Chapter 3 Multivariable Calculus

Question 1

open ball

Answer 1

B_r(x) open ball centred at a point x ∈ R^n of radius r > 0, that is
Br(x) = {y ∈ R^n : |x − y| < r}.

Question 2

open

Answer 2

A subset U ⊂ R^n is called open if for every x ∈ U there exists a ball Br(x) such that Br(x) ⊂ U

Question 3

closed

Answer 3

A subset V ⊂ R^n is called closed, if the complement R^n\V is open

Question 4

open and closed intervals

Answer 4

open intervals (a, b) ⊂ R are
open subsets: for every point x ∈ (a, b) one can find an interval (x − δ, x + δ) with an appropriate
small δ > 0 such that it is contained in (a, b). Similarly, closed intervals [a, b] ⊂ R are closed subsets.

Question 5

open closed?
empty set
[0,1)

Answer 5

The empty set ∅ and the whole space R^ n are both open and closed.

The set [0, 1) ⊂ R is neither open nor closed in R.

Question 6

An open ball Br(x) ⊂ R^n is an open set.

Answer 6

Proof. Let y ∈ Br(x) be an arbitrary point. We need to show that there exists r’ > 0 such that the
ball B_r’ (y) is contained in Br(x). Denote by ρ the distance |x − y| < r, and set r’ = r − ρ > 0. We
claim that B_r’0 (y) ⊂ Br(x). To see the latter pick an arbitrary point z ∈ B_r’ (y). Then by the triangle
inequality we obtain
|z − x| <= |z − y| + |y − x| < r’ + ρ = (r − ρ) + ρ = r, inclusion demonstrated

Question 7

closed ball

Answer 7

A closed ball B¯r(x) = {y ∈ R
n : |x − y| <= r} ⊂ R^n is a closed set.

Question 8

converging

Answer 8

A sequence (p_k) of points pk ∈ R^n, where k = 1, . . . , +∞, is called converging to
a point p ∈ R^ n, if the sequence of lengths
|p_k − p|, or equivalently the distance between p_k and p,
converges to zero, that is |pk − p| → 0 as k → +∞.

a sequence pk = (x1k, . . . , xnk) ∈ R^n converges to a point
p = (x1, . . . , xn) ∈ R^n if and only if for each i = 1, . . . , n the sequence xik converges to xi as
k → +∞.

Question 9

example: consider sequence p_k=(1/k, 0) ∈ R^2

converging?

Answer 9

limit p=(0,0)

Question 9

is this an open set:
U = R^2(∪pk) = R^2
{pk : k > 1}

Answer 9

not an open set: the
point p = (0, 0) belongs to U and no ball centred at p lies in U. Thus, the set V = (∪pk) = R^2\U is not closed. Note that if we add the limit point (0, 0) to the set V , then we obtain a closed set
V¯ = V ∪ {(0, 0)}

Question 10

PROP 3.1 closed set iff

Answer 10

A subset V ⊂ R^n is closed if and only if for any converging sequence (pk) of points pk ∈ V its limit p belongs to V .

Question 11

A subset V ⊂ R^n is closed if and only if for any converging sequence (pk) of points pk ∈ V its limit p belongs to V .

Answer 11

Proof. Suppose that V is closed, and let us show that for any converging sequence (pk) of points
p_k ∈ V its limit p belongs to V . Suppose the contrary, that is, there exists a sequence (pk) of points
pk ∈ V whose limit p does not belong to V . Then p ∈ R^n\V , and since the latter set is open, there exists r > 0 such that the ball B_r(p) lies in R^n\V . Since p is the limit of pk, there exists integer N
such that |pk − p| < r for any k > N. Hence, pk ∈ B_r(p) ⊂ R
n\V for any k > N, and in particular,
pk ∈/ V for any k > N. Contradiction.
Now we prove the converse statement. Suppose the contrary: V is not closed, and hence, R
n\V is not open. The latter means that there exists a point p ∈ R
n\V such that for any ball Br(p) there exists a point q ∈ Br(p) such that q /∈ R^n\V , i.e. q ∈ V . For any integer k > 0 choose such a point
qk ∈ B1/k(p), qk ∈ V . Then the sequence qk converges to p as k → +∞. Hence, by our hypotheses,
p ∈ V , and we arrive at a contradiction: p has been chosen from the set R^n\V .

