Submersions, immersions, and embedded submanifolds
Use derivative rank to identify local models and explain why a parametrized immersed image need not be an embedded manifold.
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The question for today
What you’ll understand
- Distinguish injective and surjective tangent maps.
- Derive a regular fiber's dimension and tangent space.
- Identify the additional topological requirement in an embedding.
Rank tells us how many independent infinitesimal directions survive a smooth map. An immersion preserves all source tangent directions, while a submersion reaches all target tangent directions. Neither definition alone describes the global topology of the image. An embedded submanifold must also carry the topology inherited from its ambient space. These distinctions make it possible to reason correctly about constraint sets, repeated parametrizations, and smooth images that cross themselves without mistaking a local rank computation for a global embedding theorem.
Injective and surjective differentials have different normal forms
For a smooth map f from an m-manifold to an n-manifold, an immersion has injective derivative at every source point, requiring m≤n. A submersion has surjective derivative everywhere, requiring m≥n. In suitable coordinates an immersion is locally the inclusion of the first m coordinates, and a submersion is locally projection onto the first n coordinates. The constant-rank theorem gives these models when the rank remains constant near the point; maximal rank persists locally because a nonzero minor stays nonzero.
A regular value y is a target point at whose every preimage the derivative is surjective. It may have no preimage, in which case the condition is vacuous. For a nonempty regular fiber, the local projection model gives an embedded submanifold of dimension m−n. Its tangent directions are exactly those killed by the derivative, because moving within the fiber leaves the target fixed. Counting equations works only after their differentials have been checked for independence.
Embedding adds the correct global topology
An embedding is a smooth immersion that is a homeomorphism onto its image with the subspace topology. It must therefore be injective, but injectivity plus immersion can still fail the topological requirement for a noncompact source. A proper injective immersion is an embedding in our Hausdorff manifold setting. Proper means inverse images of compact sets are compact. A continuous injective map from a compact source into a Hausdorff target is automatically a homeomorphism onto its image, giving a useful compact special case.
An immersion can revisit the same point with different tangent directions. The source then remains a perfectly valid smooth manifold, but the image near a crossing need not be a submanifold with the inherited topology. Treat the parametrized map, its abstract source, and the image subset as separate objects. A closed curve drawn with a crossing is not thereby a singular source circle; it is a failure of that particular map to embed the circle.
Constraints should be checked on the whole proposed fiber
A convenient way to produce embedded submanifolds is to make them regular level sets. Sphere equations, orthogonality constraints, and intersecting independent conservation laws all fit this pattern. To apply the theorem globally, verify rank at every point satisfying the constraint, not just at one representative. Symmetry can simplify that verification, but a hidden exceptional point can change the geometry and dimension of the set. A zero gradient at one point means the regular-value theorem is silent there, rather than automatically proving that the set is singular.
For a source with boundary, regularity of the full derivative is not the entire story. A neat fiber also needs regularity of the restriction to the boundary at the same target value. We postpone that condition to the transversality lesson. In the boundaryless setting here, the kernel formula links geometry directly to linear algebra: calculate the independent constraint gradients, solve their simultaneous linear equations, and interpret that kernel as the allowed infinitesimal motion.
Work it through
Intersect a sphere with an independent plane
Let . Show that F⁻¹(2,0) is a one-dimensional embedded manifold, and find its tangent line at (1,−1,0).
The constraint gradients are and (1,1,1). Dependence would require (x,y,z)=t(1,1,1). The plane constraint then gives 3t=0, contradicting the sphere constraint x²+y²+z²=2. Thus rank is 2 everywhere on the fiber.
The regular-value theorem gives dimension 3−2=1. Geometrically the fiber is the radius-√2 circle in the plane through the origin perpendicular to (1,1,1), so it is nonempty and compact.
At (1,−1,0), a tangent vector v satisfies 2v₁−2v₂=0 and v₁+v₂+v₃=0. Hence it is a scalar multiple of (1,1,−2), giving the tangent line as the kernel of DF.
The full fiber is an embedded circle, and its tangent at the chosen point is span{(1,1,−2)}. Rank was verified on the entire constraint set, not inferred from a drawing.
The living notebook
Explore the model
Interactive lab · matrix
Check that a local direction has not been lost
Pause and Reset; set a=0.25,b=1. Predict a positive determinant and two nonzero singular values. For a fresh comparison Reset and set a=1,b=0. The matrices have full rank in both cases, although their conditioning differs. This square family illustrates a derivative that is both injective and surjective; it does not compute rectangular fibers or prove global injectivity of an immersed map.
- Area scale · det A
- 1.4
- Singular values · σmax, σmin
- 1.597, 0.876
- Condition number · κ₂(A)
- 1.823
The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA local rank test describes tangent directions, while embedding requires a separate global topological argument.
Pause & reason
Check your understanding
Your turn
Try it yourself
Consider γ:S¹→R² given in angular coordinates by γ(t)=(sin t,sin 2t). Prove it is an immersion and show why it is not an embedding.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Its derivative is (cos t,2cos 2t). If the first component vanishes, inspect the second; then compare t=0 and t=π.
Compare with the solution
If cos t=0, then cos 2t=−1, so the derivative cannot vanish. At all other t the first derivative component is already nonzero. Thus the tangent map is injective everywhere and γ is an immersion. But γ(0)=γ(π)=(0,0), although these are distinct points of the circle, so the map is not injective and hence not an embedding. The abstract source circle remains smooth.
Keep the idea
Take it with you
- Immersions preserve source tangent rank; submersions reach target tangent rank.
- Regular fibers have kernel tangent spaces and the expected codimension.
- An embedded image needs the correct topology in addition to local rank.
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