9.14. INVARIANCE OF DOMAIN 229

With this lemma, the invariance of domain theorem comes right away. This remarkabletheorem states that if f : U → Rn for U an open set in Rn and if f is one to one andcontinuous, then f (U) is also an open set in Rn.

Theorem 9.14.4 Let U be an open set in Rn and let f : U → Rn be one to one andcontinuous. Then f (U) is also an open subset in Rn.

Proof: It suffices to show that if p ∈U then f (p) is an interior point of f (U). LetB(p,r)⊆U. By Lemma 9.14.3, f (U)⊇ f

(B(p,r)

)⊇ B(f (p) ,δ ) so f (p) is indeed an

interior point of f (U). ■The inverse mapping theorem assumed quite a bit about the mapping. In particular it

assumed that the mapping had a continuous derivative. The following version of the inversefunction theorem seems very interesting because it only needs an invertible derivative at apoint.

Corollary 9.14.5 Let U be an open set inRp and let f : U→Rp be one to one and con-tinuous. Then, f−1 is also continuous on the open set f (U). If f is differentiable at x1 ∈Uand if Df (x1)

−1 exists for x1 ∈U, then it follows that Df−1 (f (x1)) = Df (x1)−1.

Proof: |·| will be a norm on Rp, whichever is desired. If you like, let it be the Euclideannorm. ∥·∥ will be the operator norm. The first part of the conclusion of this corollary isfrom invariance of domain.

From the assumption that Df (x1) and Df (x1)−1 exists,

y−f (x1) = f(f−1 (y)

)−f (x1) = Df (x1)

(f−1 (y)−x1

)+o

(f−1 (y)−x1

)Since Df (x1)

−1 exists, Df (x1)−1 (y−f (x1)) = f−1 (y)−x1 + o

(f−1 (y)−x1

)by

continuity, if |y−f (x1)| is small enough, then∣∣f−1 (y)−x1

∣∣ is small enough that in theabove,

∣∣o(f−1 (y)−x1)∣∣< 1

2

∣∣f−1 (y)−x1∣∣. Hence, if |y−f (x1)| is sufficiently small,

then from the triangle inequality of the form |p−q| ≥ ||p|− |q|| ,∥∥∥Df (x1)−1∥∥∥ |(y−f (x1))| ≥

∣∣∣Df (x1)−1 (y−f (x1))

∣∣∣≥∣∣f−1 (y)−x1

∣∣− 12

∣∣f−1 (y)−x1∣∣= 1

2

∣∣f−1 (y)−x1∣∣

|y−f (x1)| ≥∥∥∥Df (x1)

−1∥∥∥−1 1

2

∣∣f−1 (y)−x1∣∣

It follows that for |y−f (x1)| small enough,∣∣∣∣∣o(f−1 (y)−x1

)y−f (x1)

∣∣∣∣∣≤∣∣∣∣∣o(f−1 (y)−x1

)f−1 (y)−x1

∣∣∣∣∣ 2∥∥∥Df (x1)−1∥∥∥−1

Then, using continuity of the inverse function again, it follows that if |y−f (x1)| is pos-sibly still smaller, then f−1 (y)−x1 is sufficiently small that the right side of the aboveinequality is no larger than ε . Since ε is arbitrary, it follows

o(f−1 (y)−x1

)= o(y−f (x1))