Homework 2 - Brown University

Brown University
Math 2410 - Algebraic Topology
September 23, 2014
Professor Sara Maloni
Problem 2. Show that change-of-basepoint homomorphism βh depends only on the homotopy class of h.
Solution. Suppose that h and h0 are two homotopic paths from x0 to x1 . We would like to prove that
βh [f ] = βh0 [f ] for every loop f based at x1 . By definition, βh [f ] = [hf h] where h is inverse path to h.
Lemma. Suppose that h ∼
= h0 .
= h0 . Then h ∼
Proof. By assumption, there exists a continuous map F : I×I → X such that F (s, 0) = h(s), F (s, 1) = h0 (s)
and F (0, t) = x0 , F (1, t) = x1 for every s, t ∈ I. Define G : I × I → X by G(s, t) = F (1 − s, t). Then,
G(s, 0) = h(1 − s) = h(s), G(s, 1) = h0 (1 − s, 1) = h0 (s), G(0, t) = x1 , G(1, t) = x0 for all s, t ∈ I. So h ∼
= h0 .
We will now use a general fact proved in Homework 1, namely: if f1 ' f2 and g1 ' g2 , then f1 · g1 ' f2 · g2 .
So h ' h0 together with f ' f implies h · f ' h0 · f , which together with h ' h0 (from Lemma) implies
h · f · h ' h0 · f · h0 . Therefore, βh [f ] = βh0 [f ] and this completes the proof.
Problem 3. For a path-connected space X, show that π1 (X) is abelian iff all basepoint-change homomorphisms βh depend only on the endpoints of the path h.
Solution. (⇒) Assume that π1 (X) is abelian. Suppose that h and h0 are any two paths from x0 to x1 . We
would like to prove that βh [f ] = βh0 [f ] for every loop f based at x1 . Note that p1 = [hf h0 ] ∈ π1 (X, x0 ) and
p2 = [h0 h] ∈ π1 (X, x0 ). Since π(X, x0 ) is abelian, we get
p1 p2 = p2 p1 =⇒ [hf h0 ] · [h0 h] = [h0 h] · [hf h0 ] =⇒ [hf h0 h0 h] = [h0 h · hf h0 ] =⇒ [hf h] = [h0 f h0 ]
Since βh [f ] = [hf h], we conclude that βh [f ] = βh0 [f ]. Hence, βh only depends on the endpoints of h.
(⇐) Assume that all basepoint-change homomorphisms βh depend only on the endpoints of the path h.
Since X is simply-connected, the fundamental group is independent of the basepoint. So it suffices to
show π(X, x0 ) is abelian for a single x0 ∈ X. Fix some x0 ∈ X, and consider [g1 ], [g2 ] ∈ π1 (X, x0 ) where
g1 and g2 are two loops based at x0 . By the hypothesis, βg1 [f ] = βg2 [f ] for any loop f based at x0 ,
because g1 and g2 share the same endpoints (namely x0 ). In particular, βg1 [g1 ] = βg2 [g1 ] or equivalently,
[g1 g1 g1 ] = [g2 g1 g2 ] =⇒ [g1 ] = [g2 g1 g2 ] =⇒ [g1 ] · [g2 ] = [g2 g1 g2 ] · [g2 ] = [g2 g1 g2 g2 ] =⇒ [g1 · g2 ] = [g2 ] · [g1 ]
Thus, [g1 · g2 ] = [g2 ] · [g1 ] and π1 (X, x0 ) is abelian.
Problem 5. Show that for a space X, the following three conditions are equivalent:
(a) Every map S 1 → X is homotopic to a constant map, with image a point.
(b) Every map S 1 → X extends to a map D2 → X.
(c) π1 (X, x0 ) = 0 for all x0 ∈ X.
Deduce that a space X is simply-connected iff all maps S 1 → X are homotopic. [In this problem, ‘homotopic’
means ‘homotopic without regard to basepoints’]
Solution. (a)⇒(b) Assume that every map S 1 → X is homotopic to a constant map. Let f : S 1 → X be
any map. By hypothesis, there exists a continuous map F : S 1 × I → X with F (s, 0) = f (s) and F (s, 1) = x0
for all s ∈ S 1 where x0 ∈ X is some point. We can realize S 1 and D2 as subsets of the complex plane:
S 1 = {z ∈ C : |z| = 1} and D2 = {z ∈ C : |z| < 1}. We will construct a continuous map from the closed disk
∼
D2 = {z ∈ C : |z| ≤ 1} to X that extends f . Consider the map f : D2 → X given by
∼
f (re2πiθ ) = F (e2πiθ , 1 − r)
for 0 ≤ r ≤ 1, 0 ≤ θ < 1
This definition is valid because every point z ∈ D2 with z 6= 0 can be represented uniquely as z = re2πiθ
for some 0 < r ≤ 1 and 0 ≤ θ < 1. But for z = 0, we have infinitely many representations,
since
∼
0 = re2πiθ where r = 0 and θ can be anything. Fortunately, F (e2πiθ , 1) = x0 for all θ, and so f is well∼
defined
at
z
=
0.
As
a
composition
of
continuous
functions,
f
is continuous. Finally, when r ∼= 1, we get
