In algebraic topology, a fiber-homotopy equivalence is a map over a space B that has homotopy inverse over B (that is we require a homotopy be a map over B for each time t.) It is a relative analog of a homotopy equivalence between spaces.
Given maps p:D→B, q:E→B, if ƒ:D→E is a fiber-homotopy equivalence, then for any b in B the restriction
f
:
p
−
1
(
b
)
→
q
−
1
(
b
)
is a homotopy equivalence. If p, q are fibrations, this is always the case for homotopy equivalences by the next proposition.
The following proof is based on the proof of Proposition in Ch. 6, § 5 of (May). We write
∼
B
for a homotopy over B.
We first note that it is enough to show that ƒ admits a left homotopy inverse over B. Indeed, if
g
f
∼
B
id
with g over B, then g is in particular a homotopy equivalence. Thus, g also admits a left homotopy inverse h over B and then formally we have
h
∼
f
; that is,
f
g
∼
B
id
.
Now, since ƒ is a homotopy equivalence, it has a homotopy inverse g. Since
f
g
∼
id
, we have:
p
g
=
q
f
g
∼
q
. Since p is a fibration, the homotopy
p
g
∼
q
lifts to a homotopy from g to, say, g' that satisfies
p
g
′
=
q
. Thus, we can assume g is over B. Then it suffices to show gƒ, which is now over B, has a left homotopy inverse over B since that would imply that ƒ has such a left inverse.
Therefore, the proof reduces to the situation where ƒ:D→D is over B via p and
f
∼
id
D
. Let
h
t
be a homotopy from ƒ to
id
D
. Then, since
p
h
0
=
p
and since p is a fibration, the homotopy
p
h
t
lifts to a homotopy
k
t
:
id
D
∼
k
1
; explicitly, we have
p
h
t
=
p
k
t
. Note also
k
1
is over B.
We show
k
1
is a left homotopy inverse of ƒ over B. Let
J
:
k
1
f
∼
h
1
=
id
D
be the homotopy given as the composition of homotopies
k
1
f
∼
f
=
h
0
∼
id
D
. Then we can find a homotopy K from the homotopy pJ to the constant homotopy
p
k
1
=
p
h
1
. Since p is a fibration, we can lift K to, say, L. We can finish by going around the edge corresponding to J:
k
1
f
=
J
0
=
L
0
,
0
∼
B
L
0
,
1
∼
B
L
1
,
1
∼
B
L
1
,
0
=
J
1
=
id
.
