10.2. INVARIANCE OF DOMAIN 219

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 10.2.3, 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 10.2.5 Let U be an open set in Rp and let f : U →Rp be one to one and contin-uous. Then, f−1 is also continuous on the open set f(U). If f is differentiable at x1 ∈U andif Df(x1)

−1 exists for x1 ∈U, then it follows that Df(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 the

above, ∣∣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∣∣− 1

2

∣∣f−1 (y)−x1∣∣

=12

∣∣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 possiblystill smaller, then f−1 (y)−x1 is sufficiently small that the right side of the above inequalityis no larger than ε . Since ε is arbitrary, it follows

o(f−1 (y)−x1

)= o(y− f(x1))