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Let $X$ be a topological space and $x_0\in X$. Let $f,g:I\to X$ be two paths starting from the same point $x_0$. Let $F:f\simeq g ~\text{rel}~x_0$ be a homotopy. We know that $h(t)=F(1,t)$ is a path from $f(1)$ to $g(1)$. My question is, are paths $f*h$ and $g$ homotopic relative to endpoints? What is the explicit formula for the homotopy?enter image description here

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The homotopy that you want can be obtained from the homotopy $F : [0,1] \to [0,1]$ in the following manner.

First, we know that $F(s,0)=f(s)$, $F(s,1)=g(s)$, $F(0,t)=x_0$, and $F(1,t)=h(t)$.

The thing to do is to define an appropriate continuous function $Q : [0,1] \to [0,1]$ and then to show that $F \circ Q : [0,1] \to [0,1]$ is a path homotopy from $f*h$ to $g$. Here are four features of $Q$ which we will need in order be able to show that.

First, $Q(0,t)=(0,t)$. That guarantees that $F \circ Q(0,t)=x_0$.

Second, $Q(1,t)=(1,1)$. That guarantees that $F \circ Q(1,t)=x_2$.

Third, $Q(s,0)=(2s,0)$ if $0 \le s \le 1/2$; and $Q(s,0) = (1,2s-1)$ if $1/2 \le s \le 1$. That guarantees that the map $s \mapsto F \circ Q(s,0)$ is the path $f*h$.

Fourth, $Q(s,1)=(s,1)$. That guarantees that $F \circ Q(s,1)=g(s)$.

So all that is needed is to show that $Q$ exists with those properties. I've already said exactly how to define $Q(b)$ for each point $b \in \partial([0,1]\times[0,1])$ meaning the boundary of $[0,1]\times[0,1]$.

Next, let $c=(1/2,1/2)$ and require that $Q(c)=c$.

Finally, extend over the "radii" connecting $c$ to $\partial([0,1]\times[0,1])$, meaning that for each $b \in \partial([0,1] \times[0,1]$ and each $0 \le u \le 1$ define $$Q((1-u)c + ub) = (1-u)Q(c) + u Q(b) = (1-u)c + u Q(b) $$

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The same homotopy will work if you parametrize it differently. Just use e.g. $$G(x,t)=\begin{cases}F(2x/(t+1),t) & x\leq (t+1)/2 \\ F(1,2x-1) & x> (t+1)/2\end{cases}$$

It might be a good exercise for you to verify that that homotopy gives you exactly what you want (and is coninuous!). To see how I came up with it, try graphing how it maps the unit square to itself.

And note that while the above is one of the easiest and simple-to-write examples of such a homotopy, any continuous map from the unit square to itself that behaves as necessary on the boundary will suffice.

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