Question 12

defn continuous

Answer 12

Let W ⊂ R^n be an open subset. A map Φ : W → R^m is called continuous at a point x ∈ W if for any ball B_ε(Φ(x)) ⊂ R
m there exists a ball Bδ(x) ⊂ W such that Φ(B_δ(x)) ⊂ B_ε(Φ(x)).

A map Φ : W → R^m is called continuous if it is continuous at every point x ∈ W.

Question 12

EXAMPLE: checking if homeomorphism
f : R → R, f(x) = x^2

Answer 12

it is not a homemorphism (since it is neither surjective nor injective, hence,not bijective).

However, the map f+ : (0, +∞) → (0, +∞), f+(x) = x^2 is a homeomorphism.

Question 13

Proposition III.2. Let W ⊂ R^n be an open set. Then for a map Φ : W → R^m the following
hypotheses are equivalent:

Answer 13

(i) Φ is continuous everywhere in W (in the sense of Definition III.5);
(ii) for any open subset U ⊂ R
m the pre-image Φ−1(U) is open;
(iii) for any closed subset V ⊂ R
m the pre-image Φ−1 (V ) is closed;
(iv) for any x ∈ W and any converging sequence xk → x, where x_k ∈ W, the sequence Φ(x_k)
converges to Φ(x)

proof: notes

Question 14

example:
f_1, . . . , f_m collection of continuous functions on R^n.

U = {x ∈ R^n: f_1(x) < 0, . . . , f_m(x) < 0}

closed? open?

Answer 14

Let f_1, . . . , f_m be a collection of continuous functions on R^n. Then by Proposition III.2 the subset

U = {x ∈ R^n: f_1(x) < 0, . . . , f_m(x) < 0} ⊂ R^n is open, and the subset

U¯ = {x ∈ R^n: f_1(x) <= 0, . . . , f_m(x) <= 0} ⊂ R^n
is closed

Question 14

HOMEOMORPHISM

Answer 14

Let U ⊂ R^n and V ⊂ R^n be two open sets. A map Φ : U → V is called a homeomorphism if it is bijective, continuous, and the inverse map Φ−1is also continuous

Question 15

EXAMPLE: checking if homeomorphism

f : R → R, f(x) = x^3:

Answer 15

(i) f : R → R, f(x) = x^3: it is a homemorphism. First, f is a continuous function. It is straightforward to see that it is bijective, and the inverse function, given by f−1(y) = y^{1/3}
, is also continuous

Question 16

are these homeomorphic?
intervals
(−1, 1) and (−n, n)

Answer 16

homemorphism exists
Φ(t) = nt, where t ∈ (−1, 1).
bijective, inverse continuous

Question 17

homeomorphism?
f(x) = x^k

where k > 0 is an integer?

Answer 17

odd or even function?

Question 17

HOMEOMORPHIC

Answer 17

The sets U ⊂ R^ n and V ⊂ R^n are called homoeomorphic if there exists a homeomorphism Φ : U → V .

Question 18

are these homeomorphic?
intervals
(−1, 1) or (−n, n)
with
X = (−∞, −1) ∪ (1, +∞).

Answer 18

neither
suppose there exists a homeomorphism
Φ : (−1, 1) → X.
Let t_1 and t_2 be points from
(−1, 1) such that
Φ(t_1) = −2 and Φ(t_2) = 2.

Then by the intermediate value theorem there exists a point t_0 ∈ (−1, 1) such that Φ(t_0) = 0. However, this is a contradiction, since Φ takes values in X and 0 ∈/ X

Question 19

IVT

Answer 19

intermediate value thm

If f is continuous on [a,b] and f(a) neg f(b) pos
Then there exists a c in the interval st f(c) equals zero

Question 19

continuous embedding meaning

Answer 19

any map satisfying is continuous

since a map Φ : U → R
m is assumed to be injective, the map
Φ : U → Φ(U) is bijective, and one can talk about the inverse map Φ−1
: Φ(U) → U. The last hypothesis in the definition says precisely that the inverse map Φ−1 is continuous in the so-called
sequential sense

Question 19

continuous embedding

Answer 19

Let U ⊂ R^n be an open subset. A map Φ : U → R^ m, where m > n, is called a continuous embedding if it is injective and any sequence x_k ∈ U converges to a point x ∈ U if and only if the sequence Φ(x_k) converges to Φ(x).

