Taking algebraic topology class this semester... and met the notion of the fundamental group days ago. Okay they are set of classes of a loop homotopy and thing were fine... but the professor proved that the group has an actual group structure and thing were confusing.

For example he showed that a constant path multiplied by a path is just the path by finding a homotopy between them. I understand they are homotopic after some math but how can I construct them without referring to the textbook? Have mathematicians done it by trial and error or is there any neat way to construct the homotopies to show the group structure on the fundamental group?

Thank you so much


For the proof you have to check:

$\mbox{1.) }\gamma\cdot e\sim \gamma \sim e\cdot \gamma$

$\mbox{2.) }\gamma\cdot \gamma^{-1}\sim e$

$\mbox{3.) }(\beta\cdot \gamma)\cdot \delta\sim \beta\cdot(\gamma\cdot\delta)$

$e$ is the constant path. I will give the explanation for the second case. For me it is helpful to draw boxes like this:

enter image description here

(Well, it's not the best quality, but I think, it's clear what I mean.)

We want to go the path $\gamma$ then $\gamma^{-1}$ in a certain time which is given by the interval $I$. So we have to double the speed for each path.

This is exactly the meaning of the definition of the product: $$\gamma\cdot \gamma^{-1}(t):=\begin{cases} \gamma(2t)\mbox{ for }0\le t\le1/2,\\ \gamma(2t-1)\mbox{ for }1/2\le t\le 1, \end{cases}$$

This is homotopic to the constant path. The idea behind it, is the following: We stay at the basepoint, then we go $\gamma$ then $\gamma^{-1}$ back and the rest of the time we stay at the basepoint, i.e. the const. path $e$.

For example: If you consider the red line, we just stay at the basepoint $x_0$ up to $1/4$ of the time. Then we go $\gamma$, but not the full way of it, but maybe $1/4$ of the way and go back by $\gamma^{-1}$. The rest of the time we stay at $x_0$.

Now the box gives us a helpful tool giving an explicit definition of the corresponding homotopy $H\colon I\times I\to X$, $(s,t)\mapsto H(s,t)$. The domain of $s$ has to shrink by growing $t$. By our foregoing explanation, we know that there are exactly 4 parts we have to define: the first part and the fourth part, where we stay at $x_0$, and the second and third part, where we go $\gamma$ and $\gamma^{-1}$ back. This gives us the following: $$H(s,t)= \begin{cases} e(s)\mbox{ for }0\le s\le \frac{1}{2}t,\\ \gamma(2s)\mbox{ for } \frac{1}{2}t \le s\le \frac{1}{2},\\ \gamma(2s-1)\mbox{ for } \frac{1}{2} \le s \le 1 - \frac{1}{2}t,\\ e(s)\mbox{ for }1-\frac{1}{2}t\le s \le 1. \end{cases}$$

We have $H(s,0)=\gamma\cdot \gamma^{-1}(t)$ and $H(s,1)=e(s)$.

To get a feeling (not explicit formulas) for the whole proof, watch this video:


In general it is important to imagine how the homotopy should work!

  • $\begingroup$ exactly what i wanted. thanks $\endgroup$ – Mathcho Oct 3 '15 at 10:44

Mathematicians tend to think in terms of intuition behind the concept of homotopy. If you think of a path as a rope with both ends fixed, then you can move the rope in space, lengthen it by stretching/shorten it by tightening, and you'll end up with a homotopy equivalent path.

A loop is a rope with both ends fixed at the same spot. A composition of two loops naturally suggests itself as taking two ropes and tying the end of the first to the beginning of the second; now these 2 points are still physically present at the loop point but are no longer tied to it - only the beginning of the first and the end of the second are tied to it.

Then an inverse of a loop is just a second rope that runs exactly over the first but retraces it back, from end to beginning. If you compose a rope with its inverse, it's like you made some complicated path in space, then retraced your steps back. Since your original end-point is now (after composition) unstuck, you can keep shortening the rope, stealing all the time from its middle (original end), watching it recede along the path towards the beginning, until there's nothing left. There, you proved the existence of the group structure.

A constant path is lots of rope piled on the same spot... you get the picture. An explicit homotopy is a nice formal exercise and of course provides the necessary underlying rigor, but you need to learn to think topologically, with pictures.

  • $\begingroup$ thanks for the good explanation. now i get the picture of it but i just wonder just how to 'construct' the explicit formula for the homotopies here... $\endgroup$ – Mathcho Sep 24 '15 at 10:41

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