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Thread: retract, inclusion, normal subgroup

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    retract, inclusion, normal subgroup

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