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Math Help - Commutative diagram

  1. #1
    Junior Member
    Joined
    Oct 2008
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    25

    Commutative diagram

    Hi everyone

    I have this diagram, which I must show commutes, i.e. that
    \hat{\beta}\circ (h_{x_0})_*=(h_{x_1})_*\circ \hat{\alpha}
    Where \alpha is a path in X and \beta=h\circ\alpha
    and h:X->Y is continous with h(x_0)=y_0 and 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 \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
    \hat{\beta}\circ (h_{x_0})_*=(h_{x_1})_*\circ \hat{\alpha}
    Where \alpha is a path in X and \beta=h\circ\alpha
    and h:X->Y is continous with h(x_0)=y_0 and 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 \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 \alpha is a path and \bar{\alpha} is the reverse of \alpha. Thus, \bar{a}(t)=\alpha(1-t). Similarly, \overline{h \circ \alpha}(t) = (h \circ \alpha)(1-t). Now we see that \overline{h \circ \alpha}(t) = h \circ \bar{\alpha}(t).
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