Kenneth Kuttler

16.3. SOMETHING WHICH IS NOT A RETRACT 431

Proof: Suppose such an h exists. Let λ ∈ [0,1] and let pλ (x)≡ x+λ (h(x)−x) . Thisfunction, pλ is a homotopy of the identity map and the retraction, h. Let

I (λ )≡∫

B(0,R)det(Dpλ (x))dx.

Then using the dominated convergence theorem,

I′ (λ ) =∫

B(0,R)∑i. j

∂ det(Dpλ (x))∂ pλ i, j

∂ pλ i j (x)∂λ

=∫

B(0,R)∑

i∑

∂ det(Dpλ (x))∂ pλ i, j

(hi (x)− xi), j dx

=∫

B(0,R)∑

i∑

jcof(Dpλ (x))i j (hi (x)− xi), j dx

Now by assumption, hi (x) = xi on ∂B(0,R) and so one can integrate by parts and write

I′ (λ ) =−∑i

∫B(0,R)

∑j

cof(Dpλ (x))i j, j (hi (x)− xi)dx = 0.

Therefore, I (λ ) equals a constant. However,

I (0) = mn

(B(0,R)

)> 0

butI (1) =

∫B(0,R)

det(Dh(x))dmn =∫

∂B(0,R)#(y)dmn = 0

because from polar coordinates or other elementary reasoning, mn (∂B(0,1)) = 0.The last formula uses the change of variables formula for functions which are not one

to one. In this formula, #(y) equals the number of x such that h(x) = y. To see this is soin case you have not seen this, note that h is C1 and so the inverse function theorem fromadvanced calculus applies. Thus∫

B(0,R)det(Dh(x))dmn =

∫[det(Dh(x))>0]

det(Dh(x))dmn

+∫[det(Dh(x))<0]

det(Dh(x))dmn

Thus h is locally one to one on the two open sets [det(Dh(x))> 0] , [det(Dh(x))< 0].Now use inverse function theorem and change of variables for one to one h to verify thatboth of these integrals equal 0. You cover [det(Dh(x))> 0] with countably many ballson which h is one to one and then use change of variables for each of these integrals over[det(Dh(x))> 0] intersected with this ball.

The following is the Brouwer fixed point theorem for C2 maps.

Lemma 16.3.4 If h ∈C2(

B(0,R))

and h : B(0,R)→ B(0,R), then h has a fixed point, xsuch that h(x) = x.

16.3. SOMETHING WHICH IS NOT A RETRACT 431Proof: Suppose such an h exists. Let A € [0,1] and let py (x) =x+A (h(x) —x). Thisfunction, py, is a homotopy of the identity map and the retraction, h. LetAj= Lam (Dp, (x)) dx.,Then using the dominated convergence theorem,; _ 0 det (Dp; (x)) 9Paij (x)ma) = Irom OPRi,j oA_ O det ( (Dpg (x7 bam 1 ae reOPI, J~ =Now by assumption, /; (x) = x; on 0B(0,R) and so one can integrate by parts and write- Y ea Leot(Pp, (x));;,j (a(x) — xi) dx = 0.x)) (hj (x) — xj) ;dxLY cof (Dy (x));; (Mi (x) — xi) jaxijTherefore, J (A) equals a constant. However,1(0) =m, (2(0.2)) >0but1(1)= | det (Dh(x)) dm, = #(y)dim, =0JB(0,R) 2B(0,R)because from polar coordinates or other elementary reasoning, m, (0B(0,1))=0. IThe last formula uses the change of variables formula for functions which are not oneto one. In this formula, #(y) equals the number of x such that h(x) = y. To see this is soin case you have not seen this, note that h is C ! and so the inverse function theorem fromadvanced calculus applies. Thus[ det(Dh(x))dm, = det (Dh (x)) dm,JBOR)Fcon )>0]+f det (Dh (x)) dm,[det(Dh(x)) <0]Thus h is locally one to one on the two open sets [det (Dh(x)) > 0], [det (Dh (x)) < 0].Now use inverse function theorem and change of variables for one to one h to verify thatboth of these integrals equal 0. You cover [det (Dh(x)) > 0] with countably many ballson which h is one to one and then use change of variables for each of these integrals over[det (Dh (x)) > 0] intersected with this ball.The following is the Brouwer fixed point theorem for C* maps.Lemma 16.3.4 [fh € C2 (B (0,R)) and h: B(0,R) —> B(0,R), then h has a fixed point, xsuch that h(x) =x.