Question 20

are these continuous embeddings?

Example III.12Consider the following maps, representing different waysof bending intervals in a plane:
(i) Φ_1 : (0, 1) → R^2 is given by Φ_1(t) = (cos 2πt,sin 2πt).

Answer 20

..
1) is a continuous embedding
Here we consider t_k and t in (o,1)

Φ(t_k) converges to Φ(t) implies t_k converges to t

Case t in [o, 1/2) y>o since y_k converges to y conclude y_k>o for all sufficiently large k
Write

Φ(t_k) equals (x_n, y_n)

Φ(t) equals (x,y)
Implies t_k in (o,1/2) in this interval cos is a continuously monotonic function and inverse t_k equals arcos(x_k) implies arccos(x) equal t

Case t in (1/2,1) similarly y>o y_k >o
x_k <o use arcsin(y_k) instead so t in (1/4,3/4)
x<o x_k <o
For sufficiently large x so t_k in (1/4,3/4) on this interval and sin is a decreasing function, invertivle t_k equals arcsin (y_k) so arcsin(y) equals t

Question 20

vector function
DIFFERENTIABLE

Answer 20

A vector-function Φ : U ⊂ R^n → R^ m is called differentiable at a point x ∈ U if there exists a linear map L : R^n → R^m such that
lim_h→0
|Φ(x + h) − Φ(x) − L(h)|/|h|
= 0,

· |length of a vector in R^m in numerator, in R^n in denominator.

Question 21

Example III.14 (Derivative and differential in one variable). Let Φ : U ⊂ R → R be a real-valued
function of one variable. Recall that any linear map L : R → R has the form L(h) = ah for some real
number a ∈ R

if Φ is differentiable at a point x ∈ U….

Answer 21

a as 1x1 matrix
0=
lim_h→0|Φ(x + h) − Φ(x) − L(h)|/|h|
= lim_h→0|Φ(x + h) − Φ(x) − ah|/|h|
= lim_h→0
|[Φ(x + h) − Φ(x)]/h− a|

limit (Φ(x+h)−Φ(x))/h exists and equals a.

Question 21

DIFFERENTIAL:
from being differentiable

Answer 21

linear map L is called the differential of Φ at x, and is denoted by D_xΦ.

linear map L is unique and satisfies:
L : R^n → R^ m is a linear operator such that
|L(h)| / |h| → 0 as h → 0, then L ≡ 0.

Question 22

if Φ is differentiable
at a point x ∈ U in the sense of then the derivative Φ’
(x) exists, and it is related to
the differential D_xΦ : R → R by the formula:

Answer 22

D_xΦ(h) = Φ’(x)h
h in R is a variable

conversely for a given x ∈ U the derivative Φ’(x) exists, then Φ is differentiable at x

Question 23

jacobian

Answer 23

Jacobian matrix of Φ at a point x ∈ U.

the matrix of the differential D_xΦ in the standard bases in the Euclidean spaces

for a map Φ : U ⊂ R^n → R^m the matrix of partial derivatives

J_xΦ =

[∂Φ1/∂x1(x) ∂Φ1/∂x2(x) · · ·∂Φ1/∂xn(x)]
[∂Φ2/∂x1(x) ∂Φ2/∂x2(x) · · ·∂Φ_2/∂x_n(x)]
[· · · · · · · · · · · ·]
[∂Φ_m/∂x1(x) ∂Φm/∂x2(x) · · ·∂Φm/∂xn(x)]

Question 24

PROPN 3.3
differential
jacobian

Answer 24

Let Φ : U ⊂ R^n → R^m be a map differentiable at a point x ∈ U. Then the matrix of the differential D_xΦ : R^n → R^m in the standard bases e1, . . . , en and ε1, . . . , εm in R^n and R^m coincides with the Jacobian matrix J_xΦ

Question 25

proof@
Let Φ : U ⊂ R^n → R^m be a map differentiable at a point x ∈ U. Then the matrix of the differential D_xΦ : R^n → R^m in the standard bases e1, . . . , en and ε1, . . . , εm in R^n and R^m coincides with the Jacobian matrix J_xΦ

Answer 25

notes

Question 26

when do partial derivs exist

Answer 26

if Φ is differentiable at a point
x ∈ U , then all partial derivatives (∂Φj/∂xi) exist at x

Question 27

smooth

Answer 27

Φ : U ⊂ R^n → R^m are
smooth, that is all its mixed partial derivatives of any order exist at every point in U.

