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Thread: Commutative diagram

  1. #1
    Junior Member
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    Oct 2008
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    25

    Commutative diagram

    Hi everyone

    I have this diagram, which I must show commutes, i.e. that
    $\displaystyle \hat{\beta}\circ (h_{x_0})_*=(h_{x_1})_*\circ \hat{\alpha}$
    Where $\displaystyle \alpha$ is a path in X and $\displaystyle \beta=h\circ\alpha$
    and $\displaystyle h:X->Y$ is continous with $\displaystyle h(x_0)=y_0$ and $\displaystyle h(x_1)=y_1$


    But I'm a little stuck. What I do is this:



    Is that correct? But I'm not sure if it's allowed to just say $\displaystyle \overline{h\circ\alpha}=h\circ \bar{\alpha}$. What is the argument for this, that h is continous?
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  2. #2
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by Carl View Post
    Hi everyone

    I have this diagram, which I must show commutes, i.e. that
    $\displaystyle \hat{\beta}\circ (h_{x_0})_*=(h_{x_1})_*\circ \hat{\alpha}$
    Where $\displaystyle \alpha$ is a path in X and $\displaystyle \beta=h\circ\alpha$
    and $\displaystyle h:X->Y$ is continous with $\displaystyle h(x_0)=y_0$ and $\displaystyle h(x_1)=y_1$


    But I'm a little stuck. What I do is this:



    Is that correct? But I'm not sure if it's allowed to just say $\displaystyle \overline{h\circ\alpha}=h\circ \bar{\alpha}$. What is the argument for this, that h is continous?
    It seems correct to me. Recall that $\displaystyle \alpha$ is a path and $\displaystyle \bar{\alpha}$ is the reverse of $\displaystyle \alpha$. Thus, $\displaystyle \bar{a}(t)=\alpha(1-t)$. Similarly, $\displaystyle \overline{h \circ \alpha}(t) = (h \circ \alpha)(1-t)$. Now we see that $\displaystyle \overline{h \circ \alpha}(t) = h \circ \bar{\alpha}(t)$.
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