Exercise 1. (A. Hatcher, Algebraic Topology, p. 38, 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 first that π1 (X) is abelian, i.e. π1 (X, x0 ) is abelian, for all x0 ∈ X. We shall
prove that basepoint-change homomorphisms induced by paths depend only on the
endpoints of those paths. Indeed, let x0 , x1 ∈ X be arbitrary and let f, g : [0, 1] → X
be arbitrary paths such that f (0) = g(0) = x0 and f (1) = g(1) = x1 . Then, it is
obvious that [f ] ? [ḡ] ∈ π1 (X, x0 ). Thus, for every loop h in X based at x0 , we have
[f ] ? [ḡ] ? [h] = [h] ? [f ] ? [ḡ],
which gives immediately
[ḡ] ? [h] ? [g] = [f¯] ? [h] ? [f ],
i.e. βg ([h]) = βf ([h]). Thus βg = βf . Conversely, assume that basepoint-change
homomorphisms induced by paths depend only on the endpoints of those paths. We
shall prove that π1 (X) is abelian, i.e. that for all x0 ∈ X, the group π1 (X, x0 ) is
abelian. Indeed, let x0 ∈ X be arbitrary. Let f, g be arbitrary loops in X based
at x0 . Then, by assumption we have βf = βg , thus βf ([f ]) = βg ([f ]), which gives
immediately [f ] ? [g] = [g] ? [f ]. Thus, π1 (X, x0 ) is abelian.
Exercise 2. (A. Hatcher, Algebraic Topology, p. 38, 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 conditions 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 one-to-one correspondence
between [S 1 , X] and the set of conjugacy classes in π1 (X), when X is path-connected.
Solution
Let q : I → S 1 be the quotient map such that q(0) = q(1) = s0 . Then if f : I → X is
a loop in X based at x0 , f is constant on the fibers of q. We denote by fe the unique
map that makes the following diagram commute.
I
f
q
fe
/X
S1
If g 'p f , let H be the path homotopy between g and f . Then H is constant on the
fibers of q × id : I × I → S 1 × I. q × id is indeed a quotient map since I is locally
e such that the following diagram commutes and it
compact. Therefore there exists H
e is a basepoint-preserving homotopy between ge and fe.
is clear that H
I ×I
H
q×id
S1 × I
e
H
1
/
#
X
We can therefore regard π1 (X, x0 ) as the set of basepoint-preserving homotopy classes
of maps (S 1 , s0 ) → (X, x0 ).
If f is a loop in X based at x0 in X, we shall denote by [f ]p the path homotopy class
of f in π1 (X, x0 ) and by [f ] its homotopy class in [S 1 , X].
Let [g] ∈ [S 1 , X]. If x1 = g(s0 ), then since X is path connected, there exists a path h
in X from x0 to x1 . For each t in I we define ht : I → X by ht (s) = h(s + (1 − s)t).
Then H : I × I → X defined by
H(s, t) = ht ? g ◦ q ? ht (s, t)
(1)
is a homotopy between h ? g ? h and ex1 ? g ? ex1 . Notice that H(0, t) = H(1, t) = h(t).
e : S 1 × I → X and we
Therefore H descends to the quotient to give a homotopy H
have Φ([h ? g ? h]p ) = [g] which proves the surjectivity of Φ.
Suppose [f ]p = [h]p ? [g]p ? [h]p where f ,g and h are loops in X. We have the same
homotopy from the first part of the question with x1 = x0 and hence [f ] = [g]. Now
suppose [f ] and [g] are loops such that Φ([f ]p ) = Φ([g]p ), that is [f ] = [g]. Then there
exists a homotopy H between f and g. Define h(s) = H(s0 , s), the path followed
by x0 under the homotopy. h is a loop because h(1) = H(s0 , 1) = g(s0 ) = x0 . Let
ht (s) = h(ts) for every t in I. Then we can define a path homotopy
F (s, t) = ht ? H ◦ (q × id) ? ht (s, t)
(2)
between ex0 ? f ? ex0 and h ? g ? h. Note that F (0, t) = F (1, t) = x0 . Hence we have
shown that [f ]p = [h]p ? [g]p ? [h]p .
Exercise 3. (A. Hatcher, Algebraic Topology, p. 38, 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
We consider the function f : S 1 × I → S 1 × I given by
f (z, s) = (ze2πis , s), ∀(z, s) ∈ S 1 × I.
Obviously, the map F : S 1 × I × I → S 1 × I given by
F (z, s, t) = (ze2πits , s), ∀(z, s, t) ∈ S 1 × I × I,
is continuous. Moreover, we have
F (z, s, 0) = (z, s), F (z, s, 1) = (ze2πis , s) = f (z, s), ∀(z, s) ∈ S 1 × I
and
F (z, 0, t) = (z, 0), ∀(z, t) ∈ S 1 × I.
Therefore, F is a homotopy from idS 1 ×I to f which is stationary on the circle S 1 ×{0}.
In a similar manner we obtain the required result for the other boundary circle.
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Suppose now that there exists a homotopy H : S 1 × I × I → S 1 × I from idS 1 ×I to f
that is stationary on both S 1 × {0} and S 1 × {1}, i.e. such that we have
H(z, s, 0) = (z, s), H(z, s, 1) = f (z, s), ∀(z, s, t) ∈ S 1 × I × I
and
H(z, 0, t) = (z, 0), H(z, 1, t) = (z, 1), ∀(z, t) ∈ S 1 × I.
Consider the path
β(s) = (1, s), s ∈ I
in S 1 × I. Then, it is obvious that H induces a path-homotopy between β and the
path α = f ◦ β in S 1 × I (simply consider H(β(s), t), (s, t) ∈ I × I). Consider also
the path γ(s) = (e2πis , 0), s ∈ I in S 1 × I. Since α, β are path-homotopic, it follows
that the path β ? ᾱ is path-homotopic to the constant loop e1 of S 1 × I based at point
β(0) = (1, 0). Moreover, considering the obvious deformation retraction of S 1 × I
onto S 1 × {0} (namely, G(z, s, t) = (z, (1 − t)s), (z, s, t) ∈ S 1 × I × I), we see that
β ? ᾱ is path-homotopic to the loop e1 ? γ̄ of S 1 × I, thus path-homotopic to the
loop γ̄ on S 1 × I. It follows that γ is path homotopic to e1 , which is a contradiction
since [γ] generates π1 (S 1 × I, (1, 0)) (as can be seen from the well-known isomorphism
(Hatcher, p. 34, Proposition 1.12) between π1 (S 1 × I, (1, 0)) and π1 (S 1 , 1) × π1 (I, 0),
and the facts that π1 (I, 0) is trivial and [(e2πis , s ∈ I)] generates π1 (S 1 , 1)) and this
group, being isomorphic to π1 (S 1 , 1), is non-trivial.
Exercise 4. (A. Hatcher, Algebraic Topology, p. 39, 10.)From the isomorphism
π1 (X × Y, (x0 , y0 )) ≈ π1 (X, x0 ) × π1 (Y, y0 ) it follows that loops in X × {y0 } and
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{x0 } × Y represent commuting elements of π1 (X × Y, (x0 , y0 )). Construct an explicit
homotopy demonstrating this.
Solution
Let f × {y0 } and {y0 } × g be loops in X × {y0 } and {x0 } × Y . Notice that (f ?
ex0 , ey0 ? g) = (f, ey0 ) ? (ex0 , g). Let H1 be the homotopy between f ? ey0 and ey0 ? f
and H2 be the homotopy between g ? ex0 and ex0 ? g. Then H1 × H2 gives the required
homotopy.
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