Question 28

gradient

Answer 28

for a smooth real-valued function Φ : U ⊂ R^n → R the gradient ∇Φ(x) is defined as the vector in R^n whose
coordinates are partial derivatives (∂Φ/∂x_i)(x), where x ∈ R

Question 28

EXAMPLE
Let Φ : R^3 → R^2 be a smooth map defined by the relation
Φ(x1, x2, x3) = (x^2_1 + x_2, x_2x_3)

find
DxΦ(h)
for h=(1,0,-1)

Answer 28

its differential D_xΦ
at a point x = (x1, x2, x3) is represented by the 2 × 3-matrix
JxΦ =
[2x_1 1 0]
[ 0 x_3 x_2]

For a vector h = (1, 0, −1) ∈ R^3
the value D_xΦ(h) is the vector (2x1, −x2) ∈ R^2; the latter is obtained
by multiplying the 2 × 3-matrix J_xΦ by the column (1, 0, −1)^t

Question 29

WHEN ARE LINEAR MAPS DIFFERENTIABLE

Answer 29

Let L : R^n → R^m be a linear map. Then D_xL = L for any x ∈ R^n,
and in particular, L is differentiable. Indeed, to show that DxL = L we compute the limit
lim_h→0
|L(x + h) − L(x) − L(h)|/|h|
= lim_h→0
0/|h|
= 0,
where the first relation follows by linearity of L.

Question 30

formula relating gradient and differential

Answer 30

D_xΦ(h) = ∇Φ(x) · h

where h ∈ R^n.

This can be seen as a consequence of Proposition (III.3):
D_xΦ(h) = J_xΦh^T
= J_xΦ^T · h
= ∇Φ(x) · h,

where we used that J_xΦ is an 1 × n-matrix, and its transpose J_xΦ^T is a vector in R^n

USEFUL

Question 31

PROP 3.4 CHAIN RULE

Answer 31

Let f : U ⊂ R^n → V ⊂ R^m be a map differentiable at a point x ∈ U,
and g : V ⊂ R^m → R^ℓ be a map differentiable at a point f(x) ∈ V . Then the map g ◦f : U ⊂ R^n → R^ ℓ
is differentiable at x and Dx(g ◦ f) = D_f(x) g ◦ D_x f.

Proof. This is a reformulation of a statement in the course of calculus, which is normally stated in
terms of Jacobian matrices, that is J_x(g ◦f) = J_f(x) g J_x f, or relationships between partial derivatives.

Question 32

thinking of vectors v in R^n as velocity vectors to curves

Answer 32

for any x, v ∈ R^n there exists a PC γ : (−ε, ε) → R^n such that
γ(0) = x and γ’(0) = v.

Question 33

corollary 3.5 differential

Answer 33

Corollary III.5. Let Φ : U ⊂ R^n → R^m be a map differentiable at x ∈ U. Then the differential DxΦ
satisfies the following relation
D_xΦ(v) = d/dt |_t=0 Φ(γ(t))
where γ : (−ε, ε) → R^n is a PC such that γ(0) = x and γ’(0) = v.

Proof. A direct consequence of Proposition III.4 with g = Φ and f = γ. Mind that the velocity γ’(0)
can be thought of as the vector D_0 γ(1).

Question 34

DIFFEOMORPHISM

Answer 34

Let U ⊂ R^n and V ⊂ R^m be open sets. A map Φ : U → V is called a
diffeomorphism if Φ is smooth, bijective, and the inverse map Φ−1
: V → U is also smooth

change of coordinate system ensures calcs equivalent in different

Question 35

proof

Answer 35

Proof. Since Φ−1 ◦ Φ = IdU , then by Proposition III.4 we obtain
DΦ(x)Φ
−1
◦ DxΦ = Dx IdU = Id,
and conclude that the differential DxΦ has a left inverse operator. Similarly, one obtains that
DΦ−1(y)Φ ◦ DyΦ
−1 = Id, where y = Φ(x), and we conclude that DxΦ has a right inverse operator also. Thus, it is a linear isomorphism.

