Restricting Maps to Submanifolds
Having built the vocabulary of
embedded
and
immersed submanifolds,
we turn to the first of the two technical questions that make that vocabulary usable: when a smooth
map, restricted so that a submanifold becomes its domain or its codomain, remains smooth. The two
cases are not symmetric. Restricting the domain is automatic. Restricting the codomain is not, and
the asymmetry is exactly the asymmetry between the two ways the inclusion of a submanifold can
behave. As on the previous page, every ambient manifold is taken without boundary until the final
section, which works throughout on a manifold with boundary. The results of the intervening sections
extend to the boundary case, but that refinement is set aside for this series.
Restricting the domain never costs anything, because the inclusion of any submanifold is smooth.
Theorem (Restricting the Domain of a Smooth Map)
Let \(M\) and \(N\) be smooth manifolds, let \(F : M \to N\) be smooth, and let \(S \subseteq M\)
be an immersed or embedded submanifold. Then the restriction \(F|_S : S \to N\) is smooth.
Proof Sketch.
The inclusion \(\iota : S \hookrightarrow M\) is a smooth immersion, hence smooth, by the very
definition of a submanifold. Since \(F|_S = F \circ \iota\) is a composition of smooth maps, it
is smooth.
Restricting the codomain is a different matter. If a smooth map \(F : N \to M\) happens to take all
its values in a submanifold \(S \subseteq M\), one would like to regard it as a smooth map into
\(S\). But the smooth structure on \(S\) is the one that makes its inclusion an immersion,
and a map into \(S\) is smooth only relative to that structure. For an immersed submanifold, that
structure can be subtler than the subspace structure suggests. In the following example the
restriction fails to be smooth in the worst possible way. It is not even continuous.
Let \(S \subseteq \mathbb{R}^2\) be the
figure-eight curve, given the
topology and smooth structure induced by the injective immersion \(\beta\) that traces it, and define
\(G : \mathbb{R} \to \mathbb{R}^2\) by \(G(t) = (\sin 2t, \sin t)\). The formula is that of
\(\beta\), but the domain is now the whole line rather than an open interval. The image of \(G\) lies
in \(S\), so one is tempted to view \(G\) as a map \(\mathbb{R} \to S\).
Yet \(\beta^{-1} \circ G\) is discontinuous at \(t = \pi\). There \(G(\pi)\) is the crossing point
\(\beta(0)\) itself, but as \(t\) approaches \(\pi\) from either side the points \(G(t)\) run along
the strand that \(\beta\) parametrizes near the ends \(\pm\pi\) of its domain. Hence
\(\beta^{-1}(G(t)) \to \pm\pi\) while \(\beta^{-1}(G(\pi)) = 0\). The two values \(0\) and \(\pm\pi\)
are the parameters that the topology of \(S\) keeps far apart, even though their images crowd
together at the crossing. So \(G\), regarded as a map into \(S\), is not even continuous, let alone
smooth. The obstruction is entirely topological, and it turns out to be the only one.
Theorem (Restricting the Codomain of a Smooth Map)
Let \(M\) be a smooth manifold, let \(S \subseteq M\) be an immersed
submanifold, and let \(F : N \to M\) be a smooth map whose image is contained in \(S\). If \(F\)
is continuous as a map from \(N\) into \(S\), then \(F : N \to S\) is smooth.
Proof Sketch.
Fix \(p \in N\) and let \(q = F(p) \in S\). Because \(S\) is immersed, it is
locally embedded.
There is a neighborhood \(V\) of \(q\) in \(S\) that is an embedded submanifold of \(M\), so by
the
local slice criterion
it admits a slice chart \((W, \psi)\) of \(M\). Writing \(\pi\) for the projection onto the slice
coordinates and setting \(V_0 = V \cap W\), the pair \((V_0, \pi \circ \psi|_{V_0})\) is a smooth
chart for \(S\) near \(q\).
