Bundle map

A morphism of fibered manifolds, i.e. a smooth map of total spaces compatible with the projections.

Bundle map

Let $\pi:E\to M$ and $\pi':E'\to M'$ be fibered manifolds . A bundle map (morphism of fibered manifolds) is a pair of smooth maps $(F,f)$ with $F:E\to E'$ and $f:M\to M'$ such that

\pi'\circ F \;=\; f\circ \pi.

Equivalently, $F$ is a fiber-preserving map for which both projections are surjective submersions; when $f$ is understood one says “ $F$ is a bundle map over $f$ ”.

Differentiating the commutative square gives, for each $e\in E$ ,

d\pi'_{F(e)}\circ dF_e \;=\; df_{\pi(e)}\circ d\pi_e.

In particular, $dF_e$ maps $\ker(d\pi_e)$ into $\ker(d\pi'_{F(e)})$ , so it sends the vertical tangent space at $e$ into the vertical tangent space at $F(e)$ . On the level of the tangent bundles , this says $dF:TE\to TE'$ is compatible with the vertical subbundles.

Examples

Differential of a smooth map: for a smooth map $\varphi:M\to N$ , the differential $d\varphi:TM\to TN$ is a bundle map over $\varphi$ .
Product-type maps: if $\pi=\mathrm{pr}_1:M\times F\to M$ and $\pi'=\mathrm{pr}_1:M'\times F'\to M'$ , then any map $(x,u)\mapsto(f(x),h(x,u))$ defines a bundle map over $f$ .
Composition: if $F:E\to E'$ is over $f$ and $G:E'\to E''$ is over $g$ , then $G\circ F$ is a bundle map over $g\circ f$ .