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Math Help - Sufficient condition for Existence of continous retraction

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    Sufficient condition for Existence of continous retraction

    If f maps X to X continuously and the range of f is A (subset of X) then is this sufficient to show that there exist a continuous retraction from X to A?

    If so, how can this be proven?
    If not, any counterexamples?

    Notes: This is the topological continuity I am speaking of. Also a I am using retraction to mean a map from X to a subspace A such that f(a)=a for all a in A.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Xian View Post
    If f maps X to X continuously and the range of f is A (subset of X) then is this sufficient to show that there exist a continuous retraction from X to A?

    If so, how can this be proven?
    If not, any counterexamples?

    Notes: This is the topological continuity I am speaking of. Also a I am using retraction to mean a map from X to a subspace A such that f(a)=a for all a in A.
    This question makes no sense. What are you asking?
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    Quote Originally Posted by Xian View Post
    If f maps X to X continuously and the range of f is A (subset of X) then is this sufficient to show that there exist a continuous retraction from X to A?

    If so, how can this be proven?
    If not, any counterexamples?

    Notes: This is the topological continuity I am speaking of. Also a I am using retraction to mean a map from X to a subspace A such that f(a)=a for all a in A.
    What I mean is that the the existence of continuous map f:X\rightarrow X where f(X)=A\subseteq X should guarantee the existence of a continuous retraction of X to A.

    Actually I'd like to make my statement even stronger:

    If there is a function f:X\rightarrow X where f(X)\subseteq A\subseteq X then there exists a continuous retraction r:X\rightarrow A such that \forall a\in A  r(a)=a.

    Is this true? Proof?
    Is it false? Counterexample?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Xian View Post
    What I mean is that the the existence of continuous map f:X\rightarrow X where f(X)=A\subseteq X should guarantee the existence of a continuous retraction of X to A.

    Actually I'd like to make my statement even stronger:

    If there is a function f:X\rightarrow X where f(X)\subseteq A\subseteq X then there exists a continuous retraction r:X\rightarrow A such that \forall a\in A  r(a)=a.

    Is this true? Proof?
    Is it false? Counterexample?
    I just am still lost. Why do we have two maps? A deformation retract is a map r:X\to A\subseteq X such that r\circ\iota_A=\text{id}_A. What does f have to do with this? Are you discussing sufficient conditions for f to bea deformation retract?
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    Quote Originally Posted by Drexel28 View Post
    I just am still lost. Why do we have two maps? A deformation retract is a map r:X\to A\subseteq X such that r\circ\iota_A=\text{id}_A. What does f have to do with this? Are you discussing sufficient conditions for f to bea deformation retract?
    I'm saying suppose such an f exists with the above properties. Does the existence of f imply the existence of r?
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