Multivariable Advanced Calculus

288 BROUWER DEGREE∞∑∫∞∑∫= det (Dh (x)) f (h (x)) dx = f (y) d (y, Ω, h) dyi=1 B i i=1 h(B i )∫∫= f (y) d (y, Ω, h) dy = f (y) d (y, Ω, h) dy∪ ∞ i=1 h(B i)h(∪ ∞ i=1 B i)∫= f (y) d (y, Ω, h) dy,h(Ω)the last line following from Lemma 10.7.2. This proves the theorem. 10.8 Exercises1. Show the Brouwer fixed point theorem is equivalent to the nonexistence of acontinuous retraction onto the boundary of B (0, r).2. Using the Jordan separation theorem, prove the invariance of domain theorem.Hint: You might consider B (x, r) and show f maps the inside to one of twocomponents of R n \ f (∂B (x, r)) . Thus an open ball goes to some open set.3. Give a version of Proposition 10.6.2 which is valid for the case where n = 1.4. Suppose n > m. Does there exists a continuous one to one map, f which maps R monto R n ? This is a very interesting question because there do exist continuous mapsof [0, 1] which cover a square for example. Hint: First show that if K is compactand f : K → f (K) is one to one, then f −1 must be continuous. Now considerthe increasing sequence of compact sets{ (increasing sequence f B (0, k))} ∞k=1{B (0, k)} ∞k=1 whose union is Rm and thewhich you might assume covers R n . Youmight use this to argue that if such a function exists, then f −1 must be continuousand then apply Corollary 10.4.4.5. Can there exists a one to one onto continuous map, f which takes the unit intervalto the unit disk? Hint: Think in terms of invariance of domain and use the hintto Problem 4.6. Consider the unit disk,{(x, y) : x 2 + y 2 ≤ 1 } ≡ Dand the annulus {(x, y) : 1 }2 ≤ x2 + y 2 ≤ 1 ≡ AIs it possible there exists a one to one onto continuous map f such that f (D) = A?Thus D has no holes and A is really like D but with one hole punched out. Canyou generalize to different numbers of holes? Hint: Consider the invariance ofdomain theorem. The interior of D would need to be mapped to the interior of A.Where do the points of the boundary of A come from? Consider Theorem 5.3.5.7. Suppose C is a compact set in R n which has empty interior and f : C → Γ ⊆ R n isone to one onto and continuous with continuous inverse. Could Γ have nonemptyinterior? Show also that if f is one to one and onto Γ then if it is continuous, sois f −1 .