Question 35

Example III.18 (Polar coordinates on Euclidean half-plane). Consider the map
Φ : R+ × (0, π) → {(x, y) : y > 0}, Φ(r, ϕ) = (r cos ϕ, r sin ϕ)

is this a diffeomorphism

Answer 35

Φ is bijective, smooth, and the inverse map Φ−1
, given by
the formula
Φ−1(x, y) = root( x^2+y^2)
arccos(x/root(x^2 + y^2)),
is also smooth. Thus, Φ is a diffeomorphism.

Question 35

polar coords

Answer 35

Note that polar coordinates can be defined on a larger set than the one we used in Example III.18.
For instance, we can consider the following map
Φ : R+ × (0, 2π) → R^2
{(x, 0) : x > 0}, Φ(r, ϕ) = (r cos ϕ, r sin ϕ).
This map is also a diffeomorphism, but writing down the inverse map is slightly more tedious

Question 35

corollary 3.6 diffeomorphism differential is a

Answer 35

Let Φ : U → V be a diffeomorphism of an open set U ⊂ R^n onto an open set V ⊂ R^m. Then for any x ∈ U the differential D_xΦ : R^n → R^m is a linear isomorphism. In
particular, the dimensions of the domain and codomain spaces coincide, n = m.

Proof. Since Φ−1 ◦ Φ = IdU , then by Proposition III.4 we obtain
DΦ(x)Φ−1◦ DxΦ = Dx IdU = Id,
and conclude that the differential DxΦ has a left inverse operator. Similarly, one obtains that
DΦ−1(y)Φ ◦ DyΦ−1 = Id, where y = Φ(x), and we conclude that DxΦ has a right inverse operator also. Thus, it is a linear isomorphism

Question 35

Theorem III.7 (Inverse Function Theorem).

Answer 35

Let Φ : U → Rn be a smooth map defined on an open
set U ⊂ R
n. Suppose that for a given point p ∈ U the differential DpΦ : Rn → R
n is a linear
isomorphism. Then there exist open subsets V , W ⊂ R
n containing points p and Φ(p) respectively
such that Φ : V → W is a diffeomorphism.
Proof. MATH5113M only.

Question 35

PROOF
Theorem III.7 (Inverse Function Theorem).

Answer 35

L5

Question 35

EXAMPLE
Φ : R^2 → R^2
Φ(x_1, x_2) = (sin(x_1 + x_2), x_1x_2)
Differential
D_pΦ at the point p = (0, π)

is this a linear isomorphism?

Answer 35

computing partial derivatives
evaluate at p
J_p Φ =
[cos π cos π
[ π 0]
=
[−1 −1]
[π 0]

This matrix has rank two, and hence, is invertible,
det J_p Φ /= 0, and we see that
D_pΦ : R^2 → R^2 is a
linear isomorphism. Thus, by Theorem III.7 there exist open sets V, W ⊂ R^2 with (0, π) ∈ V and (0, 0) ∈ W such that Φ : V → W is bijective, and its inverse Φ−1: W → V is smooth.

Note that Φ : R^2 → R^2
is clearly not a diffeomorphism, and looking for such sets V and W can be a non-trivial job.

Question 35

Corollary III.8. Φ is a diffeomorphism
and
linear isomorphism??

Answer 35

Corollary III.8. Let U, V ⊂ R
n be open subsets, and let Φ : U → V be a smooth bijective map.
Then Φ is a diffeomorphism if and only if the differential
D_xΦ : R^n → R^n is a linear isomorphism
for any x ∈ U

proof:only check inverse is smooth by inverse function thm

Question 35

HOW TO CHECK A BIJECTIVE SMOOTH MAP IS A DIFFEOMORPHISM

Answer 35

to make sure that a given bijective smooth map is a diffeomorphism, we do not really need to write down the inverse map and try to check that it is smooth; it is sufficient to study the differential only. In other words, Corollary III.8 replaces the calculus problem by the algebraic
problem of studying the differential of a given map. A

Question 35

EXAMPLE POLAR COORDS
consider map
Φ : R+ × (0, 2π) → R^2{(x, 0) : x > 0},

Φ(r, ϕ) = (r cos ϕ, r sin ϕ).
is this a diffeomorphism

Answer 35

smooth and bijective
so by corollary 3.8 its diffeomorphism if we check differential D_p Φ is linear isomorphism p = (r, ϕ) ∈ R
+ × (0, 2π).

computing partial derivatives
J(r,ϕ)Φ =
[cos ϕ −r sin ϕ]
[sin ϕ r cos ϕ]

determinant =r > 0,
thus DpΦ is an isomorphism.
Thus, by Corollary III.8, the map Φ is a diffeomorhism

Question 36

Example III.21 (Folding singularity)

linear isomorphism?
Consider the map Φ : R^2 → R^2
Φ(x, y) = (x, y^2).

