Hint. For the first part, consider a map F t : S 1 × I → S 1 × I by F t (θ, s) = (θ + 2πst, s). For the second part, consider projection p : S 1 × I → S 1 and the path s 7→ (θ 0 , s).

USTC, School of Mathematical Sciences Winter semester 2018/19 Algebraic topology by Prof. Mao Sheng Hint to exercise sheet 2 MA04311 Tutor: Lihao Huang, Han Wu Posted by Dr. Muxi Li Ex 1. (1 pt) Define f : S ¹ × I → S ¹ × I by f (θ, s) = (θ + 2πs, s), so f restricts to identity on the two bundary circles of S ¹ × I. Show that f is homotopic to the identity by a homotopy f _t that is stationary on one of the boundary circles, but not by any homotopy f _t that is stationary on both boundary circles. [Consider what f does to the path s 7→ (θ ₀ , s) for fixed θ ₀ ∈ S ¹ .]

Hint. For the first part, consider a map F _t : S ¹ × I → S ¹ × I by F _t (θ, s) = (θ + 2πst, s). For the second part, consider projection p : S ¹ × I → S ¹ and the path s 7→ (θ ₀ , s).

Ex 2. (1 pt) Does the Borsuk-Ulam theorem hold for the torus? In other words, for every map f : S ¹ × S ¹ → R ² must there exist (x, y) ∈ S ¹ × S ¹ such that f (x, y) = f (−x, −y)?

Hint. Consider the projection p ₁ : S ¹ × S ¹ → S ¹ by p ₁ (s 1 , s 2 ) = s 1 and the natural imbedding i : S ¹ ,→ R ² . Let f = i ◦ p ₁ , then Borsuk-Ulam theorem doesn’t hold in this case.

Ex 3. (1 pt) Let A ₁ , A ₂ , A ₃ be compact sets in R ³ . Use the Borsuk-Ulam theorem to show that there is one plane P ⊂ R ³ that simultaneously divides each A _i into two pieces of equal measure.

Hint. Method 1, take s ∈ S ² ⊂ R ³ , then ∃ ! one plane P ₁ ^s in R ³ with normal vector − →

0s such that P ₁ ^s divides A ₁ into two pieces of equal measure. Take p s ∈ P ₁ ^s , then define P s = {v ∈ R ³ |−→ vp s · − →

0s ≥ 0} (note: this is independent of the choice of p _s ). Let f 1 (s) (resp. f 2 (s)) be the measure of P s ∩ A ₂ (resp.

P _s ∩ A ₃ ). In this way, we get a map f : S ² → R ² by f (s) = (f ₁ (s), f ₂ (s)).

By the Borsuk-Ulam theorem, we get a s ₀ ∈ S ² such that f (s ₀ ) = f (−s 0 ), then P ₁ ^s

is just the plane we want.

Method 2, using the Borsuk-Ulam theorem for maps S ³ → R ³ .

Take s = (s ₁ , s 2 , s 3 , s 4 ) ∈ S ³ ⊂ R ⁴ , then consider P s = {(x, y, z) ∈ R ³ |xs ₁ + ys 2 + zs 3 + s 4 ≥ 0}. Let f ₁ (s) (resp. f 2 (s), f 3 (s)) be the measure of P s ∩ A ₁ (resp. P _s ∩ A ₂ , P _s ∩ A ₃ ). In this way, we get a map f : S ³ → R ³ by f (s) = (f 1 (s), f 2 (s), f 3 (s)). By the Borsuk-Ulam theorem, we get a v ∈ S ³ such that f (v) = f (−v). For (v 1 , v 2 , v 3 ) ∈ R ³ \{0}, the plane xv ₁ + yv 2 + zv 3 + v 4 = 0 is just what we want.

Ex 4. (1 pt) From the isomorphism π ₁ (X × Y, (x 0 , y 0 )) ≈ π 1 (X, x 0 ) × π ₁ (Y, y ₀ ) it follows that loops in X × {y ₀ } and {x ₀ } × Y represent commuting elements of π ₁ (X × Y, (x 0 , y 0 )). Construct an explicit homotopy demonstrat- ing this.

