With these theorems it is possible to give easy proofs of some very important and difficult theorems.
Definition 11.5.1 If f : U ⊆ ℝ^{p} → ℝ^{p} where U is an open set. Then f is locally one to one if for every x ∈ U, there exists δ > 0 such that f is one to one on B
As a first application, consider the invariance of domain theorem. This result says that a one to one continuous map takes open sets to open sets. It is an amazing result which is essential to understand if you wish to study manifolds. In fact, the following theorem only requires f to be locally one to one. First here is a lemma which has the main idea.

The symbol on the left means:
Proof: For t ∈

Then for x ∈ ∂B

and this requires x = 0 which is not the case since

Now with this lemma, it is easy to prove the very important invariance of domain theorem.
A function f is locally one to one on an open set Ω if for every x_{0} ∈ Ω, there exists B
Theorem 11.5.3 (invariance of domain)Let Ω be any open subset of ℝ^{p} and let f : Ω → ℝ^{p} be continuous and locally one to one. Then f maps open subsets of Ω to open sets in ℝ^{p}.
Proof: Let B

Then g satisfies the conditions of Lemma 11.5.2, being one to one and continuous. It follows from that lemma there exists δ > 0 such that
With the above, one gets easily the following amazing result. It is something which is clear for linear maps but this is a statement about continuous maps.
Proof: Suppose not and let f be such a continuous map,

Then let g
Corollary 11.5.5 If f is locally one to one and continuous, f : ℝ^{p} → ℝ^{p}, and

then f maps ℝ^{p} onto ℝ^{p}.
Proof: By the invariance of domain theorem, f
The next theorem is the famous Brouwer fixed point theorem.
Theorem 11.5.6 (Brouwer fixed point) Let B = B
Proof: Assume there is no fixed point. Consider h

By homotopy invariance,

is constant. But when t = 0, this is d
You can use standard stuff from Hilbert space to get this the fixed point theorem for a compact convex set. Let K be a closed bounded convex set and let f : K → K be continuous. Let P be the projection map onto K as in Problem 30 on Page 266. Then P is continuous because

Hence, subtracting the first from the last,

consequently,

and so
Now let r be so large that K ⊆ B
Theorem 11.5.7 Let f : K → K be continuous where K is compact and convex and nonempty, K ⊆ ℝ^{p}. Then f has a fixed point.
Definition 11.5.8 f is a retract of B

and f
Proof: Suppose f were such a retract. Then for all x ∈ ∂B

which is clearly impossible because f^{−1}
You should now use this theorem to give another proof of the Brouwer fixed point theorem.
The proofs of the next two theorems make use of the Tietze extension theorem, Theorem 2.3.5.
Theorem 11.5.10 Let Ω be a symmetric open set in ℝ^{n} such that 0 ∈ Ω and let f : ∂Ω → V be continuous where V is an m dimensional subspace of ℝ^{n},m < n. Then f
Proof: Suppose not. Using the Tietze extension theorem on components of the function, extend f to all of ℝ^{n}, f

because B

and this is a contradiction because V is m dimensional. ■
This theorem is called the Borsuk Ulam theorem. Note that it implies there exist two points on opposite sides of the surface of the earth which have the same atmospheric pressure and temperature, assuming the earth is symmetric and that pressure and temperature are continuous functions. The next theorem is an amusing result which is like combing hair. It gives the existence of a “cowlick”.
Theorem 11.5.11 Let n be odd and let Ω be an open bounded set in ℝ^{n} with 0 ∈ Ω. Suppose f : ∂Ω → ℝ^{n} ∖
Proof: Using the Tietze extension theorem, extend f to all of ℝ^{n}. Also denote the extended function by f. Suppose for all x ∈ ∂Ω, f


Thus there exists a homotopy of f and I and a homotopy of f and −I. Then by the homotopy invariance of degree,

But this is impossible because d