Results 1 to 3 of 3

Math Help - Two Suspensions to Mention

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    54

    Two Suspensions to Mention

    I am to show that a continuous function, f: X \rightarrow Y, induces a continuous function between their respective suspensions, \Sigma X ,\, \Sigma Y.

    Let q:Y \times [0,1] \rightarrow \Sigma Y be the standard quotient map (in particular, it is continuous). Also, let \bar{f}: X \times [0,1] \rightarrow Y \times [0,1] by \bar{f}(x,t)=(f(x), t). Note that \bar{f} is also continuous (I think this is fairly obvious, but does anyone think it necessary to include some details here?).

    Now I have q \circ \bar{f}, which is, of course, continuous (given the prior statements) from X \times [0,1] \rightarrow \Sigma Y.

    So if p denotes the canonical quotient map from X \times [0,1] to \Sigma X, I would like to say that F:\Sigma X \rightarrow \Sigma Y where F=p^{-1} \circ (q\circ \bar{f}) is the desired induced map. However, it remains to be shown (in order to apply the theorem I'm basically quoting here, Theorem 22.2 in Munkres) that q\circ \bar{f} is constant on each set p^{-1} ([x,t]), and this is where I am stuck. Looks like this means that q\circ \bar{f} takes a set of things that are equivalent in X \times [0,1] (that's p^{-1}([x,t]), of course) and says that these are also equivalent under the quotient map on Y \times [0,1]. I can't tie it together.

    I've either got the wrong approach, or I could use some intuitive enlightenment. This whole set-up seems to imply some necessary connection between the quotient maps/spaces, but I'm not seeing the "necessitude".
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by cribby View Post
    I am to show that a continuous function, f: X \rightarrow Y, induces a continuous function between their respective suspensions, \Sigma X ,\, \Sigma Y.

    Let q:Y \times [0,1] \rightarrow \Sigma Y be the standard quotient map (in particular, it is continuous). Also, let \bar{f}: X \times [0,1] \rightarrow Y \times [0,1] by \bar{f}(x,t)=(f(x), t). Note that \bar{f} is also continuous (I think this is fairly obvious, but does anyone think it necessary to include some details here?).
    See Theorem 19.6 of Munkres if you need a detailed proof. Note that an identity map is continuous.

    Now I have q \circ \bar{f}, which is, of course, continuous (given the prior statements) from X \times [0,1] \rightarrow \Sigma Y.

    So if p denotes the canonical quotient map from X \times [0,1] to \Sigma X, I would like to say that F:\Sigma X \rightarrow \Sigma Y where F=p^{-1} \circ (q\circ \bar{f}) is the desired induced map. However, it remains to be shown (in order to apply the theorem I'm basically quoting here, Theorem 22.2 in Munkres) that q\circ \bar{f} is constant on each set p^{-1} ([x,t]), and this is where I am stuck. Looks like this means that q\circ \bar{f} takes a set of things that are equivalent in X \times [0,1] (that's p^{-1}([x,t]), of course) and says that these are also equivalent under the quotient map on Y \times [0,1]. I can't tie it together.

    I've either got the wrong approach, or I could use some intuitive enlightenment. This whole set-up seems to imply some necessary connection between the quotient maps/spaces, but I'm not seeing the "necessitude".
    Figure 1.

    \begin{matrix}X \times [0,1] & & \\\downarrow_{p} & \searrow_{g} & \\\sum X & \rightarrow_{F} &\sum Y \end{matrix}

    Figure 2.

    \begin{matrix}X \times [0,1] & \rightarrow_{\bar{f}} & Y \times [0,1] \\<br />
\downarrow_{p} & & \downarrow_{q} \\<br />
\sum X & \rightarrow_{F} &\sum Y \end{matrix}

    To apply theorem 22.2 of Munkres, we need to show that p^{-1}(\{(y, t)\}) for (y, t) \in \sum X is constant by a mapping g in Figure 1. If t is not {0} or {1}, each p^{-1}(\{(y, t)\}) is a single point in X \times [0,1], so it is straightforward to see that g is constant on those points (See Figure 2)

    By the definition of the suspension map, we see that X \times \{0\} or X \times \{1\} is mapped by p to a single point in \sum X. We call it "u" and "v", respectively. To apply theorem 22.2 of Munkres, we need to show that p^{-1}(\{u\}) and p^{-1}(\{v\}) are constant within the mapping of g. Note that X \times \{0\} is mapped by \bar{f} to Y \times \{0\} ( X \times \{1\} to  Y \times \{1\}, respectively) (See Figure 2). Then they are constant by a mapping q in Figure 2 because Y \times \{0\} (or Y \times \{1\}) is mapped to a single point by a mapping q in Figure 2. Now the hypothesis of Theorem 22.2 of Munkres is satisfied and we conclude that F:\sum X \rightarrow \sum Y is continuous.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    54
    ...because they're both suspensions! On the one hand, I feel like a total tool for not tuning-in to that one. On the other hand, I'm glad I didn't have some gross misunderstanding of the problem.

    Thank you much.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum