Equivalent condition for continuity of a bundle map

Question

Let $U$ be a topological space, and let $U \times \mathbb{R}^n$ be the trivial bundle. Let $\varphi: U \times \mathbb{R}^n \to U \times \mathbb{R}^n$ be a bundle map, but assume we don't know whether it's continuous. Let $\Phi: U \to \operatorname{Hom}_{\mathbb{R}}(\mathbb{R}^n, \mathbb{R}^n)$ be the set map corresponding to $\varphi$. It is a basic lemma that $$\text{$\varphi$ is continuous iff $\Phi$ is.}$$ I was trying to show this rigorously, and my argument was a little long-winded. I wonder if I missed a tighter, or even more categorical, way of doing it.

Here's my way:

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Consider the following maps

1. $\varphi$
2. $\varphi_{jk}:U \times \mathbb{R} \to U \times \mathbb{R}$, given by $U \times \mathbb{R} \xrightarrow{1 \times i_j}{} U \times \mathbb{R}^n \xrightarrow{\varphi}{} U \times \mathbb{R}^n \xrightarrow{1 \times \pi_k}{} U \times \mathbb{R}$
3. $\Phi_{jk}:U \to \operatorname{Hom}_{\mathbb{R}}(\mathbb{R}, \mathbb{R})$, given by $\pi_{jk} \circ \Phi$, where $\pi_{jk}$ is projection onto the basis vector $e_i^* \otimes e_j \in \mathbb{R}^{n,*} \otimes \mathbb{R}^n \cong \operatorname{Hom}_{\mathbb{R}}(\mathbb{R}^n, \mathbb{R}^n)$.
4. $\Phi$.

My way is to show that $$\text{(1) cnts $\implies$ (2) cnts $\implies$ (3) cnts $\iff$ (4) cnts $\implies$ (1) cnts}.$$

(1) $\implies$ (2) is clear, and (3) $\iff$ (4) is clear since it's just the definition of the topology of a real vector space. It remains to show (2) $\implies$ (3) and (4) $\implies$ (1).

(2) $\implies$ (3) follows because when we identify $\operatorname{Hom}_\mathbb{R}(\mathbb{R}, \mathbb{R})$ with $\mathbb{R}$, $\Phi_{jk}$ is just the result of taking $x \mapsto \varphi_{ij}(x,1)$, and then projecting onto the second coordinate.

(4) $\implies$ (1) follows because $\varphi$ is the result of taking $$(x,v) \overset{(1 , \Phi) \times 1}{\mapsto} (x, \Phi(x), v) \overset{1 \times \mathrm{eval}}{\mapsto} (x, \Phi(x)\cdot v),$$ where $(-, -)$ is the notation for taking a map into $A$ and a map into $B$ and turning them into a map into $A \times B$. Also note that $\mathrm{eval}:\operatorname{Hom}_{\mathbb{R}}(\mathbb{R}^n, \mathbb{R}^n) \times \mathbb{R}^n \to \mathbb{R}^n$ is continuous because it is bilinear, and the spaces are finite-dimensional.

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So that's it! Am I missing something more obvious, or slicker?

Answer 1

This isn't quite a full answer to your question, since I have to make some (very weak!) restrictions on your space $U$, but I do want to talk about some slicker machinery that you could think about.

Remember that a bundle map is the identity on the $U$-component, so all the data of a bundle map is contained in the projection to the $\mathbb{R}^n$-component. So by basic topology $\varphi: U \times \mathbb R^n \to U \times \mathbb R^n$ is continuous iff $\pi_2 \circ \varphi: U \times \mathbb R^n \to \mathbb R^n$ is continuous.

Now for the magic: let's assume $U$ is compactly generated Hausdorff (cgH for short) (edit: see Kevin Carlson's comment for elaboration on this point). As $\mathbb R^n$ is also cgH, and as the category of cgH spaces and continuous maps between them is a Cartesian closed category, we have a bijection (which is the obvious bijection!) $$\mathrm{Maps}(U \times \mathbb{R}^n, \mathbb R^n) \cong \mathrm{Maps}(U, \mathrm{Maps}(\mathbb R^n, \mathbb R^n)),$$ where the $\mathrm{Hom}$-spaces of continuous maps (which I denote $\mathrm{Maps}$) have the compact-open topologies. If you've done any functional programming, by the way, this is just the currying isomorphism.

It seems to me that this is a very hands-off and categorical way to do it. I'm not quite sure that there's nothing I'm hiding here, but it looks like you only need the following things to conclude the lemma you want:

1. The category of cgH spaces and continuous maps is Cartesian closed; i.e., it has a terminal object (a point), products (just topology), and exponential objects (just some more-involved topology).

2. In any Cartesian closed category we have the currying bijection (just abstract nonsense).

This does seem like a bit of smoke-and-mirrors to me, but I really can't figure out where a gap would be.