The hypothesis that \(F\) is continuous into \(S\) gives an open set \(U \ni p\) in \(N\) with
\(F(U) \subseteq V_0\). On \(U\) the coordinate representation of \(F : N \to S\) is
\(\pi \circ (\psi \circ F)\), a composition of smooth maps because \(F : N \to M\) is smooth.
Hence \(F\) is smooth near each point of \(N\).
When the submanifold is embedded, the troublesome continuity hypothesis is free. A map into an
embedded submanifold is continuous into \(S\) the moment it is continuous into \(M\).
Corollary (Restricting to an Embedded Submanifold)
Let \(M\) be a smooth manifold and \(S \subseteq M\) an embedded submanifold. Then every smooth
map \(F : N \to M\) whose image is contained in \(S\) is also smooth as a map from \(N\) into
\(S\).
Proof Sketch.
Since \(S\) carries the subspace topology, the
universal property of the subspace topology
makes \(F : N \to S\) continuous as soon as \(F : N \to M\) is. The previous theorem then
upgrades continuity to smoothness.
The corollary fails for some immersed submanifolds, and the figure-eight above is the witness. It
does hold, however, for certain immersed-but-nonembedded ones, and the property is worth naming.
Definition: Weakly Embedded Submanifold
An immersed submanifold \(S \subseteq M\) is weakly embedded if every smooth map
\(F : N \to M\) whose image lies in \(S\) is automatically smooth as a map from \(N\) into \(S\).
Such submanifolds are also called initial submanifolds.
Embedded submanifolds are weakly embedded, but so are others that arise naturally. The most important
are the Lie subgroups encountered later in the manifold series, which are weakly embedded even when
they are not embedded.
Uniqueness of Smooth Structures
A question has been hanging over the theory since the slice criterion. When we recognized a subset
\(S \subseteq M\) as an embedded submanifold, we equipped it with a smooth structure, the one read
off from slice charts. Could a different choice of charts, or even a different topology, have made
the same set \(S\) into a submanifold in some other way? If so, "the smooth structure on \(S\)" would
be ambiguous, and every statement about submanifolds would have to specify which structure it meant.
The results of this section close that gap. For embedded submanifolds the structure is completely
rigid. The set alone determines it. This is what lets us speak of the sphere as a
submanifold, with no further qualification. This uniqueness is what the local slice criterion claimed
but left for later, when it first identified subsets satisfying the slice condition.
Theorem (Uniqueness for Embedded Submanifolds)
Let \(M\) be a smooth manifold and \(S \subseteq M\) an embedded submanifold. Then the subspace
topology and the smooth structure furnished by the
local slice criterion
are the only topology and smooth structure with respect to which \(S\) is a submanifold,
embedded or immersed.
Proof Sketch.
Suppose \(S\) carried a second topology and smooth structure making it an (embedded or immersed)
submanifold. Write \(\widetilde S\) for the set \(S\) equipped with this alternative structure,
and write \(\widetilde\iota : \widetilde S \to M\) for its inclusion, which by assumption is an
injective immersion. The key is to compare the two structures through the identity map of
underlying sets. Since \(\widetilde\iota(\widetilde S) = S\) lands in the embedded submanifold
\(S\), the corollary on
restricting to an embedded submanifold
guarantees that \(\widetilde\iota\), viewed as the map \(\widetilde S \to S\), is smooth. Call
this map \(j : \widetilde S \to S\). It is the identity on points, and it is a smooth bijection.
Now factor the inclusion of \(\widetilde S\) as \(\widetilde\iota = \iota \circ j\), where
\(\iota : S \to M\) is the embedding. Differentiating gives
\(d\widetilde\iota_p = d\iota_p \circ dj_p\) at each point. The left side is injective because
\(\widetilde\iota\) is an immersion, so \(dj_p\) is injective and \(j\) is itself an immersion.
An immersion has constant rank, equal to the dimension of its domain, so \(j\) is a bijective map
of constant rank, and the
global rank theorem
promotes a constant-rank bijection to a diffeomorphism. A diffeomorphism that is the identity on
points forces \(\widetilde S\) and \(S\) to carry the same topology and the same smooth
structure.
