Null-homotopic, Retracts: Review for Prelims.

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SUMMARY

Every continuous map f: CP^2 → S^1 x S^1 x S^1 is nullhomotopic due to the triviality of the induced map on the third homology group, which is zero. The wedge sum RP^2 ∨ RP^2 is not a retract of the product RP^2 x RP^2, as demonstrated by Kunneth's theorem, which shows that their homology groups differ. The discussion emphasizes the importance of using cell decomposition and fundamental group properties in these homotopy and homology arguments.

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  • Understanding of homotopy theory and nullhomotopic maps
  • Familiarity with cell decomposition in algebraic topology
  • Knowledge of homology groups and Kunneth's theorem
  • Basic concepts of fundamental groups and covering spaces
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  • Study the implications of Kunneth's theorem in algebraic topology
  • Learn about cell complexes and their applications in homotopy theory
  • Explore the properties of covering spaces and their role in lifting maps
  • Investigate the structure of fundamental groups for wedge sums and products
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Mathematicians, particularly those specializing in algebraic topology, graduate students preparing for prelim exams, and anyone interested in the properties of homotopy and homology in topological spaces.

WWGD
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Hi, Everyone:

I am trying to help someone with their prelims, but ended up needing some help myself:

1)Show that every map f: CP^2--->S^1 x S^1 x S^1 is nullhomotopic

2)Is the wedge RP^2 \/RP^2 a retract of RP^2 x RP^2 ?


For 1), I think we need to use the cell decomposition for CP^2 somehow, or just use
the fact that the 3rd homology ( 3 being odd ) is 0 , so that the induced map on
3rd homology must be trivial.

For 2, all I can think is that if the wedge were a retract, then it would be homotopic
to the product , so that all homology/homotopy groups would be equal. I think the
answer is no, since the homology of the wedge is the free product, but the
homology of the product is given by Kunneth's theorem, and we get different groups.

Does this work? Any ideas?

Thanks.
 
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I think for #1 , you can use lifting of maps ; R^3 is a covering space for the 3-torus
maps lift if these maps satisfy a condition on the pushforward of the fundamental groups;
if the maps lift into R^3, then these maps are trivial, since R^3 is contractible, and
every map into a contractible space is trivial.
 
isn't the fundamental group of the wedge of two projective planes the free product of Z/2Z with Z/2Z where the fundamental group of the product is Z/2Z x Z/2Z?

no composition of group homomorphisms Z/2ZvZ/2Z -> Z/2Z x Z/2Z -> Z/2ZvZ/2Z
can be the identity.
 

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