Download Math 131 Homework 7 Solutions

Math 131 Homework 7 Solutions
Greg Parker, Wyatt Mackey, Christian Carrick
November 3, 2015
Problem 1 (Munkres 51.3)
A space X is called contractible if the identity map is nullhomotopic. That is, if there is a homotopy of maps
H(x, t) : X ⇥ I ! X
such that H( , 1) = idX ( ) and H( , 0) = c for some fixed c 2 X.
Note that since homotopy is an equivalence relation, this is equivalent to having H( , 1) = c and
H( , 0) = idX .
(a)
I = [0, 1] and R are contractible.
Proof. Consider the map I ⇥ I ! I defined by
H(x, t) = xt
This is obviously continuous as a map on the square. At t = 1 we have H(x, 1) = x, and at t = 0 we
have H(x, 0) = 0. Thus this is a homotopy of maps between the the identity and the constant map taking
I to 0, and so I is contractible.
For R, we note the same map H(x, t) = xt suffices, as tit easily extends to a continuous map R ⇥ I ! R.
By the same argument as above, this is a homotopy from the identity map to the constant map at 0, so R
is contractible.
(b)
A contractible space X is path connected.
Proof. As path connectedness is an equivalence relation, it suffices to show each point x 2 X is connected
to a fixed c. By definition of contractibility, there is a homotopy
H(x, t) : X ⇥ I ! X
such that H(x, 1) = x and H(x, 0) = c for some fixed c. Fix x0 2 X. Then we can define a path
:I!X
(t) = H(x0 , t)
This map is continuous as it is the restriction of a continuous map to {x0 } ⇥ I. In addition, it must have
(1) = x0 and (0) = c as H is a homotopy from the identity to c. Therefore there is a path from every
x0 2 X to c, and the space is path connected.
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(c)
If Y is contractible, then for any X there is a single homotopy class of maps X ! Y .
Proof. As homotopy is an equivalence relation, it suffices to show every map is homotopic to a fixed map.
Yet Y is contractible, so there is a homotopy H(y, t) : Y ⇥ I ! Y that is idY at t = 1 and the constant map
at c 2 Y at t = 0. Therefore for some f : X ! Y the family of maps F (x, t) = H(f (x), t) is a homotopy
taking f to the constant map at c, as F (x, 1) = H(f (x), 1) = f (x) and F (x, 0) = H(f (x), 0) = c.
(d)
If X is contractible and Y is path connected, then there is a single homotopy class of maps X ! Y
Proof. First note that by path connectedness, any two constant maps are homotopic. As for constant maps
X ! {c0 } 2 Y and X ! {c1 } 2 Y , there is a homotopy H(x, t) = (t), where (t) is a path from c0 to c1 .
As homotopy is an equivalence relation, it suffices to show that every map is homotopic to a constant
map. Let f : X ! Y be a map, and let H(x, t) be a contraction on X. Define a family of maps
F (x, t) = f (H(x, t))
This is the composition of continuous maps, hence continuous. At t = 1 this is f (id(x)) = f (x) and at t = 0
this is f (c) for some fixed c, so is constant. Therefore this is a homotopy from the map f to a constant map
at f (c).
Problem 2
Let X be path connected, and h : X ! Y a continuous map. Say x0 , x1 2 X be points, and y0 = h(x0 ), y1 =
h(x1 ) are their images. If ↵ is a path from x to y and = h ↵ it image, which is a path from y0 to y1 .
Then
ˆ (hx )⇤ = (hx )⇤ ↵
ˆ
0
1
where (hz )⇤ : ⇡1 (X, z) ! ⇡1 (Y, h(z)) is the induced map of fundamental groups, and ˆ is the map as
defined on page 331.
Proof. It suffices to show a homotopy class [f ] 2 ⇡1 (X, x0 ) is taken to the same element under both sides of
the equation. The left hand side acts taking a class of loops [f ] to the image class, then attaching the path
along and back. Symbolically
ˆ (hx )⇤ [f ] = ˆ ([h f ]) = ⇤ [h f ] ⇤ [ ] = [h ↵
¯ ] ⇤ [h f ] ⇤ [h ↵]
(0.1)
0
where [ ] denotes the backwards parameterized path. Noting in the last step that the image of the backwards
parameterization is the backwards parameterization of the image (i.e. [h ↵
¯ ] = [h ↵]).
Alternatively, the right hand side acts on the class [f ] by
(hx1 )⇤ ↵
ˆ [f ] = (hx1 )⇤ ([ˆ
↵] ⇤ [f ] ⇤ [↵])
But noting that the concatenation of the images is the image of the concatenations, i.e. h([↵] ⇤ [ ]) =
[h ↵] ⇤ [h
], as at the bottom of page 333, the above becomes exactly (0.1) and the two sides agree.
Equivalently, we have shown the following diagram commutes:
⇡1 (X, x0 )
(hx0 )⇤
ˆ
↵
ˆ
⇡1 (X, x1 )
⇡1 (Y, y0 )
(hx1 )⇤
2
⇡1 (Y, y1 )
pg. 347, Problem 6. Consider the maps g, h : S 1 ! S 1 given by g(z) = z n and h(z) = z n .