Geometrically

Answer 36

takes a plane and folds it along the line y = 0 to the top half-plane. Computing partial derivatives, we obtain
J_(x,y) Φ =
[1 0]
[0 2y]

where (x, y) ∈ R^2. Note that for any point (x, 0), i.e. contained in the line y = 0, the differential D_pΦ
is not a linear isomorphism.

the line y = 0 along which the map folds is all singular points of Φ

Question 37

REGULAR

Answer 37

For a smooth map Φ : U ⊂ Rn → Rm a point p ∈ U is called regular if the
differential DpΦ : Rn → Rm has maximal rank, that is rank DpΦ = min{n, m}

Question 38

SINGULAR

Answer 38

A point p ∈ U is
called singular, if it is not regular.

Question 39

Example III.23 (Pinch point singularity).
Consider the map Φ : R2 → R3 given by the relation
Φ(x, y) = (xy, x, y^2).

Answer 39

Computing partial derivatives, we obtain the Jacobian matrix
J(x,y)Φ =
[y x]
[1 0]
[0 2y]
(x, y) ∈ R^2

AT (0, 0) differential DpΦ has rank one, and at all other
points has rank two. Thus, the point (0, 0) is the only singular point.

The image of Φ set in R^3 Whitney umbrella, pinch point singularity at origin

Question 40

regular map

Answer 40

.A smooth map Φ : U ⊂ Rn → R m is called regular if any point p ∈ U is a regular
point of Φ.

Question 41

Example III.25 (Regularity of curves).

for smooth map
γ : I → R^n, where I ⊂ R is an open interval
is this regular?

Answer 41

differential Dtγ has maximal rank (= 1) if and only if γ’(t) /= 0. Thus, the
map γ is regular if and only if γ’(t) /= 0 for any t ∈ I. This is precisely the definition of a regular parametrised curve

Viewing γ as a vector-function (γ1, . . . , γn), differential
Dtγ can be written n × 1-matrix
J_t γ =
[γ’_1(t)]
· · ·
[γ’_n(t)]

Question 42

(ii) Φ_2 : (−1/4, 3/4) → R^2
is given by Φ_2(t) = (cos 2πt,sin 4πt).
(draw the images of Φ1 and Φ2 and explain that Φ1 is a continuous embedding,
while Φ2 is not.)

Answer 42

Injective continuous map
But no continuously embedded
We need to construct sequence st Φ(t_k) converges to Φ(t) but t_k doesn’t converge to t

Consider
T_k equal to -1/4 +1/k
These converge to -1/4

Φ(t_k) converge to (o,o) but
t equal 1/4 t_k don’t converge to this they converge to -1/4

Question 43

Example III.26 (Regularity of surfaces).

DIFFERENTIAL FOR
Let Ω ⊂ R2 be an open subset of the plane R2, and let
r : Ω → R3 be a smooth map. Let (u, v) be coordinates on R2, and (x, y, z) coordinates on R3. Then
for a given point p ∈ Ω the differential D_p r : R2 → R3
can be viewed as

REGULARITY OF SURFACE

Answer 43

J_pr =
x_u x_v
y_u y_v
z_u z_v

maximal rank (= 2)
IFF
vectors
r_u = (x_u, y_u, z_u) and
r_v = (x_v, y_v, z_v) are linearly
independent,
equivalent to
r_u × r_v /= 0.

map r is regular
iff
vectors ru = (xu, yu, zu) and rv = (xv, yv, zv) are linearly independent for any point p = (u, v) ∈ Ω.

THIS IS NOT FOR OUR COURSE DEFN doesnt rule out self intersections

Question 44

MAP OF CONSTANT RANK

Answer 44

A smooth map Φ : U ⊂ R^n → V ⊂ R^m is called the map of constant rank k if rank D_ pΦ = k for any p ∈ U.

