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Thread: Two Suspensions to Mention

  1. #1
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    Two Suspensions to Mention

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

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

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

    So if $\displaystyle p$ denotes the canonical quotient map from $\displaystyle X \times [0,1]$ to $\displaystyle \Sigma X$, I would like to say that $\displaystyle F:\Sigma X \rightarrow \Sigma Y$ where $\displaystyle F=p^{-1} \circ (q\circ \bar{f})$ 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 $\displaystyle q\circ \bar{f}$ is constant on each set $\displaystyle p^{-1} ([x,t])$, and this is where I am stuck. Looks like this means that $\displaystyle q\circ \bar{f}$ takes a set of things that are equivalent in $\displaystyle X \times [0,1]$ (that's $\displaystyle p^{-1}([x,t])$, of course) and says that these are also equivalent under the quotient map on $\displaystyle Y \times [0,1]$. 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".
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  2. #2
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    Quote Originally Posted by cribby View Post
    I am to show that a continuous function, $\displaystyle f: X \rightarrow Y$, induces a continuous function between their respective suspensions, $\displaystyle \Sigma X ,\, \Sigma Y$.

    Let $\displaystyle q:Y \times [0,1] \rightarrow \Sigma Y$ be the standard quotient map (in particular, it is continuous). Also, let $\displaystyle \bar{f}: X \times [0,1] \rightarrow Y \times [0,1]$ by $\displaystyle \bar{f}(x,t)=(f(x), t)$. Note that $\displaystyle \bar{f}$ 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 $\displaystyle q \circ \bar{f}$, which is, of course, continuous (given the prior statements) from $\displaystyle X \times [0,1] \rightarrow \Sigma Y$.

    So if $\displaystyle p$ denotes the canonical quotient map from $\displaystyle X \times [0,1]$ to $\displaystyle \Sigma X$, I would like to say that $\displaystyle F:\Sigma X \rightarrow \Sigma Y$ where $\displaystyle F=p^{-1} \circ (q\circ \bar{f})$ 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 $\displaystyle q\circ \bar{f}$ is constant on each set $\displaystyle p^{-1} ([x,t])$, and this is where I am stuck. Looks like this means that $\displaystyle q\circ \bar{f}$ takes a set of things that are equivalent in $\displaystyle X \times [0,1]$ (that's $\displaystyle p^{-1}([x,t])$, of course) and says that these are also equivalent under the quotient map on $\displaystyle Y \times [0,1]$. 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.

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

    Figure 2.

    $\displaystyle \begin{matrix}X \times [0,1] & \rightarrow_{\bar{f}} & Y \times [0,1] \\
    \downarrow_{p} & & \downarrow_{q} \\
    \sum X & \rightarrow_{F} &\sum Y \end{matrix}$

    To apply theorem 22.2 of Munkres, we need to show that $\displaystyle p^{-1}(\{(y, t)\})$ for $\displaystyle (y, t) \in \sum X$ is constant by a mapping g in Figure 1. If t is not {0} or {1}, each $\displaystyle p^{-1}(\{(y, t)\})$ is a single point in $\displaystyle X \times [0,1]$, 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 $\displaystyle X \times \{0\}$ or $\displaystyle X \times \{1\}$ is mapped by p to a single point in $\displaystyle \sum X$. We call it "u" and "v", respectively. To apply theorem 22.2 of Munkres, we need to show that $\displaystyle p^{-1}(\{u\})$ and $\displaystyle p^{-1}(\{v\})$ are constant within the mapping of g. Note that $\displaystyle X \times \{0\}$ is mapped by $\displaystyle \bar{f}$ to $\displaystyle Y \times \{0\}$ ($\displaystyle X \times \{1\}$ to $\displaystyle Y \times \{1\}$, respectively) (See Figure 2). Then they are constant by a mapping q in Figure 2 because $\displaystyle Y \times \{0\}$ (or $\displaystyle Y \times \{1\}$) 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 $\displaystyle F:\sum X \rightarrow \sum Y$ is continuous.
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  3. #3
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    ...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.
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