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Thread: retract, inclusion, normal subgroup

  1. February 19th 2009, 12:01 PM #1
    GenoaTopologist
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    retract, inclusion, normal subgroup

    Prove that, if A is a retract of X, r:X \rightarrow A is a retraction, i: A \rightarrow X is the inclusion, and i_*\pi(A) is a normal subgroup of \pi(X), then \pi(X) is the direct product of the subgroups image i_* and kernel r_*.
    [So I need to show \pi_1(X) = \text{ker}(r_*) \times i_*(\pi_1(A)).]
    Last edited by GenoaTopologist; February 19th 2009 at 01:28 PM.
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