I'm familiar with the *definition* of the category of vector bundles, but I'm trying to *derive* it from some first principles about general fiber bundles. My intuition is that vector bundles should be understood as Cartan geometries (without a connection) of type $(\mathrm{Aff}(V),\mathrm{GL}(V))$. A $\mathrm{GL}$-structure on a fiber bundle obviously preserves the vector space structure of the model fiber, but I would think it should also force bundle morphisms to be fiberwise linear. Thus, starting with the definition of a bundle morphism I would like to understand how to properly constrain that definition in the context of a specific Cartan geometry $(G,H)$ in such a way that morphisms are represented, for example, by linear maps on fibers. I would like to avoid mentioning principal bundles in this process.

- The "default" structure group of a fiber bundle $E\overset{\pi}{\to}M$ with model fiber $F$ is $\mathrm{Diff}(F)$. In an atlas $\{(U_\alpha,\chi_\alpha)\}$ consisting of maps $\chi_\alpha:\pi^{-1}(U_\alpha)\to U_\alpha\times F$ we have transition functions $t_{\beta\alpha}:U_{\alpha\beta}\to \mathrm{Diff}(F)$ so that $\chi_\beta\circ\chi_\alpha^{-1}(p,f)=(p,t_{\beta\alpha}(p)\cdot f)$.
- A bundle morphism between two bundles $(E'\overset{\pi'}{\to}M',F)$ and $(E'\overset{\pi'}{\to}M',F')$ consists of two maps $\tilde{h}:E\to E'$ and $h:M\to M'$. Picking atlases $\{(U_\alpha,\chi_\alpha)\}$ and $\{(U_a',\chi_a')\}$ for these two bundles, and denoting $U_{\alpha a}=U_\alpha\cap h^{-1}(U_a')$, we can represent the morphism by local maps $\hat{h}_{\alpha a}:U_{\alpha a}\times F\to F'$ in the following way: $$\tilde{h}(\chi_\alpha^{-1}(p,f))=\chi_a^{\prime -1}(h(p),\hat{h}_{\alpha a}(p,f)).$$
- By combining these two definitions and writing out the definition of $\hat{h}_{\beta b}$ we get the transformation rule $$\boxed{\hat{h}_{\beta b}(p,t_{\beta\alpha}(p)f)=t_{ba}'(h(p))\hat{h}_{\alpha a}(p,f).}$$

**Question:** Now if I assume a choice of a $G$-structure on the first bundle and a $G'$-structure on the second, is there a way to simplify this transformation rule so as to turn it into a functor? More precisely, suppose we have two objects $(E\overset{\pi}{\to}M,G\curvearrowright F,\mathcal{G})$ and $(E'\overset{\pi'}{\to}M',G'\curvearrowright F',\mathcal{G}')$, where $\mathcal{G}$ is a $G$-structure on $E$ and $\mathcal{G}'$ is a $G'$-structure on $E'$. Furthermore, choose a map $h:M\to M'$, a map $\psi:F\to F'$, and a homomorphism $\lambda:G\to G'$. Is there a way of defining a map $\tilde{h}:E\to E'$ so that it is a bundle morphism which, in some sense, respects the given structures?

**Attempt 1:** Given the data that I provided, it would seem that we should be able to express any "reasonable" bundle morphism as
$$\hat{h}_{\alpha a}(p,f)=\lambda(\bar{g}_a(p))\psi(g_\alpha(p)f)$$
using some maps $g_\alpha,\bar{g}_a:U_{\alpha a}\to G$. The transformation law then becomes
$$\psi(f)=\lambda(\bar{g}_b)t_{ba}'(h(p))\lambda(\bar{g}_a)^{-1}\psi\left((g_\beta t_{\beta\alpha}g_\alpha^{-1})^{-1}f\right),$$
where all of the omitted arguments are $p$. Now, this turns out to quickly fail: consider for example isomorphisms of bundles, in which case $h=\mathrm{id}_M$, $\psi=\mathrm{id}_F$ and $\lambda=\mathrm{id}_G$, and we can also take $\alpha=a,\beta=b$ and $\bar{g}_a=e$. Then we get
$$t_{\beta\alpha}'=g_\beta t_{\beta\alpha}g_\alpha^{-1}.$$
This is indeed what bundle isomorphisms generally look like, but it doesn't let us make any inferences about what $g_\alpha$'s must be.