∼
∼
2πiθ
2πiθ
2πiθ
f (e
) = F (e
, 0) = f (e
) for each 0 ≤ θ < 1, and so f agrees with f on S 1 . Thus, f : D2 → X
1
extends f : S → X.
1
(b)⇒(c) Assume that every map S 1 → X extends to a map D2 → X. Let x0 ∈ X and γ be any loop in X
with base-point at x0 . We will show that γ is path homotopic to the constant loop at x0 . Since γ is a loop
(i.e. γ(0) = γ(1) = x0 ), it defines a function γ : S 1 → X where γ(s0 ) = x0 for some s0 ∈ S 1 . By hypothesis,
∼
γ extends to a map γ : D2 → X. It is clear that D2 deformation retracts onto s0 ∈ S 1 ; so there exists a
map H : D2 × I → D2 such that H(s, 0) = idD2 , H(s, 1) = s0 , H(s0 , t) = s0 for all s ∈ D2 , t ∈ I. Hence we
∼
∼
∼
∼
get a map G = γ ◦ H : D2 × I → X with H(s, 0) = γ(s), H(s, 1) = γ(s0 ) = x0 , H(s0 , t) = γ(s0 ) = x0 for all
s ∈ D2 , t ∈ I. Now, restrict G to S 1 × I to get map F = G|S 1 ×I : S 1 × I → X. It follows that F (s, 0) = γ(s),
F (s, 1) = x0 , F (s0 , t) = x0 for all s ∈ S 1 , t ∈ I. So F is a path homotopy from γ to the constant loop at x0 .
This proves that π(X, x0 ) is trivial for all x0 ∈ X.
(c)⇒(a) Assume that π1 (X, x0 ) = 0 for all x0 ∈ X. Consider any map f : S 1 → X. Fix any s0 ∈ S 1 . Then
f is a loop in X based at x0 := f (s0 ). Since π1 (X, x0 ) is trivial, it follows that f is homotopic (in fact path
homotopic!) to the constant loop at x0 .
Suppose that X is simply connected. By definition, X is path-connected and π(X, x0 ) = 0 for all x0 ∈ X,
which is (c) above. So (a) holds, which means that every map S 1 → X are homotopic to a constant map. But
X is path-connected, and so all the constant maps are homotopic. Thus, all maps S 1 → X are homotopic.
Conversely, suppose all maps S 1 → X are homotopic. In particular, the condition (a) holds, which means
that (c) holds. In addition, all the constant maps are homotopic (because constant maps are also maps
S 1 → X), so that X is path-connected. Condition (c) + “path-connectedness” = X is simply-connected. Problem 6. We can regard π1 (X, x0 ) as the set of basepoint-preserving homotopy classes of maps (S 1 , s0 ) →
(X, x0 ). Let [S 1 , X] be the set of homotopy classes of maps S 1 → X with no condition on basepoints. Thus
there is a natural map Φ : π1 (X, x0 ) → [S 1 , X] obtained by ignoring basepoints. Show that Φ is onto if X is
path-connected, and that Φ([f ]) = Φ([g]) iff [f ] and [g] are conjugate in π1 (X, x0 ). Hence, Φ induces a oneto-one correspondence between [S 1 , X] and the set of conjugacy classes in π1 (X), when X is path-connected.
Solution. Let [γ] ∈ [S 1 , X]. Pick any x1 ∈ γ(S 1 ). Since X is path-connected, there exists a path h : I → X
connecting x0 to x1 . Then βh [γ] := [hγh] ∈ π1 (X, x0 ). We claim that hγh is homotopic γ (in the weak sense
of not fixing basepoints). Define
(
h(s + t) if 0 ≤ s ≤ 1 − t
ht (s) =
h(1)
if 1 − t ≤ s ≤ 1
Note that h0 = h and ht = h(1) = x1 is the constant map. Set ψt := ht γht . Since ψ0 = hγh and ψ1 = γ, we
see that ψt is a desired homotopy between hγh and γ. Hence, Φ([hγh]) = γ and so Φ is onto.
Now suppose that [f ] and [g] are conjugate in π1 (X, x0 ). Then there exists [h] ∈ π1 (X, x0 ) such that f = hgh.
Once again, we define:
(
h(s + t) if 0 ≤ s ≤ 1 − t
ht (s) =
h(1)
if 1 − t ≤ s ≤ 1
And ψt := ht ght for each t ∈ I. Then ψ0 = hgh = f and ψ1 = g; thus, ψt is a homotopy (in the weak sense
of not fixing basepoints) from f to g. This proves Φ([f ]) = Φ([g]).
Next suppose that Φ([f ]) = Φ([g]). So there exists a homotopy F (in the weak sense of not fixing basepoints)
from f to g. More precisely, there is a continuous map F : I × I → X where F (s, 0) = f (s) and F (s, 1) = g(s)
for all s ∈ I. Recall that f (0) = f (1) = g(0) = g(1) = x0 as [f ], [g] ∈ π1 (X, x0 ). Let h(t) = F (0, t) = F (1, t);
note that h : I → X and h(0) = h(1) = x0 , so that [h] ∈ π1 (X, x0 ). We claim that [f ] = [hgh]. Define
H : I × I → X by:


