Smooth Maps, Pushforwards, and Pullbacks

A smooth map $F:M\to N$ has a differential $dF_p:T_pM\to T_{F(p)}N$ that pushes individual tangent vectors forward. Covectors and all differential forms pull back along any smooth map by feeding their arguments through $dF$. Vector fields do not generally push forward to vector fields: points in one fiber of $F$ may prescribe incompatible vectors at the same target point.

Required background. Calculus and Linear Algebra for Relativity — use Jacobians, rank, kernels, images, the chain rule, and dual maps.

Smooth maps and their differentials

A map $F:M\to N$ is smooth when its coordinate representatives are smooth. For a tangent vector represented as a curve velocity, define

$$dF_p[\gamma]=[F\circ\gamma].$$

Equivalently, if $v$ is a derivation and $h\in C^\infty(N)$,

$$(dF_pv)(h)=v(h\circ F).$$

These definitions agree and give a linear map. In coordinates $x^i$ on $M$ and $y^a$ on $N$,

$$dF_p(v)^a = \left.\frac{\partial (y^a\circ F)}{\partial x^i}\right|_p v^i.$$

The chain rule becomes the invariant identity $d(G\circ F)_p=dG_{F(p)}\circ dF_p$. This construction and the rank-based classes of maps are developed in Gualtieri 2017, §§2.1–2.5, pp. 14–23; §§6.1–6.6, pp. 53–63, PDF.

An immersion has injective differential at every point; a submersion has surjective differential; a local diffeomorphism has invertible differential. A diffeomorphism is globally bijective with smooth inverse. The qualifiers matter: an immersion can self-intersect, and a bijective smooth map need not be a diffeomorphism unless its inverse is smooth.

Pushforwards and the fiber obstruction

The differential always pushes one vector $v_p$ to a vector at $F(p)$. To push an entire vector field $X$ from $M$ to $N$, however, one would need a field $Y$ satisfying

$$dF_p(X_p)=Y_{F(p)}$$

for every $p$. Such fields are called $F$-related. They need not exist. For the projection $\pi:\mathbb R^2\to\mathbb R$, $\pi(x,y)=x$, take $X=y\partial_x$. Then $d\pi_{(x,y)}X=y\partial_x$ at the target $x$. Different points $(x,y)$ in the same fiber demand different values for one putative target field. Thus there is no well-defined $\pi_*X$.

If $F$ is a diffeomorphism, each target point has one preimage and $F_*X$ is a well-defined smooth field. Fiber compatibility is also sufficient when $F$ is a surjective submersion: local smooth sections of $F$ construct the descended field, and compatibility makes their answers agree. For a general smooth map, however, compatibility alone does not guarantee a smooth field on the target.

For example, $F:\mathbb R\to\mathbb R$, $F(x)=x^3$, is bijective, so each fiber contains one point. Yet $X=\partial_x$ would have to descend to

$$Y=3|y|^{2/3}\partial_y,$$

which is not differentiable at $y=0$. The derivative $dF_xX_x=3x^2\partial_y$ exists everywhere, but it defines a field along the map, not a smooth vector field on the target. For nonsurjective maps one must also ask whether the prescribed values on the image admit a smooth extension. These are the reasons to use $F_*X$ cautiously for arbitrary maps.

Pullbacks of covariant objects

For a covector $\alpha\in T_{F(p)}^*N$, define

$$(F^*\alpha)_p(v)=\alpha_{F(p)}(dF_pv).$$

No inverse of $F$ is needed: the arguments move forward, so the covector can move backward. For a covariant $k$-tensor $T$,

$$(F^*T)_p(v_1,\ldots,v_k) =T_{F(p)}(dF_pv_1,\ldots,dF_pv_k).$$

Pullback respects tensor products, wedge products, and composition: $(G\circ F)^*=F^*G^*$. For functions it is simply $F^*h=h\circ F$, and $F^*(dh)=d(h\circ F)$.

As a concrete check, let $F(r,\theta)=(r\cos\theta,r\sin\theta)$ on $r>0$. Then

$$F^*(dx\wedge dy)=r\,dr\wedge d\theta.$$

The factor $r$ is the determinant of the Jacobian, while the order $dr\wedge d\theta$ records orientation. At $r=0$ the differential loses rank and polar coordinates cease to be a local diffeomorphism. The pullback formula itself remains meaningful for the smooth map on a suitable coordinate domain; what fails is invertibility.

Rank, local models, and scope

The inverse function theorem says that an invertible $dF_p$ makes $F$ a diffeomorphism between neighborhoods of $p$ and $F(p)$. The constant-rank theorem gives local coordinates in which a constant-rank map is a coordinate projection followed by a coordinate inclusion. These are local conclusions. Global injectivity, properness, topology of fibers, and embedding properties require separate hypotheses; Lee 2013, Chs. 2 and 4, pp. 32–49 and 77–97 states the distinctions precisely.

This page concerns smooth maps and tensors before a metric. A metric can produce adjoints, orthogonal projections, and gradients, but none belongs to the bare pullback construction. Later pages use pullback to restrict forms to submanifolds and define integration on parametrized domains.

A useful composition check

Let a curve first map into a surface and the surface then into an ambient manifold. Applying the differentials in sequence must give the velocity of the fully composed curve. Testing this identity in coordinates catches a frequent domain error: every Jacobian must be evaluated at the point reached by the preceding map. The dual statement reverses order, so pulling a covector back through the composite means first pulling through the outer map and then through the inner one.

This reversal is the source of the word “contravariant” for functorial behavior, but it should not be confused with contravariant tensor indices. The pullback acts naturally on covariant slots because those slots accept pushed-forward vectors. Pushing a vector field to the target requires both consistent values on fibers and smooth dependence on the target point.

The constant map gives a sharp limiting case. Its differential sends every tangent vector to zero, so every positive-degree form pulls back to zero, while a function pulls back to its constant value at the image point. Nothing is singular in this construction; it simply records that the map carries no directional information. This check distinguishes “pullback always exists” from the incorrect claim that pullback preserves all information.

Exercises

Let $F:\mathbb R^2\to\mathbb R^2$ be $F(r,\theta)=(r\cos\theta,r\sin\theta)$. Compute $dF$ and $F^*(x\,dy-y\,dx)$. Explain separately what fails at $r=0$ and what fails when $\theta$ is allowed to differ by $2\pi$.

Solution

The differential sends

$$\partial_r\mapsto \cos\theta\,\partial_x+\sin\theta\,\partial_y, \qquad \partial_\theta\mapsto-r\sin\theta\,\partial_x+r\cos\theta\,\partial_y.$$

Substituting $x=r\cos\theta$, $y=r\sin\theta$ and their differentials gives

$$F^*(x\,dy-y\,dx)=r^2d\theta.$$

At $r=0$, $dF(\partial_\theta)=0$, so the map is not locally invertible. If all real $\theta$ are admitted, distinct points differing by $2\pi$ have the same image, so global injectivity fails even for $r>0$. Local rank and global bijectivity are different questions.

References

• Gualtieri, Marco. Geometry and Topology I. University of Toronto course notes, 2017, §§2.1–2.5, pp. 14–23; §§6.1–6.6, pp. 53–63, PDF.
• Lee, John M. Introduction to Smooth Manifolds. 2nd ed., Graduate Texts in Mathematics 218, Springer, 2013, Chs. 2 and 4, pp. 32–49 and 77–97.