With uniqueness secured, the recognition problem for embedded submanifolds collapses to a question
about the set itself. A subset \(S \subseteq M\) either satisfies the local slice condition or it
does not, and if it does, its submanifold structure is automatic. Because the slice condition is
local, this even globalizes. If every point of \(S\) has a neighborhood \(U\) in \(M\) with
\(U \cap S\) an embedded \(k\)-submanifold of \(U\), then \(S\) is an embedded \(k\)-submanifold of
\(M\).
For immersed submanifolds the situation is genuinely weaker, and it is worth being precise about how.
The earlier examples, the figure-eight and the dense torus curve, already show that a single set can
be made into an immersed submanifold with topologies finer than the subspace topology. What
cannot vary is the smooth structure once the topology is fixed.
Theorem (Uniqueness of the Smooth Structure on an Immersed Submanifold)
Let \(M\) be a smooth manifold and \(S \subseteq M\) an immersed submanifold. For the given
topology on \(S\), there is exactly one smooth structure making \(S\) an immersed submanifold.
Proof Sketch.
Suppose two smooth structures on the fixed topological space \(S\) both make the inclusion an
injective immersion. Write \(S_A\) and \(S_B\) for the two, and let \(j : S_A \to S_B\) be the
identity on points. Since the two structures share the same topology, the one fixed in the
statement, \(j\) is a homeomorphism. In particular it is continuous into \(S_B\). Its composite
with the inclusion of \(S_B\) is just the inclusion of \(S_A\), which is smooth into \(M\). Since
\(j\) is continuous into the immersed submanifold \(S_B\), the
codomain-restriction theorem
applies and makes \(j\) smooth. The same argument with the roles reversed makes \(j^{-1}\)
smooth, so \(j\) is a diffeomorphism that is the identity on points, and the two smooth
structures coincide.
So the only freedom for an immersed submanifold lies in the choice of topology. Pin that down and the
smooth structure follows. The intermediate class of weakly embedded submanifolds, introduced in the
previous section, recovers full rigidity. For them the topology is forced too, just as for embedded
ones.
Theorem (Uniqueness for Weakly Embedded Submanifolds)
Let \(M\) be a smooth manifold and \(S \subseteq M\) a weakly embedded submanifold. Then \(S\)
has only one topology and one smooth structure with respect to which it is an immersed
submanifold.
Proof Sketch.
Let \(S_W\) denote \(S\) with its weakly embedded structure, and let \(\widetilde S\) be the same
set with any other topology and smooth structure making it an immersed submanifold. Write
\(j : \widetilde S \to S_W\) for the identity on points. The inclusion of \(\widetilde S\) is a
smooth map into \(M\) with image in \(S\), so the defining property of weak embeddedness applied
to \(S_W\) makes that inclusion smooth into \(S_W\). In other words, \(j\) is smooth.
Factoring the inclusion of \(\widetilde S\) as the inclusion of \(S_W\) composed with \(j\) and
differentiating shows \(dj_p\) injective at each point, so \(j\) is an immersion, of constant
rank equal to the dimension of its domain. Thus \(j\) is a bijective map of constant rank, and
the
global rank theorem
makes it a diffeomorphism, identical on points, forcing the two structures, topology and smooth
structure alike, to coincide.
The three results together arrange the kinds of submanifold by how much of their structure the
ambient set determines. An embedded submanifold is rigid outright. A weakly embedded one, such as a
Lie subgroup, is equally rigid, though the proof must route around the missing subspace topology. A
general immersed submanifold retains freedom in its topology, surrendering only the smooth structure
that the topology then fixes. This hierarchy is the reason later constructions take care to record
which kind of submanifold they produce.
Extending Functions from Submanifolds
Restriction asks whether a map defined on the ambient manifold stays smooth when cut down to a
submanifold. The complementary question runs the other way: given a smooth function defined only on
the submanifold, can it be extended to a smooth function on the surrounding space? Before answering,
we must settle an ambiguity in the phrase "smooth function on \(S\)" that is easy to overlook and
important to get right.
