A path $\alpha(t)$ in a topological space X is a map $\alpha :[0,1]\to X$, starting at $x_0=\alpha(0)$ and ending at $x_1=\alpha(1)$. For any intermediate values of t, $\alpha(t)$ in general assumes other values of $x\in X$. Closed loops are those for which $\alpha(0)=\alpha(1)$. From the definition, I visualize a constant path in a topological space as a fixed point. Is that correct?
How can I geometrically visualize a constant path?
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general-topology
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3I agree. It is a path of a point that never "moves" during the time interval $[0,1]$. – 2017-01-26
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0Yes that is correct. For whatever value $t $ assumes, $\alpha(t) $ is always the same. – 2017-01-26
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0Do you think that the product of a path and its inverse remains the same as t changes? – 2017-01-26
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0Think of the path $\alpha(t)$, $0 \le t \le 1$, as recording the position at time $t$ of a moving object, perhaps you out on your daily walk. In the special case of a constant path, instead of ever moving a single inch you just sort of....... stand there, for the whole time. – 2017-01-27
1 Answers
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Yes, as others have said in the comments, there is a canonical bijection between constant paths in $X$ and points of $X$. Given a point $x \in X$, you get a constant path $\tilde{x}: I \to X$ with $\tilde{x}(t) = x$ for all $t \in I$. Conversely, given a constant path you obtain a point in $X$ just by evaluation.