Commutativity of the fundamental groups.

Question

Let $X$ be a topological space and $x_0 \in X.$ Given $[g], [h] \in \pi_1 (X, x_0)$ define a map $H : I \times I \longrightarrow X$ by

$$H (t,s) = \begin{cases} h \left (\frac {3t} {1 - s} \right ) & \text {if}\ 0 \leq t \leq \frac {1 - s} {3} \\ g \left ( \frac {3t + s -1} {1 + 2 s} \right ) & \text {if}\ \frac {1 - s} {3} \leq t \leq \frac {2 + s} {3} \\ \overline h \left (\frac {3t - s - 2} {1 - s} \right ) & \text {if}\ \frac {2 + s} {3} \leq t \leq 1 \end{cases}$$

Doesn't it mean that $h \ast g \ast \overline h$ is based homotopic to $g$ in $X\ $? But if it's the case then this is true for any two loops $g$ and $h$ in $X$ based at $x_0.$ But this in turn implies that $\pi_1 (X, x_0)$ is abelian, which is not necessarily true. Where did I go wrong?

Any help would be highly appreciated. Thanks for your time.

EDIT $:$ The homotopy diagram I mentioned is the following $:$

Remark $:$ Another point is that although $h \ast g \ast \overline h$ is not homotopic to $g$ as based loops in $X$ they are indeed homotopic as non-based loops in $X.$ The following gives one such homotopy. Consider $K : I \times I \longrightarrow X$ defined by

$$K (t,s) = \begin{cases} \overline h (s -3t) & \text {if}\ 0 \leq t \leq \frac {s} {3} \\ g \left ( \frac {3t - s} {3 - 2 s} \right ) & \text {if}\ \frac {s} {3} \leq t \leq \frac {3 - s} {3} \\ \overline h (3t + s - 3) & \text {if}\ \frac {3 - s} {3} \leq t \leq 1 \end{cases}$$

The map $H$ is not defined at points where $s = 1$. I'm guessing you want to say that $H(t, 1) = g(t)$, but then you have no guarantee that $H$ is continuous. — Michael Albanese, Nov 21 '23 at 10:24
yeah I think that is the problem @MichaelAlbanese. The map $H$ goes through $h$ and $\overline{h}$ faster and faster and would move through them "infinitely fast" for $s=1$, which then would be "equal" to $g$. — linkja, Nov 21 '23 at 10:26
@MichaelAlbanese$:$ Good remark! That's the reason they fail to be homotopic. I drew a homotopy diagram and this is what I get when rephrasing the diagram in terms of equations. So why does the homotopy diagram fail here? Any idea? I am asking this because many people prefers to describe homotopy by diagrams instead of writing down explicit formulas for the homotopy. But in my case the homotopy diagram does give a wrong intuition about the homotopy. That's why I want to know why the intuition fails to work here. — Anacardium, Nov 21 '23 at 10:31
I'm not sure what you mean by a homotopy diagram. Do you mean something like this? Maybe you could add the diagram to the question so that someone can answer the questions you ask in your comment. — Michael Albanese, Nov 21 '23 at 11:13
It turns out that the basepoints of the intermediate loops in the non-based homotopy (I defined) traces out the loop $\overline h.$ — Anacardium, Nov 21 '23 at 11:40
$K$ is not well-defined. Take $t = \frac s 3$ in the first and second line of the definition. And what is your question? — Paul Frost, Nov 21 '23 at 13:42
@PaulFrost$:$ There is a typo and there is an obvious fix to it. Edited it accordingly. The question is clearly mentioned. — Anacardium, Nov 21 '23 at 14:12
If your question is "Where did I go wrong", then it was answered in a comment by Michael Albanese: $H$ is undefined for $s=1$. Or does it concern $K$? — Paul Frost, Nov 21 '23 at 14:20
@PaulFrost$:$ If you glossed over the above comments then you can get to see that I was confused about the existence of homotopy diagram without the existence of non-based homotopy. Michael Albanese suggested me to upload the diagram and I have uploaded it which clearly indicates how the homotopy should be done. But now I realized at $s = 1$ the parametric interval of $h$ and $\overline h$ just becomes a point. So $h$ and $\overline h$ cannot be defined at $s = 1$ unless $h$ is a constant loop (and consequently so is $\overline h$). For any other $0 \leq s \lt 1$ we are fine because... contd — Anacardium, Nov 21 '23 at 14:37
...in that case the parametric intervals of $h$ and $\overline h$ are both non-degenerate closed and bounded intervals and any such interval is homeomorphic to $[0,1].$ For degenerate case we no longer have that homeomorphism. So the definition of $h$ and $\overline h$ miserably fails at $s = 1.$ — Anacardium, Nov 21 '23 at 14:40
Then I suggest you answer your own question and accept it. Just to remove it from the unanswered queue. — Paul Frost, Nov 21 '23 at 14:41
@PaulFrost$:$ I am now busy with some other job. My exams are knocking at the door and I am preparing for the same. Is it possible to remove it? — Anacardium, Nov 21 '23 at 14:46
You can delete the question. Or I can write a community wiki containing what you wrote as an explanation. Would perhaps help other users who have the same question. — Paul Frost, Nov 21 '23 at 14:56
@PaulFrost$:$ Yes then please do it that way without deleting this question. I will surely accept the answer. — Anacardium, Nov 21 '23 at 15:50
Done. Note that community wikis can be accepted and upvoted, but do not confer any reputation for authors. See https://math.stackexchange.com/help/privileges/community-wiki. Questions can be deleted only by their authors and by the moderators. See https://math.stackexchange.com/help/deleted-questions. Other users can vote for deletion or closure. — Paul Frost, Nov 21 '23 at 17:40
@PaulFrost$:$ Thanks a lot for your efforts. Accepted and upvoted the answer. — Anacardium, Nov 21 '23 at 19:02

score 2 · Accepted Answer · answered Nov 21 '23 at 17:34

The question was answered in the comments. This community wiki summarizes the relevant insights.

Formally the homotopy $H$ is not well-defined because it involves a division by $0$ for $s=1$. It is impossible to repair the definition of $H$ as we shall explain now.

The "homotopy diagram" explains what the homotopy from $h *g * \overline h$ to $h$ should do.

The problem is that at $s=1$ the parametric interval of $h$ and $\overline h$ just becomes a point. So $h$ and $\overline h$ do not fit into this degenerate interval at $s=1$ unless $h$ is a constant loop (and consequently so is $\overline h$). For any other $0≤s<1$ we are fine because in that case the parametric intervals of $h$ and $\overline h$ are both non-degenerate closed and bounded intervals and any such interval is homeomorphic to $[0,1]$. For the degenerate case we no longer have that homeomorphism. So the insertion of $h$ and $\overline h$ into the shrinked intervals miserably fails at $s=1$.

Commutativity of the fundamental groups.

1 Answers1