$a.b$ 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.

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 $s=1$.
