Pullback connection

A total space $E$ ,
A base space $B$ ,
A projection map $π : E \to B$ ,
A typical fiber $F$ (all fibers are homeomorphic to $F$ ),
and a connection $Φ \in Ω^{1} (E, V E)$ .

Consider, also, the map $f : B^{'} \to B$ and the corresponding pullback bundle $f^{*} E$ :

\begin{matrix} (1) & \begin{matrix} f^{*} E & \overset{π^{*} f}{\to} & E \\ π^{'} ↓ & ↓ π \\ B^{'} & \overset{f}{\to} & B \end{matrix} \end{matrix}

Definition. The pullback connection $f^{*} Φ$ is the unique 1-form on $f^{*} E$ such that the diagram commutes:

\begin{matrix} T (f^{*} E) & \overset{T (π^{*} f)}{\to} & T E \\ f^{*} Φ ↓ & ↓ Φ \\ V (f^{*} E) & \overset{V (π^{*} f)}{\to} & V E \end{matrix}

That is, $f^{*} Φ$ is defined so that:

V (π^{*} f) \circ f^{*} Φ = Φ \circ T (π^{*} f)

Here $V (π^{*} f)$ is the differential map restricted to the vertical bundle of the pullback bundle. It is applied onto $V E$ for the following reason. When you differentiate diagram $(1)$ , you get:

T π \circ T (π^{*} f) = T f \circ T π^{'} .

Now, if you take any vertical vector $X \in T (f^{*} E)$ such that $T π^{'} (X) = 0$ , then applying the identity above gives:

T π (T (π^{*} f) (X)) = T f (T π^{'} (X)) = T f (0) = 0.

This proves $T (π^{*} f) (X) \in \ker T π = V E$ , meaning vertical vectors get sent to vertical vectors.