There are two things one might mean. A function \(f : S \to \mathbb{R}\) on a submanifold could be
smooth in the sense intrinsic to \(S\). Intrinsic smoothness means smoothness as a function on the
manifold \(S\), so that each of its coordinate representations in the charts of \(S\) is smooth. Or
it could be smooth in the sense inherited from the ambient space. Ambient smoothness means smoothness
as a function on the
subset \(S \subseteq M\),
so that \(f\) admits a smooth extension to a neighborhood of each point. These are not the same
condition, and conflating them is a genuine source of error. We reserve the notation \(C^\infty(S)\)
for the first, intrinsic meaning.
Definition: Smooth Functions on a Submanifold
Let \(M\) be a smooth manifold and \(S \subseteq M\) an immersed or embedded submanifold. The
notation \(C^\infty(S)\) denotes the smooth functions on \(S\) in the
intrinsic sense. A function \(f : S \to \mathbb{R}\) belongs to \(C^\infty(S)\)
if it is smooth with respect to the smooth structure of \(S\) as a manifold in its own right.
This is a priori distinct from smoothness as a function on the subset \(S \subseteq M\),
which would require a local smooth extension to the ambient manifold at each point.
The relationship between the two senses is exactly the content of the extension lemma below. For an
embedded submanifold, an intrinsically smooth function does admit local ambient extensions, and a
properly embedded one admits a single global extension. The distinction matters because intrinsic
smoothness is the weaker, always-available notion, while extendability is what the lemma must work to
produce.
Lemma (Extension Lemma for Functions on Submanifolds)
Let \(M\) be a smooth manifold, let \(S \subseteq M\) be an embedded submanifold, and let
\(f \in C^\infty(S)\).
(a) There exist a neighborhood \(U\) of \(S\) in \(M\) and a smooth function
\(\widetilde f \in C^\infty(U)\) with \(\widetilde f|_S = f\).
(b) If \(S\) is properly embedded, the neighborhood \(U\) may be taken to be all of \(M\).
Proof Sketch.
For (a), work locally first. Near a point of \(S\), a slice chart presents \(S\)
as a coordinate subspace, in which the intrinsically smooth \(f\) is a smooth function of the
slice coordinates. Precomposing with the projection onto those coordinates extends it to a smooth
function on a neighborhood that restricts to \(f\). These local extensions are then glued into a
single smooth function on a neighborhood \(U\) of \(S\) by a partition of unity subordinate to
the cover, the values agreeing on \(S\) where every local piece equals \(f\).
For (b), a properly embedded submanifold is a
closed subset
of \(M\). Part (a) shows that \(f\) is smooth on \(S\) in the ambient sense, since it extends
smoothly near each point. Because \(S\) is closed, the
extension lemma for closed subsets
delivers a smooth function on all of \(M\) restricting to \(f\). Closedness is exactly the
hypothesis that lemma requires, which is why proper embeddedness is the condition that upgrades a
local extension to a global one.
Both hypotheses earn their place. Without embeddedness, an intrinsically smooth function on an
immersed submanifold may admit no ambient extension at all, because the submanifold's own topology
can be finer than the ambient trace. Without proper embeddedness, a local extension near an "open
end" of the submanifold may resist being closed up into a global one. The lemma is the precise
statement that these are the only obstructions.
The Tangent Space to a Submanifold
We come to the second technical question. A submanifold \(S\) inside \(M\) carries its own
tangent space
\(T_pS\) at each point, built from \(S\) as a manifold in its own right. Geometric intuition insists
that \(T_pS\) ought to be a subspace of the ambient \(T_pM\). The tangent plane to a surface in space
is, after all, literally a plane through the origin of the surrounding space. Making that
identification precise, and then learning to compute the subspace, is the goal of this section.
