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Let $h: I \times I \rightarrow X$ be a continuous function, and let $a, b, c, d$ be the paths in $X$ defined as follows:

  • $a(s)=h(s,0)$
  • $b(s)=h(1,s)$
  • $c(s)=h(0,s)$
  • $d(s)=h(s,1)$

Then I want to prove that $a.b$ is path homotopic to $c.d$.

I tried to write homotopy explicitly but things got messy. The idea is that treating $a.b$ and $c.d$ as paths and homotope them to the diagonal of the square. Does this idea work?


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Let $p_1(t) = \begin{cases} (2t,0) & t \in [0,\frac{1}{2}) \\ (1,2t-1) & t \in [\frac{1}{2}, 1] \end{cases}$, and let $p_2(t) = ([p_1(t)]_2,[p_1(t)]_1)$.

Define $\lambda(t,s) = (1-s)p_1(t)+s p_2(t)$. Then $h \circ \lambda$ is a suitable homotopy.

To see this, note that $a.b = (h \circ \lambda)(\cdot,0)$, $c.d = (h \circ \lambda)(\cdot,1)$, and $\lambda:I^2 \to I^2$ is continuous.

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Yes, it does. If you draw it on a coordinate system, a square (corresponding to $H$) with its diagonal with boundary paths $a\cdot b$ and $c\cdot d$.

Then draw where it should be mapped to: a rectangle of size $2\times 1$, say, where $a\cdot b$ are placed on the line $y=0$, say, and $c\cdot d$ on $y=1$, then connect them by vertical segments $x$-wise, using $H$.

While $s<1$, we connect the point $P:=(s,0)$ of the square (mapped to $a(s)=H(s,0)$) with the corresponding $Q:=(0,s)$ on $c$ by a (in this picture oblique) line segment, parametrizing with $t\in [0,1]$, it yields to $$\tilde H(s,t):=H(t\cdot Q+(1-t)\cdot P) = H((1-t)s,ts) $$ It can be given similarly for $s\in [1,2]$, then by limits (or by believing in the picture) you can argue that it will be continous also at $s=1$.

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One way to construct the homotopy explicitly is to observe that $I \times I$ is convex: So you can join two paths by using the straight line between them. In particular, define $F: I \times I \rightarrow I \times I$ by:

$$F(s,t) = \left\{\begin{array}{cc} (1-t)(2s,0) + t(0,2s) & 0 \leq s \leq 1/2 \\(1-t)(1,2s - 1) + t(2s - 1,1) & 1/2 \leq s \leq 1 \end{array} \right.$$

Then you should be able to see that $F$ is a path homotopy from the bottom right corner path to the top left corner path. Using $F$ you should be able to construct your desired homotopy.

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