Compute the induced group homomorphisms of ⇡1 (S 1 , b0 ).
We can represent the points of our circle as ei✓ , and WLOG take our baspoint to be 1.
Then since ⇡1 (S 1 , b0 ) = Z, we have g⇤ , h⇤ : Z ! Z. The only such group homomorphisms are
multiplication by d, for some d 2 Z, so it suffices to compute g⇤ (1), h⇤ (1) to find what the
group homomorphism is. Let [ ] be the homotopy class representing 1. Then g[ ] = ⇤ ... ⇤ ,
repeated n times, and h[ ] =
⇤ ... ⇤ , again repeated n times, now just with the opposite
orientation. Thus g⇤ is given by multiplication by n, and h⇤ is given by multiplication by
n.
pg. 347, Problem 7. Generalize the proof of Theorem 54.5 to show that the fundamental
group of a torus is isomorphic to Z ⇥ Z.
Let p ⇥ p : R ⇥ R ! S 1 ⇥ S 1 ⇠
= T 2 be the usual covering map, given by (x, y) 7!
(e2⇡ix , e2⇡iy ). Since R2 is simply connected, we have a bijection from the lifting correspondence
: ⇡1 (T 2 , x0 ) ! Z2 . We now must show that is a group homomorphism. This argument,
however, is unchanged from the proof of Theorem 54.5.
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1 Page 353 Problem 2
Show that if h : S 1 ! S 1 is nulhomotopic, then h has a fixed point, and h maps some point x to its antipode
°x.
A nulhomotopy on h is the same as a map B 2 ! S 1 that restricts to h on S 1 Ω B 2 . We thus have a continuous
map B 2 ! S 1 ,! B 2 , and the composition must have a fixed point by Brouwer’s theorem. But since the image
of the composition is contained in S 1 , h must have a fixed point.
Let f : S 1 ! S 1 be the antipodal map, then since h is nulhomotopic, so is f ± h (a nulhomotopy on f ± h is a
map from Cone(S 1 ) ª
= B 2 ! S 1 that restricts to f ± h on the base, but the nulhomotopy on h gives us a map
B 2 ! S 1 that restricts to h on the base, so we just postcompose that map with f ), and thus we have a map
B 2 ! S 1 ,! B 2 , which has a fixed point and thus f ± h has a fixed point, which is precisely a point x with
h(x) = °x.
⇤
2 Page 353 Problem 4
Solution by Eric Metodiev
Suppose that you are given the fact that for each n, there is no retraction r : B n+1 ! S n . Prove the following.
2.1
The identity map i : S n ! S n is not nulhomotopic.
We begin with a lemma which generalizes Lemma 55.3.
Lemma:
h : S n ! X is a continuous, nulhomotopic map iff h extends to a continuous map k : B n+1 ! X .
proof:
Let H : S n £ I ! X be a homotopy between h and a constant map c with H (x, 0) = h(x) and H (x, 1) = c. Let
º : S n £ I ! B n+1 be:
º(x, t ) = (1 ° t )x,
which for fixed t 2 [0, 1] contracts the sphere of radius 1 to a sphere of radius 1 ° t 2 [0, 1].
We note that º is continuous, closed, and surjective, so it is a quotient map. For points q 2 B n+1 aside from
0, º°1 (p) is a single point set. Similarly, º°1 (0) = S n £ {1} on which H is a constant function c. By Theorem
22.2, since H is a constant on the preimages of B n+1 under º as we have shown, we have that º induces a
continuous map k : B n+1 ! X such that k ± º = H . We note for x 2 S n that k(º(x, 0)) = H (x, 0) = h(x), so that
k extends the map h as they agree on the sphere.
Now suppose that h extends to a continuous map k : B n+1 ! X . Then let G = k ± º. We note that for x 2 S n
that G(x, 0) = k(º(x, 0)) = k(x) = h(x) as k agrees with h on the sphere and G(x, 1) = k(º(x, 1)) = k(0) which is
a constant. Thus G is a homotopy which shows that h is homotopic to a constant map, so h is nulhomotopic.
If the identity map i : S n ! S n were nulhomotopic, then by our lemma it extends to a continuous map k :
B n+1 ! S n . Then notice that k|S n is the identity of S n , so that k would be a retraction k : B n+1 ! S n , which is
a contradiction. Thus the identity map i : S n ! S n is not nulhomotopic.
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2.2
The inclusion map j : S n ! Rn+1 ° {0} is not nulhomotopic.
If the inclusion map j : S n ! Rn+1 ° {0} were nulhomotopic, then by our lemma it extends to a continuous
map k : B n+1 ! Rn+1 ° {0}. We construct a map ` : Rn+1 ° {0} ! S n that is the identity on S n , namely:
`(x) = p
x
x ·x
.
Then ` ± k : B n+1 ! S n is a map, where for x 2 S n we have ` ± k(x) = x as the extension k of the inclusion
map and the function ` are both the identity on S n . Thus ` ± k|S n is the identity of S n , so that ` ± k would
be a retraction ` ± k : B n+1 ! S n , which is a contradiction. Thus the inclusion map j : S n ! Rn+1 ° {0} is not
nulhomotopic.
