$a.b$ is path homotopic to $c.d$ 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?
Thanks.

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.
