Well definedness of the correspondence between connections on $B$ and sections of $J^1B$ Let $M$ be a smooth manifold and $\pi\colon B\to M$ a fiber bundle on $M$. A (Ehresmann) connection on $B$ is a subbundle $H$ of $TB$ such that $TB=H\oplus V$, where $V$ is the bundle of vertical vectors.
In different places I've seen claimed that there is a correspondence between connections on $B$ and sections of $J^1B$, in particular the book I'm looking at provides the following recipe to construct a section $\sigma\colon B\to J^1B$ of the jet bundle given a connection $H$:
Consider $b\in B$ and pick a local section $\sigma_b\colon U\subseteq M\to B$ such that $\sigma_b(x)=b$ and $\mathrm{Im}(T_x\sigma_b)=H_b$, then set $\sigma(b)=j^1_x\sigma_b$.
The book claims that this is obviously well defined, meaning that if I pick two different local sections $\sigma_b,\rho_b$ with $\sigma_b(x)=b=\rho_b(x)$ and $\mathrm{Im}(T_x\sigma_b)=H_b=\mathrm{Im}(T_x\rho_b)$, then $j^1_x\sigma_b=j^1_x\rho_b$, but I don't see why that's the case.
Obviously the two sections have tangent maps in $x$ with equal images, but to have the same jet I need all derivatives in local coordinates to agree in $x$, which is a stronger requirement, hence my question: why is this map well defined?
 A: Suppose you have two local sections $\sigma^1, \sigma^2 : U \subseteq M \to B$ satisfying $\sigma^i(x) = b$ and $D_x\sigma^i(T_xM) = H_b$. Since $\sigma^1$ and $\sigma^2$ are both sections of the bundle given by $\pi : B \to M$, they are injective, and so they induce isomorphisms $D_x\sigma^i : T_xM \to H_b$ (counting dimensions), which we can turn into an automorphism $(D_x\sigma^2)^{-1} \circ (D_x\sigma^1)$ of $T_xM$. The question now is, why is this automorphism the identity. In some sense, it's because $H_b$ already encodes how $\sigma^i(x)$ may vary as $x$ varies; since $x$ can only vary in $M$, $\sigma^i(x)$ cannot have a derivative along the direction of the fiber, so its derivative is guided by $H$. This is also why you should have $H$ as a vector bundle on $B$; the sections need a derivative regardless of their value!
You just need the following lemma: assume $m \le n$ are positive integers and $f,g : \mathbb R^m \to \mathbb R^n$ are injective linear maps for which $f(\mathbb R^m) = g(\mathbb R^m)$ and there exists $h : \mathbb R^n \to \mathbb R^m$ linear with $h \circ f = h \circ g$. Then $f=g$. The proof is trivial; we can assume $m=n$ without loss of generality just by replacing $\mathbb R^n$ by $f(\mathbb R^m) = g(\mathbb R^m) \simeq \mathbb R^m$, in which case it is trivial since $f$ and $g$ have the same inverse. Apply this lemma to the sections $\sigma^i : U \to B|_U$ satisfying $\pi|_U \circ \sigma^i = \mathrm{id}_U$, which implies $D_b\pi \circ D_x\sigma^i = \mathrm{id}_{T_xM}$, and you will deduce $D_x\sigma^1 = D_x\sigma^2$.
You can also see it in a less coordinate-patchy way as follows: you define the bundle of vertical vectors as the kernel of the morphism of vector bundles $V = \ker(TB \to \pi^* TM)$ on $B$. Vertical vectors go along the direction of $b \in B$ when $b$ varies along a fiber, since then $\pi(b)$ does not vary. An Ehresmann connection satisfies $TB = H \oplus V$ as vector bundles on $B$, so in particular $H \simeq TB/V \simeq \mathrm{im}(TB \to \pi^* TM) \simeq \pi^*TM$. So an Ehresmann connection is really just a global version of the locally available isomorphism $TB \simeq \pi^*TM \oplus V$ (which exists by trivializing all the bundles involved), and it is therefore not surprising that $\pi^*TM$ (and its global counterpart $H$) and $J^1B$, being two vector bundles on $B$ corresponding to derivatives on $M$, are closely related.
Hope that helps,