2.3
Every nonvanishing vector field on B n+1 points directly outward at some point of S n , and directly inward
at some point of S n .
As the vector field is nonvanishing, we have that v(x) 6= 0 for every x 2 B n+1 . In such a cases, we have that
v : B n+1 ! Rn+1 ° {0}. Suppose for the sake of contradiction that v(x) does not point directly inward at any
point x of S n . Because v|S n extends to a map of B n+1 into Rn+1 ° {0}, we have that v|S n : S n ! Rn+1 ° {0} is
nulhomotopic by our lemma. On the other hand, consider the homotopy for x 2 S n :
F (x, t ) = t x + (1 ° t )v|S n (x).
For it to be a homotopy, we must never have F (x, t ) = 0. At t = 0 and t = 1, we have that x 6= 0 and v|S n 6= 0
by construction. If F (x, t ) = 0, then v|S n (x) = °xt /(1 ° t ) which would contradict our supposition that v(x)
does not point directly inward. Thus in fact we have that F (x, t ) : S n £ I ! Rn+1 ° {0}. This shows that v|S n is
homotopic to the inclusion map j : S n ! Rn+1 ° {0}. The result of Part B shows that the inclusion map, and
thus v|S n , is not nulhomotopic. This is a contradiction, so that the vectorfield must point inward at some
point of S n .
To show that v points directly outward at some point of S n , we apply the result we have just proven to the
vector field °v(x). As °v(x) must have a point on S n where it points directly inward, v(x) must point directly
outward at that point.
2.4
Every continuous map f : B n+1 ! B n+1 has a fixed point.
For the sake of contradiction, suppose that f (x) 6= x for every x 2 B n+1 . Then defining the vector field v(x) =
f (x) ° x which is nonvanishing since f (x) 6= x, we have by Part D that there must be a y 2 S n such that v(y)
points completely outward, namely v(y) = a y for a > 0. Thus:
v(y) = f (y) ° y = a y.
Thus means that f (y) = (1+ a)y which is outside of B n+1 , which is a contradiction as f : B n+1 ! B n+1 . Therefore there must be some x 2 B n+1 such that f (x) = x.
2.5
Every n + 1 by n + 1 matrix with positive real entries has positive eigenvalue.
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We generalize the proof of Theorem 55.7. Consider a n +1 by n +1 matrix A. Let T : Rn+1 ! Rn+1 be the linear
transformation whose matrix is A. Let B be the intersection of S n with the positive part of B n+1 :
B = {(x 1 , · · · , x n+1 ) : x 1 ∏ 0, · · · , x n+1 ∏ 0}.
We can see that by projecting and stretching, B is homeomorphic to the ball B n . Thus we can apply the fixedpoint theorem of Part D for continuous maps from B to itself by making use of the homeomorphism and its
inverse.
A vector x 2 B has at least one positive component and the rest non-negative. As all the entries of A are
positive, the vector T (x) is a vector with positive components. Consider the map f : B n ! B n defined by:
T (x)
f (x) = p
,
T (x) · T (x)
which therefore has a fixed point x 0 .
p
p
Then we have that T (x 0 ) = x 0 T (x 0 ) · T (x 0 ) = ∏x 0 for ∏ ¥ T (x 0 ) · T (x 0 ) > 0. Thus we have that A has positive
eigenvalue ∏. Therefore every n + 1 by n + 1 matrix with positive real entries has positive eigenvalue.
2.6
If h : S n ! S n is nulhomotopic, then h has a fixed point and h maps some point x to its antipode °x.
As h is nulhomotopic, our lemma implies that it can be extended to a continuous map k : B n+1 ! S n . Then
composing with the inclusion map j : S n ! B n+1 yields a map j ± k : B n+1 ! B n+1 . The fixed-point theorem
of Part D implies that there exists a point x 2 B n+1 such that j (k(x)) = x. As j is the identity on S n Ω B n+1 , we
therefore have that there is an x 2 S n such that j (k(x)) = k(x) = h(x) = x, where we have used the fact that j
is an inclusion for points in S n and k equals h when restricted to S n . Thus we have shown that h must have
a fixed point.
Consider the map r : S n ! S n which reflects points about the origin to their antipodes. As h is nulhomotopic,
there exists some homotopy F : S n £ I ! S n with F (x, 0) = h(x) and F (x, 1) = c for some c 2 S n . Note that
G(x, t ) = r (F (x, t )) is a continuous function with G(x, 0) = r (F (x, 0)) = r (h(x)) = °x and G(x, 1) = r (F (x, 0)) =
r (c) = °c, which is a homotopy that shows that r ± h is homotopic to the constant map °c. Thus r ± h is a
nulhomotopic so, by the above argument, it has a fixed point y with r ±h(y) = y. Since r ±r is the identity, this
shows that h(y) = r ± r ± h(y) = r ± y = °y. Thus there is also a point y such that h maps y to its antipode °y.
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