Associativity of concatenation of path homotopies

Question

Suppose that X is a topological space. Suppose that $f,g,h: I\to X$ are given paths such that their concatenation is defined. Then I want to prove that $([f][g])([h])=[f]([g][h]),$ where [.] denotes the path homotopy class of .

I drew an image here and using that I arrived at $F:I\times I\to X$, which I expected to be continuous. $F$ is as follows:

$F(s,t)=\begin{cases}f\left(\frac s{t+1}\right); t\ge 4s-1\\ g\left(s-\frac t4 \right); 4s-2\le t\le 4s-1\\ h\left (-\frac st+\left (1+\frac 1t\right)\right); t\le 4s-2\end{cases}$

But as it can be seen from the above, $F(s,0)$ blows up when $s>\frac 12$. Can anyone please explain what went wrong in my calculation and suggest how to fix it? Thanks.

From the linked image:

$s\mapsto \frac s{t+1}$ is the map that takes $0\mapsto 0, \frac{t+1}4\mapsto \frac 14$.

$s\mapsto s-\frac t4$ is the map that takes $\frac{t+1}4\mapsto \frac 14, \frac{t+2}4\mapsto\frac 12.$

$s\mapsto -\frac st+1+\frac 1t$ is the map that takes $\frac{t+2}4\mapsto\frac 12, 1\mapsto 1$

Answer 1

From your linked image, I would rather take (given $t\in[0,1]$):

• $f$ applied to the affine function of $s$ which goes from $0$ to $1$ when $s$ varies from $0$ to $\frac{1+t}4,$ followed by
• $g$ applied to the affine function of $s$ which goes from $0$ to $1$ when $s$ varies from $\frac{1+t}4$ to $\frac{2+t}4,$ followed by
• $h$ applied to the affine function of $s$ which goes from $0$ to $1$ when $s$ varies from $\frac{2+t}4$ to $1,$ i.e. $$F(s,t)=\begin{cases}f\left(\frac{4s}{1+t}\right)&\text{if }4s\le1+t\\ g\left(4s-t-1\right)&\text{if }1+t\le4s\le2+t\\ h\left (\frac{4s-2-t}{2-t}\right)&\text{if }4s\ge2+t.\end{cases}$$