The identification is supplied by the inclusion. Since \(\iota : S \hookrightarrow M\) is a smooth
immersion, its
differential
\(d\iota_p : T_pS \to T_pM\) is injective at every point. We therefore identify \(T_pS\) with its
image \(d\iota_p(T_pS)\), a linear subspace of \(T_pM\), and henceforth regard a tangent vector to
\(S\) as a tangent vector to \(M\) that happens to be tangent to \(S\). Under this identification a
vector \(v \in T_pS\) acts on a smooth function \(f\) defined near \(p\) in \(M\) exactly as it acts
on the restriction \(f|_S\), which is intrinsically smooth on \(S\) by domain restriction. Writing
\(\widetilde v = d\iota_p(v)\), one has \(\widetilde v f = v(f|_S)\). The identification is available
whether \(S\) is embedded or merely immersed, since it uses only that the inclusion is an immersion.
The first concrete description of this subspace is kinematic. The vectors tangent to \(S\) are
precisely the velocities of curves that stay in \(S\).
Proposition: Tangent Vectors as Velocities in a Submanifold
Let \(M\) be a smooth manifold, \(S \subseteq M\) an immersed or embedded submanifold, and
\(p \in S\). A vector \(v \in T_pM\) lies in the subspace \(T_pS\) if and only if there is a
smooth curve \(\gamma : J \to M\) whose image is contained in \(S\), which is also smooth as a
map into \(S\), with \(0 \in J\), \(\gamma(0) = p\), and \(\gamma'(0) = v\).
Proof Sketch.
If \(v \in T_pS\), then since
every tangent vector is a velocity,
there is a curve \(\sigma\) in \(S\) with \(\sigma'(0) = v\) computed in \(S\). Composing with
the inclusion gives a curve \(\gamma = \iota \circ \sigma\) in \(M\) lying in \(S\), and
computing the differential through this velocity
shows \(\gamma'(0) = d\iota_p(v)\), which is \(v\) under the identification. Conversely, a curve
\(\gamma\) that lies in \(S\) and is smooth into \(S\) has a velocity \(\gamma'(0)\) computed in
\(S\), an element of \(T_pS\). Its image in \(T_pM\) is the ambient velocity, so that ambient
velocity lies in \(T_pS\).
For embedded submanifolds there is a second description, dual to the first and often far more useful
in practice. Where the velocity criterion builds \(T_pS\) from curves inside \(S\), this one
carves it out by the functions that vanish on \(S\). A vector is tangent to \(S\) exactly
when it annihilates every function that is constant, indeed zero, along \(S\).
Proposition: The Tangent Space as an Annihilator
Let \(M\) be a smooth manifold, \(S \subseteq M\) an embedded submanifold, and \(p \in S\). As a
subspace of \(T_pM\),
\[
T_pS = \{\, v \in T_pM : vf = 0 \text{ whenever } f \in C^\infty(M) \text{ and } f|_S = 0 \,\}.
\]
Proof Sketch.
One inclusion is immediate. If \(v \in T_pS\), write \(v = d\iota_p(w)\) for \(w\) in the
intrinsic tangent space of \(S\), and suppose \(f \in C^\infty(M)\) vanishes on \(S\). Then
\(f \circ \iota \equiv 0\), so \(vf = d\iota_p(w)f = w(f \circ \iota) = 0\).
For the reverse inclusion, suppose \(v \in T_pM\) annihilates every such \(f\). We show that
\(v \in T_pS\). Choose slice coordinates \((x^1, \dots, x^n)\) about \(p\) in which \(S\) is the
slice \(\{x^{k+1} = \dots = x^n = 0\}\), so that \(T_pS\) is spanned by
\(\partial/\partial x^1, \dots, \partial/\partial x^k\). Writing
\(v = \sum_i v^i\, \partial/\partial x^i|_p\), the claim is that \(v^i = 0\) for \(i \gt k\).
