21F (Algebraic Topology)

f g ≃ 1_{Y}

and

h f ≃ 1_{X}

, composing homotopies gives

h ≃ h (f g) ≃ (h f) g ≃ g .

(1)

Thus

g f ≃ h f ≃ 1_{X}

, while

f g ≃ 1_{Y}

already holds, so

g

is a homotopy inverse of

f

More generally let

a : Y \to Y

be a homotopy inverse of

f g

and

b : X \to X

one of

h f

. Then

g^{'} = g a

satisfies

f g^{'} ≃ 1_{Y}

, and

h^{'} = bh

satisfies

h^{'} f ≃ 1_{X}

. The first argument applies. Hence the separate one-sided hypotheses imply that $f$ is a homotopy equivalence, as expressed by separate homotopy inverses combine into a homotopy equivalence.

Solution

 0  0

Take a homotopy equivalence

f : X \to Y

with inverse

g

, and universal covering maps

p_{X}, p_{Y}

. Since the covering spaces are simply connected, the path lifting theorem constructs lifts

f

f p_{X}

and

g

g p_{Y}

, after specifying one value in the appropriate fiber. Explicitly, define the lift at a point by lifting the image of a path from the chosen base point; independence from the path follows from simple connectedness and homotopy lifting.

Lift a homotopy

g f ≃ 1_{X}

starting at

g f

. The homotopy lifting theorem for a covering map makes its endpoint a map

d_{X} : X \to X

with

p_{X} d_{X} = p_{X}

. Such a lift of the identity is a deck transformation and is a homeomorphism: lift the identity again with the base point image reversed, and uniqueness of path lifts makes the two compositions identities. Likewise

f g ≃ d_{Y}

for a deck transformation of

Y

Thus both compositions are homotopy equivalences, even though the lifted homotopies need not end at the particular identity lift. Part (i) then proves