8.7. HARD TOPOLOGY THEOREMS 205

want to consider manifolds with boundary for example. The need for this theorem occursin many other places as well in addition to being extremely interesting for its own sake. Theinverse function theorem gives conditions under which a differentiable function maps opensets to open sets. The following lemma, depending on the Brouwer fixed point theorem isthe thing which will allow this to be extended to continuous one to one functions. It saysroughly that if a continuous function does not move points near p very far, then the imageof a ball centered at p contains an open set.

Lemma 8.7.10 Let f be continuous and map B(p,r) ⊆ Rn to Rn. Suppose that for all

x ∈ B(p,r), |f(x)−x|< εr Then it follows that f(

B(p,r))⊇ B(p,(1− ε)r)

Proof: This is from the Brouwer fixed point theorem, Theorem 8.7.9. Consider for y ∈B(p,(1− ε)r) the function h(x)≡ x− f(x)+y Then h is continuous and for x ∈ B(p,r),

|h(x)−p|= |x− f(x)+y−p|< εr+ |y−p|< εr+(1− ε)r = r

Hence h : B(p,r)→ B(p,r) and so it has a fixed point x by Theorem 8.7.9. Thus x− f(x)+y = x so f(x) = y. ■

The notation ∥f∥K will mean supx∈K |f(x)|. If you have a continuous function h definedon a compact set K, then the Stone Weierstrass theorem implies you can uniformly approx-imate it with a polynomial g. That is ∥h−g∥K is small. The following lemma says that youcan also have g(z) = h(z) and Dg(z)−1 exists so that near z, the function g will map opensets to open sets as claimed by the inverse function theorem. First is a little observationabout approximating.

Lemma 8.7.11 Let K be a compact set in Rn and let h : K→ Rn be continuous, z ∈ Kis fixed. Let δ > 0. Then there exists a polynomial g (each component a polynomial) suchthat

∥g−h∥K < δ , g(z) = h(z) , Dg(z)−1 exists

Proof: By the Weierstrass approximation theorem, Theorem 3.2.4, there exists a poly-nomial ĝ such that ∥ĝ−h∥K < δ

3 . Then define for y ∈ K

g(y)≡ ĝ(y)+h(z)− ĝ(z)

Then g(z) = ĝ(z)+h(z)− ĝ(z) = h(z). Also

|g(y)−h(y)| ≤ |(ĝ(y)+h(z)− ĝ(z))−h(y)|

≤ |ĝ(y)−h(y)|+ |h(z)− ĝ(z)|< 2δ

3

and so since y was arbitrary, ∥g−h∥K ≤ 2δ

3 < δ . If Dg(z)−1 exists, then this is what iswanted. If not, use Lemma 4.7.1 and note that for all η small enough, you could replaceg with y→ g(y) + η (y− z) and it will still be the case that ∥g−h∥K < δ along withg(z) = h(z) but now Dg(z)−1 exists. Simply use the modified g. ■

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