8.10. INVARIANCE OF DOMAIN 229

If Dg (z)−1 exists, then this is what is wanted. If not, use Lemma 8.10.2 and note that forall η small enough, you could replace g with y→ g (y)+η (y−z) and it will still be thecase that ∥g−h∥K < δ along with g (z) = h(z) but now Dg (z)−1 exists. Simply use themodified g. ■

The main result is essentially the following lemma which combines the conclusions ofthe above.

Lemma 8.10.4 Let f : B(p,r)→ Rn where the ball is also in Rn. Let f be one to one,f continuous. Then there exists δ > 0 such that

f(

B(p,r))⊇ B(f (p) ,δ ) .

In other words, f (p) is an interior point of f(

B(p,r))

.

Proof: Since f(

B(p,r))

is compact, it follows that f−1 : f(

B(p,r))→ B(p,r) is

continuous. By Lemma 8.10.3, there exists a polynomial g : f(

B(p,r))→ Rn such that∥∥g−f−1∥∥

f(B(p,r)) < εr, ε < 1, Dg (f (p))−1

exists, and g (f (p)) = f−1 (f (p)) = p

From the first inequality in the above,

|g (f (x))−x|=∣∣g (f (x))−f−1 (f (x))

∣∣≤ ∥∥g−f−1∥∥f(B(p,r)) < εr

By Lemma 8.10.1,

g ◦f(

B(p,r))⊇ B(p,(1− ε)r) = B(g (f (p)) ,(1− ε)r)

Since Dg (f (p))−1 exists, it follows from the inverse function theorem that g−1 also existsand that g,g−1 are open maps on small open sets containing f (p) and p respectively. Thusthere exists η < (1− ε)r such that g−1 is an open map on B(p,η)⊆ B(p,(1− ε)r). Thus

g ◦f(

B(p,r))⊇ B(p,(1− ε)r)⊇ B(p,η)

So do g−1‘ to both ends. Then you have g−1 (p) = f (p) is in the open set g−1 (B(p,η)) .Thus

f(

B(p,r))⊇ g−1 (B(p,η))⊇ B

(g−1 (p) ,δ

)= B(f (p) ,δ ) ■

pq ◦f

(B(p,r)

)B(p,(1− ε)r))

p= q(f(p))

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.