I'm having trouble really making sense of these homotopy schematics from May's A Concise Course in Algebraic Topology

Homotopy schematic

Here's what I understand so far: - The top and bottom rows are different compositions of paths, and the schematics are aiming to show equivalence.

  • The schematic represents the "run time" of the loop with longer or shorter lengths of the square's edge. For example, in the top row first loop $f$ is run at double time, $g$ and $h$ are run at double-double time.
  • The left side of the square shows $c_x$ as the constant path through point $x$, and the right side shows that both compositions of loops end up at point $w$.

Here is what I don't understand:

  • Is this just a statement of equivalence? Or do these schematics go some way to explain why these compositions of loops are equivalent?
  • I don't fully understand what the vertical lines represent. This isn't like a mapping diagram, is it?
  • Why is this called a 'domain square'?
  • Do the diagonals show how a potential path homotopy between the two compositions could go? Is it meant to suggest one?
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    $\begingroup$ if you draw a horizontal line through any of these squares, its broken into segments by the (slanted) vertical lines. Each of these segments is supposed to be mapped homeomorphically to the unit interval, and then on into the space $X$ via whatever map is written above (or below) the segment. This gives a homotopy between two paths. So the diagrams are supposed to actually define a homotopy (not just state its existence). $\endgroup$ – Tim kinsella Jan 31 '18 at 4:05
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    $\begingroup$ I always found it helpful to think of such a diagram as a homotopy of two homeomorphisms of the unit interval (say from the identity to the homeomorphism that runs at double speed on $[0,1/4]$, unit speed on $[1/4,1/2]$ and half speed on $[1/2,1]$). Then the diagram just represents the post-composition of this homotopy with some path $[0,1]\rightarrow X$ given by a concatenation of loops. $\endgroup$ – Tim kinsella Jan 31 '18 at 4:16

I had a lot of difficulty with these schematics when I first saw them as well - I could roughly understand what they were trying to show, but not how they provided any proof for me. It was helpful for me to actually write out an equation.

So, let's take a look at the first diagram. I do want to point out that you're referring to $f$, $g$, and $h$ as loops, but (unless I'm missing something about the context in May) they are generally not. Note that to concatenate these paths you'll have $x = f(0)$, $f(1) = g(0)$, $g(1) = h(0)$, and $h(1) = w$. If these were loops you'd have that all of these were equal, so in particular we'd have x=w - which might be true, but isn't necessarily so.

The square is illustrating the domain of a function, $H: I \times I \to X$. We have three paths, $f: x \mapsto y$, $g: y \mapsto z$, $h: z \mapsto w$. Every point $(s,t)$ on the left diagonal line maps to $y$ and every point $(s,t)$ on the right diagonal line maps to $z$.

In order to have a path homotopy between $(f \cdot g) \cdot h$ and $f \cdot (g \cdot h)$, we need to have that for all $t$, $H(0,t) = x$ and $H(1,t) = w$, so that for all $t$, $H( - , t)$ is a path from $x$ to $w$. We also need that $H$ is a homotopy between our desired paths; ie that $H(-,0) = (f \cdot g) \cdot h$, and $H(-,1) = f\cdot(g\cdot h)$.

We define $$H(s,t) = \left\{ \begin{array}\\ f\left(\frac{4}{t+1}s\right) & s \in \left[0, \frac{1}{4}t + \frac{1}{4} \right] \\ g(4s-t-1) & s \in \left[\frac{1}{4}t + \frac{1}{4}, \frac{1}{4}t + \frac{1}{2} \right] \\ h\left(\frac{4}{2-t}(s-1)+1\right) & s \in \left[\frac{1}{4}t+ \frac{1}{2}, 1\right] \\ \end{array} \right.$$

You can verify that this $H$ fulfills all the conditions mentioned above, so $H$ gives us a path homotopy between $(f \cdot g) \cdot h$ and $f \cdot (g \cdot h)$.

The thing is that given the domain diagram, there was only one way to write this equation. It helped me to see that I could come up with the equation that matched the diagram, but as you can see it is an ugly equation and admittedly it took a while. That's why it's left just as the domain diagram without much else explanation, but it's a little opaque at first.

It might be worth trying to write these equations for the other two diagrams (they're much simpler) to convince yourself of these ideas.

  • $\begingroup$ Thanks for the point about paths v. loops! $\endgroup$ – MBP Jan 31 '18 at 19:19
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    $\begingroup$ And I think that this helps me a bunch. I'm going to try to write some of the equations, as you recommend. It's also helpful for me to think of this as the domain I x I, and the point that all the points on the diagonal get mapped to y (or w) is helpful too. I think I won't really understand this until I can write the equations, so that's what I have to do. Thanks! $\endgroup$ – MBP Jan 31 '18 at 19:38
  • $\begingroup$ @MBP I'm glad you found it helpful! If you run into trouble with the other equations let me know. $\endgroup$ – emma Jan 31 '18 at 22:34

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