Higher Order π Groups, Fiber Map, Hurewicz, Serre
Fiber Map, Hurewicz, Serre
Let S~ be a space over S,
with a continuous function p mapping S~ onto S.
This is a fiber map for a space Z
if it possesses the homotopy lifting property for Z into S.
This means every continuous function f from Z cross I into S,
with a lift g0 over f0,
lifts to a compatible homotopy g(Z:I) into S~,
such that f = gp.
The lifte need not be unique.
If this holds for every space Z, then {S~,p,S} is hurewicz.
This is named for the mathematician Hurewicz
(biography),
who should not be confused with the similar sounding Hurwitz
(biography).
Don't confuse this with a covering space; it need not be.
However, if S~ is a covering space, it is automatically hurewicz.
By assumption, f0(Z) has a lift,
and that means the
lifting criterion is satisfied.
The loops in the image of f0 lie in a certain subgroup of the fundamental group of S.
Now consider a loop in Z cross I, and apply the homotopic function f.
Contract the unit interval down to 0, thus contracting Z:I onto Z.
This pulls our loop in f(Z:I) down to a loop in f0(Z).
This is one of our original loops.
The lifting criterion is satisfied, and the entire homotopy lifts.
Howeverr, as mentioned earlier, S~ is not usually a covering space.
If each homotopy on Z can be lifted, for every finite explicit cw complex Z, then S~ is serre
(biography).
Product Space
Let S~ be the product space of S and T,
and let p map S~ back onto S.
Assume f0(Z) lifts to f0(Z) cross g0(Z) in S cross T.
This is continuous, hence f0 and g0 are continuous.
Let f run through its homotopy, and let g be constant at g0.
Both f and g are continuous, hence f cross g is continuous.
This gives a continuous homotopy upstairs that projects back to f downstairs.
Therefore S*T is hurewicz.
In this case every image, and hence every homotopy, has a lift.
(A fiber map doesn't guarantee an initial lift; only that each lift leads to a lifted homotopy.)
Cross the image of Z in S with any point q in T for the initial lift.
Once this is done, the entire homotopy can be crossed with q to get a homotopy upstairs.
Fiber
The fiber of a fiber map is the preimage of the base point.
Since one often wants to avoid pointed spaces,
it is worth showing the fibers of any two base points a and b are homotopy equivalent,
provided {S~,p,S} is hurewicz.
Assume a and b have fibers a~ and b~ respectively.
(These are not individual points, but subspaces of S~ that map to a and b.)
Let the space a~ act as the third space Z,
and let f0 = p.
Let a~ be the initial lift of a~.
Since S is path connected,
this map is homotopic to the map that takes a~ onto b.
In other words, f slides the image of a~ from a to b.
Lift this homotopy into S~ to get a homotopy between the identity map on a~
and a map that takes a~ into b~.
Call this latter map g1(a~), the end of the homotopy upstairs.
In the same way, h1 takes b~ into a~, and is homotopic to the identity map on b~.
Let y be the image g1(a~).
Apply h1 to y and find a function that is homotopic to the identity map on y.
Compose this with g1 and g1(a~) is homotopic to h1(g1(a~)).
The former is homotopic to the identity map on a~, so by transitivity,
g1 followed by h1 is homotopic to the identity map on a~.
This works in both directions,
hence the spaces a~ and b~ are
homotopically equivalent.
It is somewhat less common, but sometimes a fiber map projects S~ into S,
rather than onto S.
This is hurewicz, or serre, if homotopies can be lifted for the appropriate spaces,
provided an initial lift is given,
and the entire homotopy lives in the image of S~.
An example is S~ embedded in S.
This is a bit unusual, since the base space S is larger than the total space S~,
but it works.
The homotopy downstairs becomes the homotopy upstairs, with no change at all.
Fibers are only defined for points in the image of S~.
If S~/S is hurewicz, and the image of S~ is path connected, fibers are homotopically equivalent by the above proof.
Let S be the union of one or more component spaces Si.
Since p is continuous, the preimage of each component is both open and closed.
Hence S~ separates into components, where each component projects into its corresponding component downstairs.
When Z maps continuously into S, it too separates into components.
The lift maps these components into the corresponding components upstairs.
Finally, consider a homotopy on Z, which draws a path for each c in Z.
This path cannot jump to a new component, hence it remains in the component of S where it started.
The homotopy keeps each component of Z in its corresponding component in S,
and the lifted homotopy lies in the corresponding component upstairs.
The homotopy can be lifted iff it can be lifted per component.
Withe this in mind, S~/S is hurewicz, or serre, iff it is same per component.