Fix such an index \(j \gt k\). The coordinate function \(x^j\) vanishes on \(S\) within the
chart, but to apply the hypothesis we need a function defined on all of \(M\) that vanishes on
\(S\). A
smooth bump function
\(\varphi\) supported in the chart domain and equal to \(1\) near \(p\) lets us extend
\(\varphi\, x^j\) by zero to a function \(f \in C^\infty(M)\). This function vanishes on \(S\),
inside the chart because \(x^j\) does and outside because it is identically zero, and it agrees
with \(x^j\) near \(p\). Then \(0 = vf = v(x^j) = v^j\). As \(j \gt k\) was arbitrary, \(v\) has
no components transverse to the slice, so \(v \in T_pS\).
When the embedded submanifold is presented as a regular level set, this annihilator description
becomes a clean computation. The tangent space is simply the kernel of the differential of any
defining map.
Proposition: The Tangent Space of a Level Set
Let \(M\) be a smooth manifold and \(S \subseteq M\) an embedded submanifold. Let
\(U \subseteq M\) be open and let \(\Phi : U \to N\) be a smooth submersion having \(S \cap U\)
as a level set, which is a
local defining map
for \(S\) when \(N\) is Euclidean. Then for each \(p \in S \cap U\),
\[
T_pS = \ker\bigl(d\Phi_p : T_pM \to T_{\Phi(p)}N\bigr).
\]
In particular, if \(S\) is the regular level set of a submersion
\(\Phi = (\Phi^1, \dots, \Phi^r) : M \to \mathbb{R}^r\), then \(v \in T_pM\) is tangent to \(S\)
if and only if \(v\Phi^1 = \dots = v\Phi^r = 0\).
Proof Sketch.
Since \(\Phi\) is constant on \(S\), the composition \(\Phi \circ \iota\) is constant, so
\(d\Phi_p \circ d\iota_p = 0\). Hence
\(T_pS = \operatorname{im} d\iota_p \subseteq \ker d\Phi_p\). Both sides have the same dimension.
On the left, \(\dim T_pS = \dim S = \dim M - \dim N\), since a level set of a submersion has
codimension \(\dim N\).
On the right, \(d\Phi_p\) is surjective, so reading it as a matrix in any bases for \(T_pM\) and
\(T_{\Phi(p)}N\) and applying the
rank-nullity law
leaves a kernel of the same dimension. A subspace contained in another of equal finite dimension
fills it, so \(T_pS = \ker d\Phi_p\). The component form is the statement that the kernel of
\(d\Phi_p\) is the common zero set of the differentials \(d\Phi^i_p\).
These descriptions also give a clean way to prove that a set is not a submanifold. That
question is otherwise awkward, since one must rule out every possible topology and smooth structure.
The strategy is to assume the set is a submanifold and derive a contradiction from properties every
submanifold must have: at each point the tangent space is a subspace of fixed dimension, every
tangent vector is the velocity of a curve in the set, and every tangent vector annihilates functions
vanishing on the set.
The graph of the absolute value function \(\{(x, y) : y = |x|\}\) in the plane, for instance, cannot
be a smooth submanifold. Away from the corner it is a one-dimensional submanifold, so it would have
to be one-dimensional throughout, and in particular at the corner its tangent space would be a line
spanned by the velocity of some curve running through it. But a smooth curve that stays in the set
and passes through the corner is forced to have zero velocity there. Its two coordinate functions are
tied by \(y = |x|\), which is not differentiable at \(0\) unless the \(x\)-component has vanishing
derivative. Hence no nonzero tangent vector can arise, and a one-dimensional tangent space is
impossible. The tangent space, made ambient, is what supplies the contradiction.
Pointing Vectors at the Boundary
When \(M\) is a smooth manifold with boundary and \(p\) is a
boundary point,
the tangent space \(T_pM\) still has the full dimension \(n\). The
tangent space at a boundary point
is an honest \(n\)-dimensional vector space, with no half-space restriction.
The boundary \(\partial M\) is an embedded hypersurface, so its tangent space \(T_p\partial M\) sits
inside \(T_pM\) as a hyperplane, splitting the complement into two open half-spaces. Vectors in one
half point into the manifold and vectors in the other point out of it. The vectors lying in the
hyperplane itself are tangent to the boundary. Making this trichotomy precise is what allows an
orientation, a flow, or a volume to be related correctly to the boundary, and it is the geometric
content behind the sign conventions of integration over a region.
