pull-back connection I have a question related to the definition of the pull-back connection, more specifically about its  uniqueness or the canonical way to induce it. 
The definition that one finds in general goes along the following lines: let $F:P\rightarrow B$ be map between differentiable manifolds, let $\pi:E\rightarrow B$ be a vector bundle and $\nabla$ a connection on $E$. Then the connection $F^*\nabla$ is uniquely determined by the following relation
$(F^*\nabla)_X(F^* s):=F^*(\nabla_{df(X)} s)$.
This should uniquely determine the connection right?
Let us start with the most trivial case, when $B$ is a point. Then $E$ is a vector space and the pull-back of $E$ is a trivial vector space over $P$. A connection on $E\rightarrow pt$ is an endomorphism of $E$ (which trivially satisfies the Leibniz relation). Could anybody explain how does that canonically induce a connection on $P\times E$, presumably the trivial connection $d$ if one starts with the zero endomorphism of $E$? Leibniz relation does not suffice. One could say let us make a convention here. 
But the more general question is how does one define $(F^*\nabla)_X$ when $X\in\ker{dF}$, in general?
 A: Just to put what others have said into an explicit formula, note that any section of $F^*E$ is of the form
$\sum_j \varphi_j F^*s_j$,
for certain $\varphi_1, \ldots, \varphi_n \in C^{\infty}(P)$, and $s_1, \ldots, s_n \in \Gamma^{\infty}(E)$. Then in the notation of the question (with $X \in T_pP$ a single tangent vector), at $p \in P$ we have by the Leibniz rule and the defining relation of $F^*\nabla$,
$(F^*\nabla)_X \left(\sum_j \varphi_j F^*s_j\right) = \sum_j \left( d\varphi_j(X)   F^*s_j  + \varphi_j (F^*\nabla)_X F^*s_j \right) 
= \sum_j \left( d\varphi_j(X) F^*s_j + \varphi_j F^*(\nabla_{dF(X)}s_j) \right) 
$
(with everything evaluated at $p$ where appropriate). This gives back $F^*\nabla = d$ in your example where $B$ is a point and you take the zero endomorphism of $E$.  
A: Hi, I have seen the equation you gave as a definition many times. For example, I think it is also used in a corresponding Wikipedia article. Nevertheless, as you correctly pointed out, it does not give a reasonable/unique description. A better formulation/definition of the pullback connection can be found for example in Milnor's and Stasheff's book 'Characteristic classes' on p. 292, Lemma 3 and its proof (definition by universal property/commutative diagram; proof: computation in local coordinates; it's the precise version of what the equation you gave tries to capture). I hope this helps more or less.
A: See 19.12.6 (page 250) in this book.
A: To Matt:
The relation $(F^*\nabla)_X=0$ does not satisfy Leibniz relation. 
Meanwhile I found out the answer to my question (a friend clarified it for me).
It turns out that there is an isomorphism $\Gamma(P;\pi^*E)\simeq \Gamma(P;\mathbb{C})\otimes \Gamma(B;E)$ where the tensor product is over $\Gamma(B;\mathbb{C})$.  It is obviously true when $E$ is a vector space. Now use Leibniz relation
$\nabla_X(f\otimes s)=X(f)\otimes s+f\otimes\nabla_Xs$
to extend the connection from $\Gamma(B;E)$ to $\Gamma (P;\pi^*E)$.
In the trivial case one gets indeed that $\nabla_X(f\otimes s)=X(f)\otimes s$.
A: Here's my summary of the situation:
1) First, observe that the space of local sections of the pullback bundle is generated by the space of sections of the original bundle composed with the map $F$. (This is better stated using sheaf language)
2) So, using the Leibniz rule, it suffices to define the pullback connection on a section obtained by composing a section of the original bundle with $F$.
3) The formula given above accomplishes this. It is worth noting that in this formula you should view $X$ as a single vector and not as a vector field.
A: OP's example is not correct for a point already.
First thing, connection is a differential operator $\nabla: \Gamma(E) \otimes TX \rightarrow \Gamma (E)$ of first order with a symbol $\partial$.
This amounts to saying that it satisfies Leibniz rule in the following sense:
$\nabla_{v} (f \gamma) = f\nabla_v(\gamma) + \partial_v (f) \gamma$
For a point, there is no need to even check Leibniz rule, because $TX = 0$, so there is no place from where you could obtain an automorphism! Connection allows you to derive the sections along a vector field, and every vector field on a point is $0$.
Moreover, pullback of the connection (which is defined correctly by the formula in the post) is a trivial connection on a trivial bundle (and it is canonical, because all fibers are canonically identified).