if 0 ≤ s ≤ (2t)/(1 + 3t)
h(s(1 + 3t)/2)
H(s, t) = F (s(1 + 3t) − 2t, t) if (2t)/(1 + 3t) ≤ s ≤ (1 + 2t)/(1 + 3t)


h(s(1 + 3t) − 3t)
if (1 + 2t)/(1 + 3t) ≤ s ≤ 1
2
We claim that H provides path-homotopy between f and hgh.
• First, let’s show that H is continuous. Since H is a piecewise function of continuous functions, it suffices to
check that H(s, t) is continuous at the “boundary” points, namely at s = (2t)/(1+3t) and s = (1+2t)/(1+3t).
At s = (2t)/(1 + 3t), we have
1 + 3t
2t
·
= h(t)
h(s(1 + 3t)/2) = h
1 + 3t
2
2t
F (s(1 + 3t) − 2t, t) = F
· (1 + 3t) − 2t, t = F (0, t) = h(t)
1 + 3t
So h(s(1 + 3t)/2) and F (s(1 + 3t) − 2t, t) agree at s = (2t)/(1 + 3t). Similarly, at s = (1 + 2t)/(1 + 3t),
1 + 2t
F (s(1 + 3t) − 2t, t) = F
· (1 + 3t) − 2t, t = F (1, t) = h(t)
1 + 3t
1 + 2t
h(s(1 + 3t) − 3t) = h
· (1 + 3t) − 3t = h(1 + 2t − 3t) = h(1 − t) = h(t)
1 + 3t
So F (s(1 + 3t) − 2t, t) and h(s(1 + 3t) − 3t) agree at s = (1 + 2t)/(1 + 3t). Thus, H(s, t) is continuous.
• Next, we need to show that homotopy fixes the endpoint x0 for all time t, i.e. we need to show that
H(0, t) = H(1, t) = x0 for each t ∈ I. This is clear from the definition of H; indeed, H(0, t) = h(0·(1+3t)/2) =
h(0) = x0 and H(1, t) = h(1 · (1 + 3t) − 3t) = h(1) = h(0) = x0 , as desired.
• Finally, let’s verify that H(s, 0) = f (s) and H(s, 1) = g(s) for all s ∈ I. The former is easily established
from the definition:


