Topology Take Home Final Exam Due on January 13, 2016 1. (a) Let f : X → Y be an identification map (or a quotient map), let A be a subspace of X , and give f (A) the induced topology from Y. Show that the restriction f |A: A → f (A) need not be an identification map.
Solution: Consider the function f : R → S1defined by f (x) = (cos 2πx, sin 2πx). By defini- tion, one can check that f is an identification map.
Now, if we let A = [0, 1) then the restriction f |A: A → f (A) = S1is not an identification map because if we let B = {(x, y) ∈ R2| x > 0, y ≥ 0} ∩ S1and U = [0, 1/2), then
Bis not an open subset of S1while ( f |A)−1(B) = U is an open subset of A = [0, 1).
Definition. Let X and Y be topological spaces; let p : X → Y be a surjective map. The map p is called an identification map (or a quotient map) provided
a subset B of Y is open in Y ⇔ p−1(B) is open in X . Since B = p(p−1(B)), p is an identification map provided
a subset p(p−1(B)) of Y is open in Y ⇔ p−1(B) is open in X .
Alternative Solution: If we let A = [0, 1) and suppose that f |A: A → f (A) = S1is an identi- fication map. Let S1∗ denote the identification space associated with the partition of A whose members are the subsets {( f |A)−1(y)}, where y ∈ S1.
Since f |A is 1 − 1, A = [0, 1) = S1∗and it is homeomorphic to S1by the Theorem (4.2)(a) on page 67 of Basic Topology by Armstrong.
However, the inverse map ( f |A)−1 : S1→ [0, 1) is not continuous at (0, 1) ∈ S1, [0, 1) is not homeomorphic to S1.
Thus, the restriction f |A: A → f (A) = S1is not an identification map.
(b) With the terminology as above, show that if A is open in X and if f takes open sets to open sets, or if A is closed in X and f takes closed sets to closed sets, then f |A: A → f (A) is an identification map.
Solution: Assume that A is an open suset of X and f takes open sets in X to open sets in Y.
Since every open set in A is an intersection of an open set in X and the set A, open sets in A will also be open sets in X .
Also, since f |A(a) = a for all a ∈ A and f takes open sets in X to open sets in Y, f |Atakes open sets in A to open sets in f (A).
Thus, f |A: A → f (A) is an identification map by the Theorem 4.3 on page 67 of Basic Topol- ogy by Armstrong.
Using a similar argument, one can also prove that if A is closed in X and f takes closed sets to closed sets, then f |A: A → f (A) is an identification map.
(c) Give an example of an identification map which is neither open nor closed.
Solution: Let X = [0, 1] ∪ (2, 3], Y = [0, 2] and let f : X → Y be defined by f(x) =
(x if x ∈ [0, 1]
x− 1 if x ∈ (2, 3]
Since X is compact, Y is Hausdorff and f : X → Y is an onto continuous map, f is an identifi-
Topology Topology Take Home Final Exam (Continued) Fall Semester 2016
cation map by the Corollary (4.4)on page 67 of Basic Topology by Armstrong.
Observe that [0, 1] is open in X while f ([0, 1]) = [0, 1] is not open in Y and (2, 3] is closed in X while f ((2, 3]) = (1, 2] is not closed in Y, f is neither open nor closed.
2. (a) Let C denote the unit circle in the plane. Suppose f : C → C is a cntinuous map which is not homotopic to the identity map i : C → C which is defined by i(x) = x for each x ∈ C. Prove that
f(x) = −x for some x of C.
Solution:
Claim: If f (x) 6= −x for all x ∈ C then f and the identity map i are never antipodal, i.e.
(1 − t) f (z) + ti(z) = (1 − t) f (z) + tz 6= (0, 0) for any t ∈ [0, 1] and z ∈ C.
Proof of Claim: Suppose there exist t ∈ [0, 1] and x ∈ C such that (1 − t) f (x) + tx = (0, 0).
Then (1 − t)k f (x)k = tkxk =⇒ 1 − t = t since kxk = k f (x)k = 1 and 0 ≤ t ≤ 1.
Thus we have t = 1/2 and 1/2( f (x) + x) = (0, 0) =⇒ f (x) = −x, a contradiction.
Hece f (z) 6= −z for all z ∈ C.
Using the Claim and suppose that f (x) 6= −x for all x of C, the map F : [0, 1] × C → C de- fined by
F(t, x) = (1 − t) f (x) + ti(x) k(1 − t) f (x) + ti(x)k
is an homotopy between f and the identity map i which contradicts to the hypothesis.
Hence, f (x) = −x for some x of C.
Alternative Solution: By identifying the unit circle C in the plane with U = {z ∈ C | kzk = 1}, the set of complex numbers of unit modulus, we may write x = eiθ for each x ∈ C.
Suppose that f (x) 6= −x for each x ∈ C, f(x)
x is a map from C into C \ {(0, −1)} when we consider f(x)
x as a division of complex numbers f (x) and x in U.
Since C \ {(0, −1)} is homeomorphic to an open interval, say (−1/2, 1/2), the map f(x)
x : C → C \ {(0, −1)} is homotopic to a trivial map, say i(0,1): C → C defined by i(0,1)(x) = (0, 1) for all x ∈ C.
This implies that f (x) = x f(x)
x = i(x) f(x)
x is homotopic to the identity map i : C → C which contradicts to the hypothesis.
Hence, there exists an x ∈ C such that f (x) = −x.
