2

Let $F:I \times I \to X$ be a continuous map, and let $f,h$ and $k,g$ be paths in $X$ defined by $$f(s) = F(s,0)$$ $$g(s) = F(1,s)$$ $$h(s) = F(0,s)$$ $$k(s) = F(s,1)$$ Then $f \cdot g$ is homotopic to $h \cdot k$

I tried like $F(s,t)F(1−t,s)$, but it clearly will not work(not even well defined). Then I thought about finding a way to sort of relableling $I×I$ to $I$, letting the homotopy starting from the lower left of the square. But I cannot find an explicit way to write down this.

Keith
  • 1,383

2 Answers2

6

You are looking at the composition of $F$ with four different paths which I will denote by $\gamma_f(s) = (s,0)$ and so on.

There is an obvious homotopy $H$ of $\gamma_f*\gamma_g$ with $\gamma_h*\gamma_k$ given by $$H(\tau,s) = \cases{\tau\gamma_h(2s) + (1-\tau)\gamma_f(2s)&if $0\le s\le\tfrac{1}{2}$,\\\tau\gamma_k(2s-1) + (1-\tau)\gamma_g(2s-1)&if $\tfrac{1}{2}\le s\le1$.}$$

Now consider the composition $F\circ H$.

amWhy
  • 209,954
Daniel Robert-Nicoud
  • 29,795
  • I don't think the OP will think your formula is obvious. – Rob Arthan Mar 19 '16 at 00:09
  • @RobArthan Well, the main idea is to move the problem from a deformation on $X$ to a deformation on $I^2$, and I think that's what more important. And if the OP has already seen some examples of homotopies, I think that he should be able to recognize a straight-line homotopy when he sees one. – Daniel Robert-Nicoud Mar 19 '16 at 01:19
  • So is $r_g(s) = (1,s)$ ? – Keith Mar 19 '16 at 02:20
  • 1
    Yes, I recognize this as a straight-line homtopy. So essentially we are talking about H(s,t) = (1-t) $r_f \ast r_g$ + t $r_h \ast r_k$ right? – Keith Mar 19 '16 at 02:25
  • @Keith Yes, exactly. – Daniel Robert-Nicoud Mar 19 '16 at 02:50
  • I am sorry, but why this question is off the topic of this site? – Keith Mar 19 '16 at 02:59
  • @Keith The question is not "off-topic" because it is not about mathematics, but because we usually ask that a person asking a question shows his or her effort in solving it (saying for example what they have tried and where the problem is exactly), and you didn't do that. – Daniel Robert-Nicoud Mar 19 '16 at 11:40
  • To rephrase Daniel's comment, if you think in terms of groupoids rather than groups, you can consider first 4 paths $I \to I^2$ as giving the 4 edges of $I^2$ and then that the compositions going either way round the sides from one corner to the opposite are homotopic rel end points as $I^2$ is convex. Then map $I^2$ into $X$. – Ronnie Brown Mar 19 '16 at 15:10
1

Here is another "constructive" proof.

Let $G(t,s)$ be defined by

$$ G(t,s) = \begin{cases} F\left(2t,\dfrac{ts}{1-s}\right) & \text{if}~~~~0\leq t\leq\dfrac{1}{2},~~~~0\leq s<1;\\ \\ F\left(0,2t\right) & \text{if}~~~~0\leq t\leq\dfrac{1}{2},~~~~s=1;\\ \\ F\left(2t-1,\dfrac{t-1+s(2-t)}{s}\right) & \text{if}~~~~\dfrac{1}{2}\leq t\leq 1,~~~~0<s\leq 1;\\ \\ F(1,2t-1) & \text{if}~~~~\dfrac{1}{2}\leq t\leq 1,~~~~s=0. \end{cases} $$ Then we observe that

$$G(t,0) = \begin{cases} F\left(2t,0\right) & \text{if}~~~~0\leq t\leq\dfrac{1}{2};\\ \\ F(1,2t-1) & \text{if}~~~~\dfrac{1}{2}\leq t\leq 1. \end{cases}$$

Also observe that

$$ G(t,1) = \begin{cases} F\left(0,2t\right) & \text{if}~~~~0\leq t\leq\dfrac{1}{2};\\ \\ F(2t-1,1) & \text{if}~~~~\dfrac{1}{2}\leq t\leq 1. \end{cases} $$

So $G(t,0)=f\ast g$ and that $G(t,1)=h\ast k$. It should be clear that $G$ is continuous, hence $f\ast g\sim h\ast k$.

math-physicist
  • 979
user284331
  • 55,591