Topological retraction

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Proof

Note that if [ilmath]r\circ i_A\eq \text{Id}_A[/ilmath] then [ilmath]r_*\circ(i_A)_*\eq (\text{Id}_A)_*[/ilmath]

Alec's thought: can we use the first group isomorphism theorem on [ilmath]r_*[/ilmath] to get [ilmath]\pi_1(A,a)[/ilmath] from [ilmath]\pi_1(X,a)[/ilmath] or something?

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Demote to grade A once tidied up. Find other sources. Be sure to link to deformation retraction and strong deformation retraction

Definition

Let [ilmath](X,\mathcal{ J })[/ilmath] be a topological space and let [ilmath]A\in\mathcal{P}(X)[/ilmath] be considered a s subspace of [ilmath]X[/ilmath]. A continuous map, [ilmath]r:X\rightarrow A[/ilmath] is called a retraction if[1]:

  • The restriction of [ilmath]r[/ilmath] to [ilmath]A[/ilmath] (the map [ilmath]r\vert_A:A\rightarrow A[/ilmath] given by [ilmath]r\vert_A:a\mapsto r(a)[/ilmath]) is the identity map, [ilmath]\text{Id}_A:A\rightarrow A[/ilmath] given by [ilmath]\text{Id}_A:a\mapsto a[/ilmath]

If there is such a retraction, we say that: [ilmath]A[/ilmath] is a retract[1] of [ilmath]X[/ilmath].
Claim 1:

  • This is equivalent to the condition: [ilmath]r\circ i_A=\text{Id}_A[/ilmath] where [ilmath]i_A[/ilmath] denotes the inclusion map, [ilmath]i_A:A\hookrightarrow X[/ilmath] given by [ilmath]i_A:a\mapsto x[/ilmath]

TODO: In the case of [ilmath]A=\emptyset[/ilmath] - does it matter? I don't think so, but check there is nothing noteworthy about it. Also proof of claims


See also

Important theorems

Lesser theorems

References

  1. 1.0 1.1 Introduction to Topological Manifolds - John M. Lee