Pullback connection is well defined and functorial

Pull back the identity in [F2]. Composition preserves matrix products and inverses, and the chain rule gives $d(A\circ f)=f^{\ast }(dA)$. Thus $f^{\ast }{\omega }^{\mathrm{\prime }}=(A\circ f{)}^{-1}(f^{\ast }\omega )(A\circ f)+(A\circ f{)}^{-1}d(A\circ f)$. These are exactly the transition matrices of the pulled-back frames, so the prescription glues uniquely and is frame independent.

For a one-form entry $\eta$ and $v\in T_{x}P$, $(g^{\ast }{f}^{\ast }\eta {)}_x(v)={\eta }_{f(g(x))}(d{f}_{g(x)}d{g}_{x}v)=((f\circ g{)}^{\ast }\eta {)}_x(v)$ by the chain rule. Hence the connection matrices coincide in the frames identified by the canonical bundle isomorphism. Gluing uniqueness gives composite functoriality; the identity case is the same evaluation with $d\mathrm{id}=\mathrm{id}$.

If $s=eu$, then $f^{\ast }s=(f^{\ast }e)(u\circ f)$. Its coefficient derivative is $X(u\circ f)=du(dfX)$, giving the displayed section identity. Constant $f$ has $df=0$, so pulled-back sections from $E$ are parallel, whereas general coefficients on $N$ still differentiate as prescribed. Empty bases and rank-zero bundles give unique zero operators; no rank condition on $df$ was used and rank one follows entrywise. All local values are specified uniquely without AC.