Download Midterm 1 solutions

KOÇ UNIVERSITY, Spring 2013, MATH 402/571, Midterm-1, March 28, 75 Minutes
SOLUTIONS BY KARATUG OZAN BIRCAN
PROBLEM 1 a (5 points): Write the definition of a homotopy between two continuous maps f, g : X → Y .
Let X and Y be topological spaces, and let f, g : X → Y be continuous maps. We say that f and
g are homotopic and denoted by f ' g if there exists a continuous map H : X × I → Y (which is called
a homotopy from f to g) such that for all x ∈ X,
H(x, 0) = f (x),
H(x, 1) = g(x).
PROBLEM 1 b (5 points): Explain the difference between free homotopy versus stationary homotopy.
Let f, g : X → Y be continuous maps and let A be a subset of X. A homotopy H : X × [0, 1] → Y
between f and g is said to be stationary on A if H(a, t) = f (a) = g(a) for all a ∈ A and t ∈ [0, 1]. In
other words, the stationary homotopy keeps the elements of A fixed.
PROBLEM 1 c (5 points): What is homotopy equivalence of topological spaces?
Let φ : X → Y be a continuous map. Another continuous map ψ : Y → X is called homotopy inverse for φ if ψ ◦ φ ' IdX and φ ◦ ψ ' IdY . If there exists a homotopy inverse for φ, then φ is called
homotopy equivalence. In that case, we say that X is homotopy equivalent to Y (or X has the same
homotopy type of Y ).
PROBLEM 1 d (10 points): Give an example of a pair of topological spaces which are not homotopy
equivalent. (prove your answer)
Consider R2 and the circle S 1 . Since π1 (R2 ) is trivial and π1 (S 1 ) ∼
= Z, we conclude that these two
spaces are not homotopy equivalent.
PROBLEM 2 a (5 points): What is a retraction?
Let X be a space and let A be a subspace of X. A continuous map r : X → A is called a retraction if r(x) = x for all x ∈ A (or equivalently, r ◦ ıA is the identity map on A, where ıA : A → X is the
inclusion map).
PROBLEM 2 b (5 points): What is a deformation retraction?
Let X be a space and let A be a subspace of X. A is a deformation retract of X if and only if there
exists a homotopy H : X × I → X (which is called deformation retraction) such that
1
2
H(x, 0) = x for all x ∈ X,
H(x, 1) ∈ A for all x ∈ X,
H(a, 1) = a for all a ∈ A.
PROBLEM 2 c (5 points): What is a strong deformation retraction?
Let X be a space and let A be a subspace of X. A is a strong deformation retract of X if and only if
there exists a homotopy H : X × I → X (which is called strong deformation retraction) such that
H(x, 0) = x for all x ∈ X,
H(x, 1) ∈ A for all x ∈ X,
H(a, t) = a for all a ∈ A and t ∈ I.
PROBLEM 2 d (10 points): Prove or disprove: S 1 is a strong deformation retraction of R3 \ {0}.
Consider the straight line homotopy H : (R3 \ {0}) × I → R3 \ {0}, sending (x, t) 7→ (1 − t)x + t(x/|x|).
This proves that S 2 is a strong deformation retract of R3 \ {0}, that is, these spaces have the same
homotopy type. Therefore since π1 (S 2 ) is trivial and π1 (S 1 ) ∼
= Z, S 1 does not have the same homotopy
2
1
type of S , and hence S is not a strong deformation retract of R3 \ {0}.
PROBLEM 3 a (10 points): State the theorem about the homotopy invariance of the fundamental group.
If f : X → Y is a homotopy equivalence, then for any point p ∈ X, f∗ : π1 (X, p) → π1 (Y, f (p)) is an
isomorphism.
PROBLEM 3 b (5 points): What is the definition of a convex subset of Rn ?
A subset X of Rn is said to be convex if for all x, y ∈ X, tx + (1 − t)y ∈ X for 0 ≤ t ≤ 1.
PROBLEM 3 c (10 points): Prove that the fundamental group of any convex subset of Rn is trivial.
Note that a convex subset X of Rn is contractible via straight line homotopy. Since the fundamental group of a contractible space is trivial, we conclude that the fundamental group of X is trivial.
PROBLEM 4 a (5 points): Give the definition of a covering space.
Let X be a topological space. A covering space of X is a space E with a continuous surjective map
p : E → X such that for every x ∈ X, there exists an open neighnorhood U of x such that the inverse
image of U under the map p is a disjoint union of connected open sets of E, each of which is mapped
homeomorphically onto U by p.
PROBLEM 4 b (10 points): State the path-lifting theorem.
Let p : E → X be a covering map and let f : I → X be any path in X. Let p(y) = f (0) for some
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y ∈ E. Then there exists a unique lift f˜ : I → E such that f˜(0) = y.
PROBLEM 4 c (10 points): State the homotopy lifting theorem (for homotopy of paths).
Let p : E → X be a covering map and let H : I × I → X be a homotopy between paths f and g.
Then if f˜ is a lift of f , then there exists a unique homotopy H̃ : I × I → E of f˜ which lifts H.
PROBLEM 5 (20 points): Prove or disprove: There is a covering space of the figure 8 where each point
has 3 preimages in the cover.
Let E be the union of three disconnected copies of figure 8, each of which is mapped with the identity map. Then each point of figure 8 has three preimages in E.