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Math Help - Homotopic Functions...

  1. #1
    TTB
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    Homotopic Functions...

    Hi,

    Suppose f & g are both functions mapping from A to B. (A & B are topological spaces.)

    Also assume that f & g are homotopic.

    Then how does one prove that the Mapping Cylinders of f & g are homotopic spaces?

    Thanks x
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  2. #2
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    Quote Originally Posted by TTB View Post
    Hi,

    Suppose f & g are both functions mapping from A to B. (A & B are topological spaces.)

    Also assume that f & g are homotopic.

    Then how does one prove that the Mapping Cylinders of f & g are homotopic spaces?

    Thanks x
    Let f,g:A \rightarrow B; let F:A \times I \rightarrow B be their homotopy such that F(a, 0) = f(a) and F(a, 1) = g(a) for each a in A. The mapping cylinder M_f for a map f:A \rightarrow B is the quotient space of disjoint union (A \times I) \coprod B obtained by identifying each (a, 1) \in A \times I with f(a) \in B. The mapping cylinder M_f deformation retracts to the subspace B by sliding each point (a, t) along the segment \{a\} \times I \subset M_f to the endpoint f(a) \in B. We can see easily that B is a deformation retract of both M_f and M_g (We can have f(A) as either F(A,0) or F(A, 1) and g(A) as either F(A,1) or F(A, 0), respectively). Thus M_f and M_g are homotopy equivalent spaces.
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  3. #3
    TTB
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    Is it THAT simple? Y is a deformation retract of both Mf & Mg, so Y is homotopic to Mf & Mg & therefore, as homotopy is an equivalence relation, Mf is homotopic to Mg?

    Cool!

    Thank you!

    Are you sure there's nothing else to it? I spent aaaaaaaaaaaaages playing around with homtopies like F:Mfx[0,1] ---> Mg, etc. Do you not need to do ANY of that?? x
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  4. #4
    TTB
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    ...but wait, that doesn't seem to require the condition that f & g are homotopic.
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    Quote Originally Posted by TTB View Post
    Is it THAT simple? Y is a deformation retract of both Mf & Mg, so Y is homotopic to Mf & Mg & therefore, as homotopy is an equivalence relation, Mf is homotopic to Mg?

    Cool!

    Thank you!

    Are you sure there's nothing else to it? I spent aaaaaaaaaaaaages playing around with homtopies like F:Mfx[0,1] ---> Mg, etc. Do you not need to do ANY of that?? x
    By convention, we say a map f is "homotopic" to a map g and a topological space A is "homotopy equivalent to" a topological space B. Rather than saying "Y is homotopic to M_f", it is more appropriate to say "Y is homotopy equivalent to M_f" (Take M_f as a quotient topological space as mentioned in the previous post).

    Quote Originally Posted by TTB View Post
    ...but wait, that doesn't seem to require the condition that f & g are homotopic.
    I think that if f and g are topological continuous mappings from A to B, then both f(A)=g(A)=B are ensured and we don't need a homotopy condition between f and g. Meanwhile, if there is a homotopy condition between f and g, then topological continuity of f and g are ensured, so f(A)=g(A)=B.
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  6. #6
    TTB
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    Awesome! Thank you!
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