1

Hint. Let [f ], [g] be elements in π ₁ (X, x ₀ ), π ₁ (Y, y ₀ ) respectively. We need to construct an explicit homotopy in X × Y from f · g to g · f with base point (x 0 , y 0 ). By definition

f · g(s) =

( (f (2s), y ₀ ) for 0 ≤ s ≤ 1/2 (x 0 , g(2s − 1)) for 1/2 < s ≤ 1

g · f (s) =

( (x ₀ , g(2s)) for 0 ≤ s ≤ 1/2 (f (2s − 1), y 0 ) for 1/2 < s ≤ 1 Let

F 1t (s) =



 

 

x ₀ for 0 ≤ s ≤ t/2

f (2s − t) for t/2 < s ≤ (t + 1)/2 x 0 for (t + 1)/2 < s ≤ 2

F _2t (s) =



 

 

y ₀ for 0 ≤ s ≤ (1 − t)/2

g(2s + t − 1) for (1 − t)/2 < s ≤ (2 − t)/2 y ₀ for (2 − t)/2 < s ≤ 1

.

Then F (t, s) : I ×I → X ×Y given by F (t, s) = (F _1t (s), F _2t (s)) is a homotopy from f · g to g · f with base point (x ₀ , y ₀ ).

Alternatively, we can construct an explicit homotopy in X × Y from f · g to (f, g) with base point (x ₀ , y ₀ ). Let

F 1t (s) =

( f (2s/(1 + t)) for 0 ≤ s ≤ (1 + t)/2 x 0 for (1 + t)/2 < s ≤ 1

F 2t (s) =

( y ₀ for 0 ≤ s ≤ (1 − t)/2

g(2(s − 1)/(1 + t) + 1) for (1 − t)/2 < s ≤ 1

Then F (t, s) : I ×I → X ×Y given by F (t, s) = (F _1t (s), F 2t (s)) is a homotopy from f · g to (f, g) with base point (x ₀ , y 0 ).

Similarly, we can construct an explicit homotopy in X × Y from g · f to (f, g) with base point (x ₀ , y 0 ).

Ex 5. (3 pts) Show that there are no retractions r : X → A in the following cases:

(a) X = R ³ with A any subspace homeomorphic to S ¹ . (b) X = S ¹ × D ² with A its boundary torus S ¹ × S ¹ . (c) X = S ¹ × D ² and A the circle shown in the figure.

(d) X = D ² ∨ D ² with A its boundary S ¹ ∨ S ¹ .

(e) X a disk with two points on its boundary identified and A its boundary S ¹ ∨ S ¹ . (f ) X the Möbius band and A its boundary circle.

2

Hint. If there exists retraction r : X → A, then the inclusion i : A → X induces an isomorphism i ∗ : π 1 (A) → π 1 (X).

(a) π ₁ (A) = Z, π 1 (X) = 0.

(b) π ₁ (A) = Z × Z, π 1 (X) = Z.

(c) π ₁ (A) = Z, π 1 (X) = Z, i ∗ = 0 (d) π ₁ (A) = Z ∗ Z, π 1 (X) = 0.

(e) π ₁ (A) = Z ∗ Z, π 1 (X) = Z.

(f ) π ₁ (A) = Z, π 1 (X) = Z, i ∗ = 2 × .

Ex 6. (2 pts) Using the technique in the proof of Proposition 1.14, show that if a space X is obtained from a path-connected subspace A by attaching a cell e ⁿ with n ≥ 2, then the inclusion A ,→ X induces a surjection on π ₁ . Apply this to show:

(a) The wedge sum S ¹ ∨ S ² has fundamental group Z.

(b) For a path-connected CW complex X the inclusion map X ¹ ,→ X of its 1-skeleton induces a surjection π ₁ (X ¹ ) → π 1 (X). [For the case that X has infinitely many cells, see Proposition A.1 in Appendix.]

Hint. The same as in the proof of Proposition 1.14, every path in X, is homotopy to a path in A.

(a) Consider the natural inclusion S ¹ ,→ S ¹ ∨ S ² ,→ S ¹ ∨ D ³ , the second injection using ∂(D ³ ) = S ² .

(b) If dim (X) < ∞, inductively use the result we get.

If dim (X) = ∞, consider a path f in X. By compactness of f (I), this f (I) ⊂ X ⁿ for some n.

3