§ 4. Differential Manifolds
239
Exercise 2. Let Z be a manifold, Y a submanifold of Z and X a subset of
Y . Show that X is a submanifold of Z if and only if its a submanifold of Y .
Exercise 3. Let f : X −→ Y be a homomorphism mapping a submanifold
X’ of X to a submanifold Y
of Y . Show that the map X
−→ Y
induced
by f is a homomorphism.
Exercise 4. Let X be the union in R
2 of the coordinate half-axes {x ≥ 0}
and {y ≥ 0} and P the point (−1, −1). For all M ∈ X, let t the slope of
the line P M, so that M → t is a homeomorphism from X to R
∗
+ . For any
open subset U of X, let C
∞ (U ) be the set of functions on U that are C
∞
as functions of t, which give a C
∞ manifold structure on X diffeomorphic
(under M → t) to R
∗
+ . Show that X is not a submanifold of R
2 .
Exercise 5. A submanifold is open in its closure. With the help of examples, show that it is not necessarily a submanifold.
(ii) Submanifolds defined by a subimmersion. Let X and Y be two manifolds of dimensions p and q and f : X −→ Y a C
r map. First consider the
set Z of solutions of f (x) = b for some given b ∈ f (X) and suppose that the
rank or of f is constant in an open subset containing Z (but not necessarily
in all of X); Z is then a submanifold of X.
Indeed, for all a ∈ X, there exist charts (U, ϕ) and (V, ψ) of Xand Y at a
and b for which ϕ(a) = 0, ψ(b) = 0, f (U ) ⊂ V and in which the map
F = ψ ◦ f ◦ ϕ
−1 .
(13.3)
Its rank which is constant in the neighbourhood of 0, is given by
F
ξ
1 , . . . , ξ
p
=
ξ
1 , . . . , ξ
r , 0, . . . , 0
.
(13.4)
ϕ(U ∩ Z) is then defined by the relations ξ
1 = . . . = ξ
r = 0, so the result of
section (i) implies that Z is a submanifold of dimension p − r of X .
If, moreover, as at the end of section (i), the space Z
(a) is identified with
its image in X
(a) under id
(a), where id : Z −→ X, then
57
Z
(a) = Ker f
(a) .
(13.5)
This is the subspace of h ∈ X
(a) such that f
(a)h = 0. Indeed, by definition
of Z, the map f ◦ id from Z to Y is the constant map z → b. Hence f
(a) ◦
id
(a) = 0, and so Z
(a) ⊂ Ker f
(a). As f
(a) : X
(a) −→ Y
(b) has rank r
and X
(a) dimension p,
dim Ker f
(a) = p − r = dim Z
(a) .
57 If u : E −→ F is a linear map, Ker u denotes the set of h ∈ E such u(h) = 0,
and Im u the set of u(h) ∈ F . Then
rg(u) = dim Im u = dim E − dim Ker u .
239
Exercise 2. Let Z be a manifold, Y a submanifold of Z and X a subset of
Y . Show that X is a submanifold of Z if and only if its a submanifold of Y .
Exercise 3. Let f : X −→ Y be a homomorphism mapping a submanifold
X’ of X to a submanifold Y
of Y . Show that the map X
−→ Y
induced
by f is a homomorphism.
Exercise 4. Let X be the union in R
2 of the coordinate half-axes {x ≥ 0}
and {y ≥ 0} and P the point (−1, −1). For all M ∈ X, let t the slope of
the line P M, so that M → t is a homeomorphism from X to R
∗
+ . For any
open subset U of X, let C
∞ (U ) be the set of functions on U that are C
∞
as functions of t, which give a C
∞ manifold structure on X diffeomorphic
(under M → t) to R
∗
+ . Show that X is not a submanifold of R
2 .
Exercise 5. A submanifold is open in its closure. With the help of examples, show that it is not necessarily a submanifold.
(ii) Submanifolds defined by a subimmersion. Let X and Y be two manifolds of dimensions p and q and f : X −→ Y a C
r map. First consider the
set Z of solutions of f (x) = b for some given b ∈ f (X) and suppose that the
rank or of f is constant in an open subset containing Z (but not necessarily
in all of X); Z is then a submanifold of X.
Indeed, for all a ∈ X, there exist charts (U, ϕ) and (V, ψ) of Xand Y at a
and b for which ϕ(a) = 0, ψ(b) = 0, f (U ) ⊂ V and in which the map
F = ψ ◦ f ◦ ϕ
−1 .
(13.3)
Its rank which is constant in the neighbourhood of 0, is given by
F
ξ
1 , . . . , ξ
p
=
ξ
1 , . . . , ξ
r , 0, . . . , 0
.
(13.4)
ϕ(U ∩ Z) is then defined by the relations ξ
1 = . . . = ξ
r = 0, so the result of
section (i) implies that Z is a submanifold of dimension p − r of X .
If, moreover, as at the end of section (i), the space Z
(a) is identified with
its image in X
(a) under id
(a), where id : Z −→ X, then
57
Z
(a) = Ker f
(a) .
(13.5)
This is the subspace of h ∈ X
(a) such that f
(a)h = 0. Indeed, by definition
of Z, the map f ◦ id from Z to Y is the constant map z → b. Hence f
(a) ◦
id
(a) = 0, and so Z
(a) ⊂ Ker f
(a). As f
(a) : X
(a) −→ Y
(b) has rank r
and X
(a) dimension p,
dim Ker f
(a) = p − r = dim Z
(a) .
57 If u : E −→ F is a linear map, Ker u denotes the set of h ∈ E such u(h) = 0,
and Im u the set of u(h) ∈ F . Then
rg(u) = dim Im u = dim E − dim Ker u .