Smooth map between manifolds

A map between smooth manifolds is smooth when its expression in every pair of coordinate charts is a smooth function.

A pushforward of a curve composes a parametrized curve in the source with a smooth map between manifolds. Its tangent is the image of the original tangent under the differential of a smooth map. A constant image curve has zero tangent, so a general smooth map need not preserve regularity of curves.

A pullback of a smooth function composes a function on the target with a smooth map between manifolds, producing a function on the source. It needs no inverse map. Acting on the pulled-back function defines the differential of a smooth map intrinsically as a map of derivations.

For a smooth map between manifolds $f:M\to N$, the differential $d{f}_x:T_{x}M\to T_{f(x)}N$ is the linear map obtained by differentiating in charts. Intrinsically it sends the tangent vector of a curve $\gamma$ through $x$ to the tangent vector of $f\circ \gamma$. The chain rule makes it independent of charts and gives $d(g\circ f{)}_x=d{g}_{f(x)}\circ d{f}_x$.

A smooth map between smooth manifolds with zero differential is locally constant: each target coordinate function has zero derivatives on a sufficiently small source coordinate ball. Integration along straight segments proves constancy there. Its fibers are open and closed, so a connected source makes the map constant globally.

The pushforward of a contravariant tensor applies the differential of a smooth map to each vector factor of a type $(r,0)$ tensor. It is defined pointwise at a specified source point. A field generally gives a field along the map, not necessarily a unique tensor field on the image when several points have the same image. A diffeomorphism allows the usual transport of arbitrary mixed tensors through its inverse as well.

The pullback of a covariant tensor composes each of its vector arguments with the differential of a smooth map. It takes a type $(0,s)$ tensor on the target to one on the source. Antisymmetry is not required; the pullback of a differential form is the alternating special case. If the target vectors are projected onto the image tangent space first, evaluating the pullback is unchanged.

The pullback of a covector is the dual linear map to the differential of a smooth map. In coordinates its components are $({\varphi }^{\ast }\eta {)}_i=({\partial }_i{y}^{\alpha }){\eta }_{\alpha }$. The same rectangular Jacobian defines vector pushforward, with its other index contracted. No invertibility is needed.

For a vector field $X$ on $M$, its pushforward by a smooth map between manifolds $f:M\to N$ is the vector field along a map defined by $(f_{\ast }X{)}_p=d{f}_p(X_p)$. Acting on a function $h$ on $N$, it satisfies $(f_{\ast }X{)}_p(h)=X_p(h\circ f)$. For a diffeomorphism this defines a vector field on $N$ by evaluation at $p=f^{-1}(q)$. For a general map there need not be a single vector at each image point: $f(x)=x^2$, $X={\partial }_x$ gives $2x{\partial }_y$, with opposite values over $y>0$.

The chain rule shows that $\alpha {\varphi }_t{\alpha }^{-1}$ is the local flow of the pushforward of a vector field ${\alpha }_{\ast }X$. If this equals $X$, uniqueness of integral curves of a vector field gives commutation. Conversely, differentiate the commuting identity at $t=0$ to obtain $d{\alpha }_p{X}_p=X_{\alpha (p)}$. All identities hold on their common domains; the vector field need not be complete.

A vector field $X$ is projectable through a smooth map between manifolds $f:M\to N$ if there is a smooth vector field $Y$ on $N$ with $d{f}_p{X}_p=Y_{f(p)}$ for every $p$. Agreement on fibres is necessary. For a surjective submersion it is also sufficient: smooth local sections of the submersion express $Y$ locally as $df(X)$ and prove its smoothness. An arbitrary non-surjective map may instead give a field only along its image, with extension to $N$ requiring additional choices.