nLab
basic problems of algebraic topology

In algebraic topology one defines and uses functors from some category of topological spaces to some category of algebraic objects to help solve some existence or uniqueness problem for spaces or maps.

There are 4 basic problems of algebraic topology for the existence of maps: extension problems, retraction problems, lifting problems and section problems.

Regarding that these problems make sense in any category, we will talk about objects and morphisms and not spaces and maps.

Extension problem

Given morphisms i:AXi:A\to X, f:AYf:A\to Y find an extension of ff to XX, i.e. a morphism f˜:XY\tilde{f}:X\to Y such that if˜=fi\circ\tilde{f}=f. Notice that if i:AXi:A\hookrightarrow X is a subobject, then if˜i\circ\tilde{f} is the restriction f˜ A\tilde{f}{|_A}, and the condition is f˜ A=f\tilde{f}{|_A} = f.

Retraction problem

Let f:AYf:A\to Y be a morphism. Find a retraction of ff, that is a morphism r:YAr:Y\to A such that rf=id Ar\circ f = id_A.

The retraction problem is a special case of the extension problem for A=XA=X and i=id Ai=id_A. Conversely, the general extension problem may (in Top and many other categories) be reduced to a retraction problem:

Proposition (Reducing an extension to a retraction)

If the pushout Y AXY\coprod_A X exists (for ii, ff as above) then the extensions f˜\tilde{f} of ff along ii are in 1–1 correspondence with the retractions of i *(f):YY AXi_*(f) : Y\to Y\coprod_A X.

Lifting problem

Given morphisms p:EBp:E\to B and g:ZBg:Z\to B, find a lifting? of gg to EE, i.e. a morphism g˜:ZE\tilde{g}:Z\to E such that pg˜=gp\circ\tilde{g}=g.

Section problem

For any p:EBp:E\to B find a section s:BEs: B\to E, i.e. a morphism ss such that ps=id Bp\circ s = id_B.

The section problem is a special case of a lifting problem where g=id E:EEg = id_E : E\to E. Then the lifting is the section: g˜=s\tilde{g} = s. A converse is true in the sense

Proposition (Reducing a lifting to a section)

If the pullback Z× BEZ\times_B E exists then the general liftings for of GG along pp as above are in a bijection with the section of g *(p)=Z× Bp:Z× BEZg^*(p)=Z\times_B p : Z\times_B E\to Z.

Revised on December 9, 2009 04:00:55 by Toby Bartels (173.60.119.197)