(G) \to EH/H$. – Craig Westerland Jul 2 '14 at 3:28

It is then obvious how to construct a $\rho$-equivariant map $EG\to EH$.