11.1. MANIFOLDS 263

where here there is α > 0 such that∣∣DS−1 (S (R−1 (x)))(

S(R−1 (x+h)

)−S

(R−1 (x)

))∣∣≥ α

∣∣(S (R−1 (x+h))−S

(R−1 (x)

))∣∣thanks to the assumption that DS−1 (S (R−1 (x)

))is one to one. Thus from 11.2

α

2

∣∣(S (R−1 (x+h))−S

(R−1 (x)

))∣∣≤ ∣∣DR−1 (x)h+o(h)∣∣ (11.3)

Now ∣∣o(S (R−1 (x+h))−S

(R−1 (x)

))∣∣|h|

≤∣∣o(S (R−1 (x+h)

)−S

(R−1 (x)

))∣∣∣∣S (R−1 (x+h))−S

(R−1 (x)

)∣∣∣∣S (R−1 (x+h)

)−S

(R−1 (x)

)∣∣|h|

From 11.3, the second factor in the above is bounded. Now continuity of S ◦R−1 impliesthat as h→ 0, the first factor also converges to 0. Thus

o(S(R−1 (x+h)

)−S

(R−1 (x)

))= o(h)

Returning to 11.2,

DR−1 (x)h+o(h) = DS−1 (S (R−1 (x)))(

S ◦R−1 (x+h)−S ◦R−1 (x))

Thus if h= tv,

limt→0

DS−1 (S (R−1 (x)))((S ◦R−1 (x+ tv)−S ◦R−1 (x)

)t

)

= DR−1 (x)v+ limt→0

o(tv)t

= DR−1 (x)v

By the above lemma, limt→0(S◦R−1(x+tv)−S◦R−1(x))

t = Dv

(S ◦R−1)(x) exists. Also

DS−1 (S (R−1 (x)))

Dv

(S ◦R−1)(x) = DR−1 (x)v

Let A(x) ≡ DS−1 (S (R−1 (x)))

. Then A∗A is invertible and x→ A(x) is continuous.Then

A(x)∗A(x)Dv

(S ◦R−1)(x) = A(x)∗DR−1 (x)v

Dv

(S ◦R−1)(x) =

(A(x)∗A(x)

)−1 A(x)∗DR−1 (x)v

so Dv

(S ◦R−1)(x) is continuous. It follows from Theorem 6.6.1 that S ◦R−1 is a func-

tion in C1(R(U ∩V

))because the Gateaux derivatives exist and are continuous. ■

Saying DR−1 (x) is one to one is the analog of the situation in calculus with a smoothcurve in which we assume the derivative is non zero and that the parametrization has con-tinuous derivative.

I will assume in what follows that Ω is a compact subset of Rq, q ≥ p. You could getby with less using Stone’s theorem about paracompactness but this is enough for what willbe used here.