Definition: Inward- and Outward-Pointing Vectors
Let \(M\) be a smooth manifold with boundary, let \(p \in \partial M\), and let
\(v \in T_pM \setminus T_p\partial M\). The vector \(v\) is inward-pointing if
for some \(\varepsilon \gt 0\) there is a smooth curve \(\gamma : [0, \varepsilon) \to M\) with
\(\gamma(0) = p\) and
velocity
\(\gamma'(0) = v\). It is outward-pointing if there is such a curve whose domain
is \((-\varepsilon, 0]\) instead. Vectors tangent to \(\partial M\) are excluded from both
classes at the outset. The next proposition establishes that the remaining vectors split into
exactly these two kinds, with none left unrealized by any curve and none admitting curves of both
types. That split is not evident from the definition itself.
The defining curves are geometrically vivid but awkward to test. A boundary chart converts the
trichotomy into a single sign, by reading off the last coordinate.
Proposition: Pointing Detected by the Last Coordinate
Let \(M\) be a smooth \(n\)-manifold with boundary, let \(p \in \partial M\), and let
\((x^i)\) be the coordinates of any
boundary chart
defined near \(p\). Writing a vector \(v \in T_pM\) as
\(v = v^i\, \partial / \partial x^i\big|_p\), the vector \(v\) is inward-pointing precisely when
its last component \(v^n\) is positive, outward-pointing precisely when \(v^n\) is negative, and
tangent to \(\partial M\) precisely when \(v^n = 0\). Consequently \(T_pM\) is the disjoint
union of \(T_p\partial M\), the inward-pointing set, and the outward-pointing set, and \(v\) is
inward-pointing if and only if \(-v\) is outward-pointing.
Proof Sketch.
In the boundary chart the slice \(\partial M\) is \(\{x^n = 0\}\) and the manifold occupies
\(\{x^n \geq 0\}\), so a tangent vector lies in \(T_p\partial M\) if and only if its last
component \(v^n\) vanishes. This settles the tangential case and shows that any
\(v \in T_pM \setminus T_p\partial M\) has \(v^n \neq 0\). It remains to match the sign of a
nonzero \(v^n\) with the two pointing classes.
A smooth curve \(\gamma\) with \(\gamma(0) = p\) and image in \(M\) has \(x^n(\gamma(t)) \geq 0\)
throughout its domain, with equality at \(t = 0\). If the domain is \([0, \varepsilon)\), then
\(t = 0\) is a minimum of the nonnegative smooth function \(t \mapsto x^n(\gamma(t))\), so its
one-sided derivative there, namely \(v^n = \tfrac{d}{dt}\big|_{0} x^n(\gamma(t))\), is
\(\geq 0\). Since \(v \notin T_p\partial M\) forces \(v^n \neq 0\), an inward-pointing vector has
\(v^n \gt 0\). Conversely, given any \(v\) with \(v^n \gt 0\), the coordinate ray
\(t \mapsto \varphi^{-1}\bigl(\varphi(p) + t\, d\varphi_p(v)\bigr)\) is a curve on
\([0, \varepsilon)\) realizing \(v\) and staying in \(M\) for small \(t\) (its last coordinate
\(t\, v^n\) is \(\geq 0\)), so \(v\) is inward-pointing. Thus the inward-pointing vectors are
exactly those with \(v^n \gt 0\).
Replacing the domain by \((-\varepsilon, 0]\) reverses the inequality, so the outward-pointing
vectors are exactly those with \(v^n \lt 0\). The three sign conditions \(v^n \gt 0\),
\(v^n \lt 0\), \(v^n = 0\) are mutually exclusive and exhaustive on \(T_pM\), giving the claimed
disjoint union, and negating \(v\) flips the sign of \(v^n\), which yields the final equivalence.
