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Math Help - quick question on paths

  1. #1
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    quick question on paths

    If f: X \rightarrow Y is a map and if g and h are paths in X with g(1)=g(0), then f(g \cdot h)=fg \cdot fh.

    I was told that this follows directly from the definition of path composition. This is where f:X \rightarrow Y so the first is a path in Y and the second is a path in Y. Is this completely justified by this, or do I need to justify this more than what is above?
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  2. #2
    Senior Member
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    Hi

    g(1)=h(0) I suppose.

    Yes that equality is quite natural. If you want to write and see, well, by definition:
    g.h(t)=\begin{cases} g(2t) & 0\leq t\leq \frac{1}{2}  \\ h(2t-1) & \frac{1}{2}\leq t\leq 1\\\end{cases}
    Therefore
    f(g.h(t))=\begin{cases} f(g(2t)) & 0\leq t\leq \frac{1}{2}  \\ f(h(2t-1)) & \frac{1}{2}\leq t\leq 1\\\end{cases}

    If you consider the two paths fg and fh in Y, then by definition:

    fg.fh(t)=\begin{cases} fg(2t) & 0\leq t\leq \frac{1}{2}  \\ fh(2t-1) & \frac{1}{2}\leq t\leq 1\\\end{cases}

    So the equality comes from the fact that \forall t\in [0,1],\ f(g(t))=fg(t)\ \text{and}\ f(h(t))=fh(t), i.e. from the definition.
    Last edited by clic-clac; March 19th 2009 at 12:09 AM. Reason: useless parentheses
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