if 0 ≤ s ≤ (2 · 0)/(1 + 3 · 0)
h(s(1 + 3 · 0)/2)
H(s, 0) = F (s(1 + 3 · 0) − 2 · 0, 0) if (2 · 0)/(1 + 3 · 0) ≤ s ≤ (1 + 2 · 0)/(1 + 3 · 0)


h(s(1 + 3 · 0) − 3 · 0)
if (1 + 2 · 0)/(1 + 3 · 0) ≤ s ≤ 1
which is simply F (s, 0) = f (s) on the interval 0 ≤ s ≤ 1. As for the latter, recall that for two paths γ1 , γ2 ,
we concatenated them as follows:
(
γ1 (2s)
if 0 ≤ s ≤ 1/2
γ1 · γ2 (s) =
γ2 (2s − 1) if 1/2 ≤ s ≤ 1
Using this procedure twice, h · (g · h) can be written as:


if 0 ≤ s ≤ 1/2
h(2s)
(*)
h · (g · h)(s) = g(4s − 2) if 1/2 ≤ s ≤ 3/4


h(4s − 3) if 3/4 ≤ s ≤ 1
Looking at the defining equation for H, we see that indeed H(s, 1) has the same definition as (∗):


if 0 ≤ s ≤ (2 · 1)/(1 + 3 · 1)
h(s(1 + 3 · 1)/2)
H(s, 1) = F (s(1 + 3 · 1) − 2 · 1, 1) if (2 · 1)/(1 + 3 · 1) ≤ s ≤ (1 + 2 · 1)/(1 + 3 · 1)


h(s(1 + 3 · 1) − 3 · 1)
if (1 + 2 · 0)/(1 + 3 · 0) ≤ s ≤ 1




if 0 ≤ s ≤ 1/2
if 0 ≤ s ≤ 1/2
h(2s)
h(2s)
= F (4s − 2, 1) if 1/2 ≤ s ≤ 3/4 = g(4s − 2) if 1/2 ≤ s ≤ 3/4




