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Math Help - Contractibility of Finite Topological Space

  1. #1
    Junior Member
    Joined
    May 2008
    Posts
    37

    Contractibility of Finite Topological Space

    I'm looking for a way to show that a finite topological space with some prescribed topology is a contractible space. I understand the definition well enough: I need two continuous functions f,g such that their compositions are homotopic to the identity map on each space, where one of these spaces is just a 1-element finite topological space.

    So as an example, say:
    x = {a,b}
    Tx = {empty set, {a}, x}
    y = {a}
    Ty = {empty set, {a}}

    f: X -> Y is continuous
    g: Y -> X is continuous

    What would f and g be in this case to make the homotopies work? It seems like one of these functions would have to map a in Y to both a and b in X in order to make them homotopic to the identity functions? Am I not seeing something obvious here? Thanks.
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  2. #2
    Senior Member
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    Nov 2008
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    Quote Originally Posted by joeyjoejoe View Post
    I'm looking for a way to show that a finite topological space with some prescribed topology is a contractible space. I understand the definition well enough: I need two continuous functions f,g such that their compositions are homotopic to the identity map on each space, where one of these spaces is just a 1-element finite topological space.

    So as an example, say:
    x = {a,b}
    Tx = {empty set, {a}, x}
    y = {a}
    Ty = {empty set, {a}}

    f: X -> Y is continuous
    g: Y -> X is continuous

    What would f and g be in this case to make the homotopies work? It seems like one of these functions would have to map a in Y to both a and b in X in order to make them homotopic to the identity functions? Am I not seeing something obvious here? Thanks.
    If I understand your question correctly, the composition maps of f and g can be constant maps. You might already know that an identity map is nulhomotophic for a contractible space. Simply put, a contractible space X is homotopic equivalent to a point such that there exists a continuous map F:X \times [0,1] \rightarrow X with  F(x,0) = id_x(x)=x and F(x,1) = c(x)=c, for all x \in X and a fixed c \in X.

    For f \circ g:Y \rightarrow Y, it maps from Y to its single element because Y is a contractible space and an identity map is homotopic to a constant map. It is similar to g \circ f.
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  3. #3
    Junior Member
    Joined
    May 2008
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    Yeah that's what I was thinking, I guess I was just stuck on how to define f and g explicitly.

    Since f: X -> Y, it would have to map both b and a in X to a in Y.
    Similarly, g: Y -> X, it would have to map a in Y to both a and b in X.

    I'm just confused as how to build these continuous maps, but I have an idea:

    If you define f and g parametrically, say:

    f(x) = {x, 0 <= t < 1}
    {a, t = 1}

    g would be defined similarly. I have already proved that this f is continuous, so I think I'm set.
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