**Attempt 2:** Now let's take some inspiration from Cartan geometry, where the model fiber is actually a homogeneous space $F=G/H=\{Hg\}_{g\in G}$ and the structure group is $H$. This means that there is a transitive left action of $G$ with stabilizers isomorphic to $H$: $G\curvearrowright F$. Then $t_{\beta\alpha}$'s can be assumed to take values in $H$, whereas $g_\alpha$'s can still take values in all of $G$. In this case the last formula obtained for bundle isomorphisms gets replaced with
$$t'_{\beta\alpha}g_\alpha t_{\beta\alpha}^{-1}g_\beta^{-1}\in\mathrm{Stab}_G(f),\text{ for all }f\in F$$
but the intersection of all stabilizers is still trivial, so we get $t'_{\beta\alpha}g_\alpha t_{\beta\alpha}^{-1}g_\beta^{-1}=e$. This should *somehow* imply that $g_\alpha$'s actually commute with some subgroup of $G$ (the group of translations in the vector bundle case), but I don't see how that would happen.

**Analysis:** my approach so far gives me absolutely no information about what the values of $g$ should actually be, namely elements of a $\mathrm{GL}_k$ subgroup of $G=\mathrm{Aff}_k$. My goal is to come up with a definition of morphisms between bundles with $\mathrm{Aff}$-structures such that it *implies* that these morphisms must be equivariant with respect to the normal subgroup $\mathbb{R}^k$ of translations, i.e. actual fiberwise linear maps. Perhaps I need to be smart about requiring equivariance of $\psi$, but it's not clear how to do that considering that we need to be able to have different Lie groups acting on the two bundles (e.g. imagine embedding a vector bundle into another vector bundle of another rank, in which case $(G,H)\neq(G',H')$).

**Attempt 3 (in response to Ben)**: every one of these maps $\hat{h}_{\alpha a}$ can be identified with a $G\times G'$-equivariant map $\Phi_{\alpha a}:U_{\alpha a}\times(G\times G')\to C^{\infty}(F,F')$ as follows:
$$\Phi_{\alpha a}:\quad (p,g,g')\mapsto \left(f\mapsto g' \hat{h}_{\alpha a}\left(h(p),g^{-1}f\right)\right).$$
The action of $G\times G'$ on $C^{\infty}(F,F')$ is by composition $(g,g',\varphi)\mapsto g'\circ\varphi\circ g^{-1}$. The transformation rule becomes very simple:
$$\boxed{\Phi_{\beta b}(p,g,g')=\Phi_{\alpha a}\left(p,t_{\beta\alpha}(p)g,t_{ba}'(h(p))g'\right).}$$
This clearly defines a global equivariant function on a principal $(G\times G')$-bundle but I want to omit that. I would like to set some restriction on these maps so that I could say that $h$ respects the $G$- and $G'$-structures in some sense. I can do it by analogy with the interpretation of a $G$-structure as a way of selecting an orbit inside the set of all isomorphisms: **require all $\Phi_{\alpha a}$'s to take values in a single orbit of the action of $G\times G'$ on $C^\infty(F,F')$**. This is a well-defined restriction because transition functions take values in $G\times G'$. It means that, up to a choice of local charts (compatible with the structures), the morphism "looks the same" in every fiber. Now, if $F$ and $F'$ have additional structure which selects in a natural way a subclass of maps $\mathrm{Hom}(F,F')\subset C^\infty(F,F')$ (and usually such that $G$ and $G'$ are the groups of automorphisms preserving those structures), then we also require those orbits to be orbits of maps from $\mathrm{Hom}(F,F')$.

Let's see what effect this has in the context of vector bundles. Here $G$ and $G'$ are some $\mathrm{GL}$-groups, say $G=\mathrm{GL}(F)$ and $G'=\mathrm{GL}(F')$. Then I'd like to say that a collection of maps $\varphi_p:F\to F'$ belonging to an orbit of a linear map under the action of $\mathrm{GL}(F)\times \mathrm{GL}(F')$ on $C^\infty(F,F')$ is equivalent to these maps being linear. This then leads to the standard definition of vector bundles. So the solution is that the additional structure of a vector space on the model fiber canonically selects of class of fiberwise maps, and these generate bundle morphisms.