Question 44

true or false
regular maps are example of constant rank maps

Answer 44

true an example of

Question 45

Example III.29 (Straightening a curve).

Answer 45

Let γ : I → R^m be an RPC. Theorem III.9 says that for any point t_0 ∈ I one can find an interval J ⊂ I containing t_0 and a diffeomorphism Φ defined in a neighbourhood of the point γ(t_0), which should be thought as a change of variables on R^m in a neighbourhood of γ(t0), such that the RPC Φ ◦ γ : J → R
n is an arc of a straight line. Moreover,
after re-parametrisation it takes the form
Φ◦ γ(τ ) wiggly overline = (τ, 0, . . . , 0), where the symbol (·) above signifies that the PC Φ ◦ γ is re-parametrised.

Question 45

Theorem III.9 (Constant Rank Theorem).

Answer 45

Let Φ : U ⊂ Rn → Rm be a smooth map of constant rank k. Then for any point x ∈ U there exist open subsets V ⊂ Rn and W ⊂ R m containing points x and Φ(x) respectively, and diffeomorphisms
F : V → V˜ ⊂ R n
and
G : W → W˜ ⊂ Rm
such that
F(x) = 0 ∈ R^n,
G(Φ(x)) = 0 ∈ R^m, and the map
Φ =˜ G ◦ Φ ◦ F−1: V˜ → W˜ has the form

Φ( ˜ x1, . . . , xn) = (x1, . . . , xk, 0, . . . , 0).

Question 45

FIND RANK
Example III.28. Let n, m, and k be positive integers such that k 6 min{n, m}, and let Φ : Rn → Rm
be a linear operator that forgets the last n − k coordinates, that is
Φ : (x1, . . . , xk, xk+1, . . . , xn) −→ (x1, . . . , xk, 0, . . . , 0).

Answer 45

by prev e.g
DpΦ = Φ for any point p ∈ R
n, and hence rank DpΦ = k.

Question 46

Example III.30 (Spheres as pre-images)

Consider the function f : R^3 → R defined by the relation
f(x, y, z) = x^2 + y^2 + z^2
where (x, y, z) ∈ R^3
For a real number r ∈ R consider the pre-image f−1(r):

Answer 46

• if r < 0, then f−1(r) = ∅, since f(x, y, z) > 0;
• if r = 0, then f−1(r) = {0}, since f(x, y, z) = 0 ⇐⇒ (x, y, z) = (0, 0, 0);
• if r > 0, then f−1(r) = {(x, y, z) ∈ R^3
  : x^2 + y^2 + z^2 = r} is a sphere of radius √r.

cases r = 0 pre-image f−1 (r) contains a singular point of the function f,
r > 0 are no singular points in f−1(r).

Question 47

Example III.31 (Whitney’s umbrella and its deformations).

Answer 47

Consider the function g : R3 → R
defined by the relation
g(x, y, z) = x^2 − y^2z, where (x, y, z) ∈ R^3
Computing the derivatives, we obtain
J(x,y,z)g = (2x, −2yz, −y^2),

and see that the set of singular points consists of the z axis, that is when x = 0 and y = 0.

pre-image g−1(0) consists of the Whitney umbrella in Example III.23 together with the negative part of the z axis.

pre-image g−1(δ), where δ ∈ R{0}, does not contain any singular points, and analysing it one can
speculate that it looks like a nice surface.

Question 48

morse theory

Answer 48

pre-images f−1(c) of various functions on surfaces yields a lot of information about surfaces themselves. Even topological properties of surfaces can be described in terms
of properties of pre-images f
−1(c), and certain information about singular points of f. This is the
subject of the so-called Morse Theory

Question 49

L5: proving the inverse funct thm

Lemma III.2 (Contraction Mapping Principle).

Answer 49

Let Φ : B¯_r → B¯_r be a map such that |Φ(x) − Φ(y)| <= q |x − y| for any x, y ∈ B¯r,
where 0 < q < 1. Then there exists a unique fixed point x_0 ∈ B¯r for Φ, that is Φ(x0) = x0.
Moreover, for any x ∈ B¯
r the sequence xk = Φk
(x) converges to x0.

proof:show its cauchy , geometric series