(b) With C as above, show that the map which takes each point of C to the point diametrically opposite is homotopic to the identity.
Solution: By identifying the unit circle C in the plane with U = {z ∈ C | kzk = 1}, the set of complex numbers of unit modulus, it is easy to see the map F : [0, 1] × C → C defined by F(t, x) = eiπti(x) is a continuous map satisfying that F(0, x) = x and F(1, x) = −x for each x∈ C, i.e F is an homotopy of the identity map and the antepodal map.
Topology Topology Take Home Final Exam (Continued) Fall Semester 2016 3. Let f : X → Snbe a continuous map which is not onto. Prove that f is null homotopic, that is to say f
is homotopic to a map which takes all of X to a single point of Sn.
Solution: Let p be an point in Snsuch that −p /∈ f (X) = { f (x) | x ∈ X}, and let g : X → Snbe the constant map defined by g(x) = −p for all x ∈ X . Since f (x) and g(x) are not antipodal points for each x ∈ X , the map F : [0, 1] × X → Sndefined by
F(t, x) = (1 − t) f (x) + tg(x) k(1 − t) f (x) + tgi(x)k is an homotopy between f and g. This proves that f is null homotopic.
Alternative Solution: Since f : X → Snis not onto, there exist a p ∈ Snsuch that f (A) ⊂ Sn\ {p}.
Let q be a point in f (X ). For each x ∈ X , join f (x) to q by a great circle γx: [0, 1] → Sn\ {p} such that γx(0) = f (x) and γx(1) = q. Then the map F : [0, 1] × X :→ Sndefined by F(t, x) = γx(t) is a hompotpy of f and the trivial map taking all of X to the point q ∈ Sn.
4. (a) Let α be a path in X from x0to x1; let β be a path in X from x1to x2. Show that if γ = α ∗ β , then γ = ˆˆ β ◦ ˆα .
Solution: For each [ f ] ∈ π1(X , x0), since
α : πˆ 1(X , x0) → π1(X , x1) is defined by
α ([ f ]) = [α ] ∗ [ f ] ∗ [α ] = [α ∗ ( f ∗ α )],ˆ we have
γ ([ f ])ˆ = [γ] ∗ [ f ] ∗ [γ] = [α ∗ β ] ∗ [ f ] ∗ [α ∗ β ] = [β ∗ α][ f ] ∗ [α ∗ β ]
= [β ] ∗ [α] ∗ [ f ] ∗ [α] ∗ [β ] = [β ] ∗ ˆα([ f ]) ∗ [β ] = ˆβ ( ˆα ([ f ])
= β ◦ ˆˆ α ([ f ]) This implies that ˆγ = ˆβ ◦ ˆα .
(b) Let x0and x1be points of path-connected space X . Show that π1(X , x0) is abelian if and only if for each pair α and β of paths from x0to x1, we have ˆα = ˆβ .
Solution: For each [ f ] ∈ π1(X , x0) and for each pair α and β of paths from x0to x1, Π1(X , x0) is abelian ⇔ [α ∗ β ] ∗ [ f ] = [ f ] ∗ [α ∗ [β ]
⇔ [α] ∗ [β ] ∗ [ f ] = [ f ] ∗ [α] ∗ [β ]
⇔ [α] ∗
[α] ∗ [β ] ∗ [ f ]
∗ [β ] = [α] ∗
[ f ] ∗ [α] ∗ [β ]
∗ [β ]
⇔ [β ] ∗ [ f ] ∗ [β ] = [α] ∗ [ f ] ∗ [α]
⇔ β ([ f ]) = ˆˆ α ([ f ]) ∀ [ f ] ∈ π1(X , x0)
⇔ β = ˆˆ α
Topology Topology Take Home Final Exam (Continued) Fall Semester 2016 5. Let A ⊂ X ; suppose r : X → A is a continuous map such that r(a) = a for each a ∈ A. (The map r is
called a retraction of X onto A.) If a0∈ A, show that
r∗ : π1(X , a0) → π1(A, a0) is surjective.
Solution: Since A is a subset of X and the retraction r : X → A is an identity map when it is restricted to the subset A, a loop α : [0, 1] → A in A based at a0= α(0) = α(1) is also a loop in X based at a0. Hence, the homormorphism r∗: π1(X , a0) → π1(A, a0) is an identity map when it is restricted to the subgroup π1(A, a0) of π1(X , a0), i.e. r∗([α]) = [α] for each [α] ∈ π1(A, a0), and r∗: π1(X , a0) → π1(A, a0) is surjective.
6. Let p : E → B be a covering map. Let α and β be paths in B with α(1) = β (0); let ˜α and ˜β be liftings of them such that ˜α (1) = ˜β (0). Show that ˜α ∗ ˜β is a lifting of α ∗ β .
Solution: Since p( ˜α ∗ ˜β ) = p( ˜α ) ∗ p( ˜β ) = α ∗ β and p( ]α ∗ β ) = α ∗ β , ˜α ∗ ˜β and ]α ∗ β are lift- ings of α ∗ β beginning at ˜α (0). By the (54.1) Path-Lifting lemma (on page 342 of Topology by J.
Munkres), we have ]α ∗ β = ˜α ∗ ˜β .
Lemma 54.1. (Path-lifting lemma) Let p : E → B be a covering map, let p(e0) = b0. Any path f : [0, 1] → B beginning at b0has a unique lifting to a path ˜f in E beginning at e0.
