# Is the associated bundle construction a bifunctor?

Let $\mathsf{Prin}_G$ be the category of (right) $G$-principal bundles, with a morphism from the bundle $p: P \to M$ to the bundle $p': P' \to M'$ being a pair of arrows $\chi: P \to P'$ and $\bar{\chi}: M \to M'$ such that $\chi$ is $G$-equivariant and the obvious diagram commutes. We only need to specify the morphism $\chi: P \to P'$ since the corresponding morphism down on the base is uniquely determined by $\chi$.

Let $\mathsf{Space}_G$ be the category of (left) $G$-spaces, with a morphism from $S$ to $S'$ being a $G$-equivariant arrow $f: S \to S'$.

Let $\mathsf{Bund}$ be the category of fibre bundles, with morphisms from $E \to M$, $E' \to M'$ being a pair of arrows such that we get a commutative square.

I'm being a bit ambiguous about whether we're working with topological spaces or smooth manifolds here, because I'm fairly certain it doesn't matter in this situation. Just assume that the arrows are continuous or smooth as appropriate.

Then the assignment $$\mathsf{Prin}_G \times \mathsf{Space}_G \to \mathsf{Bund}$$ sending a pair of morphisms $\chi: P \to P'$ and $f: S \to S'$ to the morphism
$$\chi \times_G f: P \times_G S \to P' \times_G S'$$ given by $$(\chi \times_G f) [u, s] = [\chi(u), f(s)]$$ is a bifunctor.

I checked the appropriate commutative diagrams and it seems to be true, but I just wanted to make sure since I haven't seen this explicitly stated before. I have seen it mentioned that if we vary the $G$-space then the associated bundle construction is functorial (on the category of $G$-spaces), but I don't think I've seen it mentioned that we can vary both the $G$-space and the principal bundle (including the base) to get a bifunctor.

## 1 Answer

Looks like Natural Operations in Differential Geometry by Kolar, Michor, and Slovak confirms that this is indeed the case on page 92 in section 10.9. I should have probably checked there before posting my question, but maybe it will help someone wondering about the same thing in the future.