# intuition behind the construction of the map for showing associativity of $\pi_1(X,x_0)$.

I am studying algebraic topology.I have started with the chapter on fundamental group.Fundamental group at $$x_0$$ is defined to be the set of all equivalence classes of loops based at $$x_0$$ together with composition of two classes of loops.Now I am trying to show that this indeed forms a group.Now for associativity of this set,one has to show that $$f.(g.h)\simeq(f.g).h$$.Now we have,

$$(f.g).h(s)=\begin{cases} f(4s),\text{if } 0\leq s\leq 1/4\\g(4s-1),\text{if }1/4\leq s\leq 1/2\\h(2s-1),\text{if }1/2\leq s\leq 1\end{cases}$$

and, $$f.(g.h)(s)=\begin{cases} f(2s),\text{if } 0\leq s\leq 1/2\\g(4s-2),\text{if }1/2\leq s\leq 3/4\\h(4s-3),\text{if }3/4\leq s\leq 1\end{cases}$$

Now the author of the text gives a homotopy between these two loops as follows:

$$H:[0,1]\times [0,1]\to X$$ given by,

$$H(s,t)=\begin{cases} f(\frac{4s}{1+t}),\text{if } 0\leq s\leq (1+t)/4\\g(4s-t-1),\text{if }(1+t)/4\leq s\leq (2+t)/4\\h(\frac{4s-2-t}{2-t}),\text{if }(2+t)/4\leq s\leq 1\end{cases}$$

Now this seems to have come out of blue.I tried to think but could not find a way how this can be constructed.I am looking for a clear intuition behind this construction.

• Did the author include a picture? If not, perhaps there are betters books to read. If you draw what that homotopy is doing, things are absolutely clear.
– Pedro
May 21 at 15:01
• Draw a picture of $[0,1]\times[0,1]$. You have the domain of $(f*g)*h$ on one end and $f*(g*h)$ on the other. Connect the points in between the $f,g,h$ parts on both ends with straight lines. This is often illustrated in textbooks. May 21 at 15:07

The difference between $$(f\cdot g)\cdot h$$ and $$f\cdot(g\cdot h)$$ is not the "path" they follow, but the "speed" in which they follow said paths. So we want a homotopy that (continuously) change the speed in some parts of the loop. For instance, observe that $$(f\cdot g)\cdot h$$ does the full loop of $$f$$ in $$[0,\frac{1}{4}]$$, by means of $$f(4s)$$, but $$f\cdot(g\cdot h)$$ does the full lop of $$f$$ in $$[0,\frac{1}{2}]$$, by means of $$f(2s)$$ (you can see this as if the speed of $$(f\cdot g)\cdot h$$ is double the speed of $$f\cdot(g\cdot h)$$, in the first part of the path). So we want to change $$4s$$ continuously into $$2s$$. One way to do that is, as in the homotopy you presented: $$[0,1]\ni t\mapsto \frac{4s}{1+t}.$$ But we also need to change the interval in wich $$f$$ is traversed (the interval we need to do the full loop of $$f$$), initially it is $$[0,\frac{1}{4}]$$ and we change it (continuously) to $$[0,\frac{1}{2}]$$, in such way that in every step of the way (every $$t\in[0,1]$$) $$f$$ is traversed. Can you see why the interval is $$[0,\frac{1+t}{4}]$$?
Next, we can se that $$g$$ is traversed whit the same speed, what changes is when we begin to traverse $$g$$. So we want to change $$4s-1$$ into $$4s-2$$ and $$[\frac{1}{4},\frac{1}{2}]$$ into $$[\frac{1}{2},\frac{3}{4}]$$.
The reasoning for $$h$$ is a bit more complex because not only it changes when $$h$$ starts to be traversed, but it also change the speed in which we do that. But it's done just as in the previous cases: we need to change $$2s-1$$ to $$4s-3$$ and $$[\frac{1}{2},1]$$ to $$[\frac{3}{4},1]$$ in a way that we traverse $$h$$ completely in every step of the way.