h(4s − 3)
if 3/4 ≤ s ≤ 1
h(4s − 3) if 3/4 ≤ s ≤ 1
Thus, H(s, 1) = hgh(s). This completes the proof that H is a path-homotopy from f to hgh. As a result,
[f ] = [hgh] implying that [f ] and [g] are conjugate in the fundamental group π1 (X, x0 ).
Conclusion. The map Φ induces one-to-one correspondence between the conjugacy classes in π1 (X) and
the homotopy classes (in the weak sense of not fixing basepoints) in [S 1 , X].
3
Problem 7. Define f : S 1 × I → S 1 × I by f (θ, s) = (θ + 2πs, s), so f restricts to the identity on the two
boundary circles of S 1 × I. Show that f is homotopic to the identity by a homotopy ft that is stationary
on one of the boundary circles, but not by any homotopy ft that is stationary on both boundary circles.
[Consider what f does to the path s 7→ (θ0 , s) for fixed θ0 ∈ S 1 ].
Solution. Define ft : S 1 × I → S 1 × I by ft (θ, s) = (θ + 2πs(1 − t), s). Then f0 = f , and f1 = identity map
on S 1 × I; furthermore, ft (θ, 0) = (θ, 0) for all θ. Hence, ft is a homotopy from f to the identity, and ft is
stationary on one of the boundary circle, namely on S 1 × {0}.
Next, we will show that f is not homotopic to the identity by any homotopy ft that is stationary on both
boundary circles. Assume, to the contrary, that such a homotopy ft : S 1 × I → S 1 × I exists. Fix θ0 , and
observe that γ(s) := f (θ0 , s) : I → S 1 × I is given by s 7→ (θ0 + 2πs, s), and so the image of γ “winds” once
around the cylinder from (θ0 , 0) to (θ0 , 1), as depicted below.
Let p : S 1 ×I → S 1 be the projection map given by (θ, s) 7→ θ. Now observe that gt (s) := p(ft (θ0 , s)) : I → S 1
is a homotopy from g0 (s) = p(f0 (θ0 , s)) = p(γ(s)) to the constant loop (Important: gt is a path homotopy
– this is precisely because ft is a homotopy that fixes the boundary circles). But g0 (s) = p(γ(s)) is a loop
going once around the circle; in Hatcher’s notation, g0 = ω1 . This means ω1 is homotopic to the constant
loop, which contradicts the fact that [ω1 ] = 1 is a non-trivial element of π1 (S 1 ) = Z.
Problem 10. From the isomorphism π1 (X × Y, (x0 , y0 )) ∼
= π1 (X, x0 ) × π2 (Y, y0 ) it follows that loops in
X × {y0 } and {x0 } × Y represent commuting elements of π1 (X × Y, (x0 , y0 )). Construct an explicit homotopy
demonstrating this.
Solution. Suppose we have two loops f : I → X × {y0 } and g : I → {x0 } × Y where f (0) = f (1) = (x0 , y0 ),
g(0) = g(1) = (x0 , y0 ). We need to construct an explicit homotopy from f · g to g · f . First, for each
t ∈ I, define ft (s) : I → X × Y by ft (s) = (p1 (f (s)), p2 (g(t))) where p1 : X × Y → X, (x, y) 7→ x, and
p2 : X × Y → Y , (x, y) 7→ y are the projection maps. Intuitively, ft is a loop just like f , except the second
coordinate of ft is fixed at p2 (g(t)), while the second coordinate of f is fixed at y0 . It is clear from the
definition that f0 = f1 = f . Now define ϕt : I → X × Y by


if 0 ≤ s ≤ t/2
g(2s)
ϕt (s) = ft (2s − t) if t/2 ≤ s ≤ (t + 1)/2


g(2s − 1) if (t + 1)/2 ≤ s ≤ 1
We claim that ϕt provides the desired homotopy from f · g to g · f . First, ϕt is continuous: to check this, it
suffices to show that ϕt is continuous at the “boundary” points s = t/2 and s = (t + 1)/2.
• At s = t/2, g(2s) = g(t) which agrees with ft (2s − t) = ft (0) = (p1 (f (0)), p2 (g(t))) = (x0 , p2 (g(t))) = g(t).
• At s = (t + 1)/2, ft (2s − t) = ft (1) = (p1 (f (1)), p2 (g(t))) = (x0 , p2 (g(t))) = g(t) which agrees with
g(2s − 1) = g(2[(t + 1)/2] − 1) = g(t).
Next, we show
( that ϕt is indeed a homotopy between the loops f · g and g · f .
f0 (2s)
if 0 ≤ s ≤ 1/2
• ϕ0 (s) =
, which is exactly (f · g)(s), since f0 = f . Thus, ϕ0 = f · g.
g(2s − 1) if 1/2 ≤ s ≤ 1
(
g(2s)
if 0 ≤ s ≤ 1/2
• ϕ0 (s) =
, which is exactly (g · f )(s), since f1 = f . Thus, ϕ1 = g · f .
f1 (2s − 1) if 1/2 ≤ s ≤ 1
Finally, let’s check that ϕt fixes the endpoint (x0 , y0 ) for all t ∈ [0, 1]. By definition, ϕt (0) = g(2·0) = (x0 , y0 ),
and ϕt (1) = g(2 · 1 − 1) = g(1) = (x0 , y0 ).
Conclusion: ϕt is a path homotopy from f · g to g · f . Hence, in the fundamental group π1 (X × Y, x0 × y0 ),
the elements [f ] and [g] commutes, [f ] · [g] = [g] · [f ].
4

Download Report