Because inward- and outward-pointing were defined intrinsically through curves, the
classification is independent of the chosen boundary chart, even though the number \(v^n\) itself
is not.
There is a companion description of the boundary as the zero set of a single function, which is
often the most convenient way to produce vectors pointing the right way and, later, an outward unit
normal.
Definition: Boundary Defining Function
Let \(M\) be a smooth manifold with boundary. A boundary defining function for
\(M\) is a smooth function \(f : M \to [0, \infty)\) such that \(f^{-1}(0) = \partial M\) and
\(df_p \neq 0\) for every \(p \in \partial M\). For example, \(f(x) = 1 - |x|^2\) is a boundary
defining function for the closed unit ball \(\overline{\mathbb{B}}^n\). This is unrelated to the
defining function of a regular level set discussed earlier. There the function cuts out a
submanifold as a level set in a manifold without boundary, whereas here it cuts out the boundary
of a manifold that already has one.
Proposition: Existence of a Boundary Defining Function
Every smooth manifold with boundary admits a boundary defining function. Moreover, for such a
function \(f\) and a boundary point \(p\), a vector \(v \in T_pM\) is inward-pointing if and only
if \(vf \gt 0\), outward-pointing if and only if \(vf \lt 0\), and tangent to \(\partial M\) if
and only if \(vf = 0\).
Proof Sketch.
Choose a cover of \(M\) by charts, each either an interior chart or a boundary chart. On the
domain of the \(\alpha\)-th chart define a smooth function
\(f_\alpha : U_\alpha \to [0, \infty)\) as follows. On an interior chart set
\(f_\alpha \equiv 1\), and on a boundary chart set \(f_\alpha = x^n\), the last coordinate, which
is nonnegative there and vanishes exactly on \(U_\alpha \cap \partial M\). Let
\(\{\psi_\alpha\}\) be a
partition of unity
subordinate to the cover, and set \(f = \sum_\alpha \psi_\alpha f_\alpha\).
Because the family of supports of the \(\psi_\alpha\) is locally finite, only finitely many terms
are nonzero near any given point, so the sum is finite on a neighborhood of every point. Hence
\(f\) is well-defined, smooth, and may be differentiated term by term. The function is
nonnegative. It is positive at every interior point (where some \(f_\alpha\) equals \(1\) and the
others are nonnegative) and zero at every boundary point (where every \(f_\alpha\) that
contributes vanishes), so \(f^{-1}(0) = \partial M\).
To see that \(df_p \neq 0\) at a boundary point \(p\), note that only boundary charts contain
\(p\), so the sum runs over boundary indices alone near \(p\). For such \(\alpha\),
\(f_\alpha(p) = x^n(p) = 0\), and differentiating the product gives
\[
df_p = \sum_\alpha \bigl(f_\alpha(p)\, d\psi_\alpha|_p + \psi_\alpha(p)\, df_\alpha|_p\bigr)
= \sum_\alpha \psi_\alpha(p)\, dx^n_\alpha\big|_p,
\]
the first term in each summand vanishing because \(f_\alpha(p) = 0\). Evaluating on a vector
\(v\) that is inward-pointing, each \(dx^n_\alpha|_p(v) = v^n_\alpha\) is positive by the
previous proposition (the pointing class is independent of the boundary chart), and the
coefficients \(\psi_\alpha(p) \geq 0\) sum to \(1\). Hence \(df_p(v) \gt 0\), so in particular
\(df_p \neq 0\).
The same computation gives the pointing criterion. For any \(v \in T_pM\),
\(vf = df_p(v) = \sum_\alpha \psi_\alpha(p)\, v^n_\alpha\) is a convex combination of the last
components \(v^n_\alpha\) of \(v\) in the various boundary charts at \(p\). By the previous
proposition these components all share the sign of the pointing class of \(v\): all positive when
\(v\) is inward-pointing, all negative when outward-pointing, and all zero when \(v\) is tangent
to \(\partial M\). Therefore \(vf \gt 0\), \(vf \lt 0\), or \(vf = 0\) according to the three
cases.