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A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

A vector bundle morphism is an isomorphism if it admits an inverse of the same kind. A smooth fiberwise bijective bundle morphism automatically has a smooth inverse, because inverse matrices in vector bundle trivializations depend smoothly on their entries wherever their determinants are nonzero.