# Is the functor associating a bundle with a structure group to a principal bundle faithful?

Consider a (Cartan) principal G-bundle $\xi: X \to B$, and a left $G$-space $F$. We construct the bundle $\xi[F]: X_F \to B$ associated with $\xi$ with a fiber $F$ as usual. Now for each morphism $(u, f)$ between principal bundles $\xi: X \to A$ and $\eta: Y \to B$ we can associate a morphism $(u_F, f): \xi[F] \to \eta[F]$, where $u_F((x, y)G) = (u(x), y)G$. Thus we can construct a functor from the category of principal $G$-bundles to some (not full) subcategory of the category of bundles ($\mathbf{Top}^\to$), which we call the category of bundles with the structure group $G$ and the fiber $F$.

Is this functor faithful?

For any two morphisms $(u, f), (v, f): \xi \to \eta$ $u_F = v_F$ is equivalent to the identity $\{(u(x)s, s^{-1}y) \mid s \in G\} = \{(v(x)s, s^{-1}y \mid s \in G\}$ for any $x \in X$ and for any $y \in F$. In particular, this implies that $u(x) G_y = v(x) G_y$, where $G_y$ is the stabilizer of $y \in F$, so if $G_y = 1$ for some $y \in F$, then the functor is faithful.

Is this correct? I am not comfortable with this argument because of the massive amount of quotients and factorizations involved in construction of associated bundles, can someone please check if I'm missing something?

Also, while the action of $GL_n(\mathbb{R})$ on $\mathbb{R}^n$ is faithful, every vector has a non-trivial stabilizer. Is the functor in question faithful in case of vector bundles?

If I need to clarify something or ask a more concrete question, please inform me in the comments.

• What's a free functor? If you mean it's the left adjoint of some forgetful functor, say that instead. Oct 17, 2011 at 6:30
• @ZhenLin: I meant faithful, sorry :) Oct 17, 2011 at 12:03
• What if $F$ is a point? Oct 17, 2011 at 12:24
• @AkhilMathew: then for any trivial principal bundle $\pi_B: G \times B \to B$ any non-trivial automorphism of $G$ induces the identity automorphism on $\pi_B[*]$. But then the stabilizer of any $y \in *$ is the whole $G$ so the functor in question is only faithful when $G = 1$. Oct 17, 2011 at 14:41