Suppose Y is obtained by attaching a zero-cell, so Y=Xcupast. Then Y2 is (XtimesX)cup(Xtimesast)cup(asttimesX)cup(asttimesast)
and so Y2/X2 is homeomorphic to
astcupXcupXcupast.
This can be arbitrarily complicated depending on X.
ADDENDUM: By request, a connected example is the inclusion mathbbCP1subsetmathbbCP2. The quotient Y2/X2, in this case, has a nonzero cohomology operation Sq2 from H6 to H8 with ℤ/2-coefficients, and there is no wedge of products of spheres that can have this cohomology. (You should work out the details.)
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